3.227 \(\int \frac {(c e+d e x)^4}{(a+b \sin ^{-1}(c+d x))^3} \, dx\)

Optimal. Leaf size=322 \[ -\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac {27 e^4 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {25 e^4 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {e^4 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{16 b^3 d}+\frac {27 e^4 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}-\frac {25 e^4 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{32 b^3 d}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^4 \sqrt {1-(c+d x)^2} (c+d x)^4}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

[Out]

-2*e^4*(d*x+c)^3/b^2/d/(a+b*arcsin(d*x+c))+5/2*e^4*(d*x+c)^5/b^2/d/(a+b*arcsin(d*x+c))-1/16*e^4*Ci((a+b*arcsin
(d*x+c))/b)*cos(a/b)/b^3/d+27/32*e^4*Ci(3*(a+b*arcsin(d*x+c))/b)*cos(3*a/b)/b^3/d-25/32*e^4*Ci(5*(a+b*arcsin(d
*x+c))/b)*cos(5*a/b)/b^3/d-1/16*e^4*Si((a+b*arcsin(d*x+c))/b)*sin(a/b)/b^3/d+27/32*e^4*Si(3*(a+b*arcsin(d*x+c)
)/b)*sin(3*a/b)/b^3/d-25/32*e^4*Si(5*(a+b*arcsin(d*x+c))/b)*sin(5*a/b)/b^3/d-1/2*e^4*(d*x+c)^4*(1-(d*x+c)^2)^(
1/2)/b/d/(a+b*arcsin(d*x+c))^2

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Rubi [A]  time = 0.86, antiderivative size = 318, normalized size of antiderivative = 0.99, number of steps used = 26, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {4805, 12, 4633, 4719, 4635, 4406, 3303, 3299, 3302} \[ -\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \cos \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {e^4 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^4 \sqrt {1-(c+d x)^2} (c+d x)^4}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*ArcSin[c + d*x])^3,x]

[Out]

-(e^4*(c + d*x)^4*Sqrt[1 - (c + d*x)^2])/(2*b*d*(a + b*ArcSin[c + d*x])^2) - (2*e^4*(c + d*x)^3)/(b^2*d*(a + b
*ArcSin[c + d*x])) + (5*e^4*(c + d*x)^5)/(2*b^2*d*(a + b*ArcSin[c + d*x])) - (e^4*Cos[a/b]*CosIntegral[a/b + A
rcSin[c + d*x]])/(16*b^3*d) + (27*e^4*Cos[(3*a)/b]*CosIntegral[(3*a)/b + 3*ArcSin[c + d*x]])/(32*b^3*d) - (25*
e^4*Cos[(5*a)/b]*CosIntegral[(5*a)/b + 5*ArcSin[c + d*x]])/(32*b^3*d) - (e^4*Sin[a/b]*SinIntegral[a/b + ArcSin
[c + d*x]])/(16*b^3*d) + (27*e^4*Sin[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSin[c + d*x]])/(32*b^3*d) - (25*e^4*S
in[(5*a)/b]*SinIntegral[(5*a)/b + 5*ArcSin[c + d*x]])/(32*b^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 4633

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Simp[(x^m*Sqrt[1 - c^2*x^2]*(a + b*ArcSin
[c*x])^(n + 1))/(b*c*(n + 1)), x] + (Dist[(c*(m + 1))/(b*(n + 1)), Int[(x^(m + 1)*(a + b*ArcSin[c*x])^(n + 1))
/Sqrt[1 - c^2*x^2], x], x] - Dist[m/(b*c*(n + 1)), Int[(x^(m - 1)*(a + b*ArcSin[c*x])^(n + 1))/Sqrt[1 - c^2*x^
2], x], x]) /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && LtQ[n, -2]

Rule 4635

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[(a + b*x)^n*S
in[x]^m*Cos[x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[m, 0]

Rule 4719

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_)*((f_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[
((f*x)^m*(a + b*ArcSin[c*x])^(n + 1))/(b*c*Sqrt[d]*(n + 1)), x] - Dist[(f*m)/(b*c*Sqrt[d]*(n + 1)), Int[(f*x)^
(m - 1)*(a + b*ArcSin[c*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && LtQ[n,
-1] && GtQ[d, 0]

Rule 4805

Int[((a_.) + ArcSin[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSin[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {(c e+d e x)^4}{\left (a+b \sin ^{-1}(c+d x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sin ^{-1}(x)\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \frac {x^3}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{b d}-\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {x^5}{\sqrt {1-x^2} \left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{2 b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (6 e^4\right ) \operatorname {Subst}\left (\int \frac {x^2}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {x^4}{a+b \sin ^{-1}(x)} \, dx,x,c+d x\right )}{2 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (6 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^4(x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (6 e^4\right ) \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{4 (a+b x)}-\frac {\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{8 (a+b x)}-\frac {3 \cos (3 x)}{16 (a+b x)}+\frac {\cos (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}+\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (25 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b^2 d}+\frac {\left (75 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {\left (3 e^4 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (25 e^4 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {\left (3 e^4 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {\left (75 e^4 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}-\frac {\left (25 e^4 \cos \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}+\frac {\left (3 e^4 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}-\frac {\left (25 e^4 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {\left (3 e^4 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{2 b^2 d}+\frac {\left (75 e^4 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}-\frac {\left (25 e^4 \sin \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{32 b^2 d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{2 b d \left (a+b \sin ^{-1}(c+d x)\right )^2}-\frac {2 e^4 (c+d x)^3}{b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {5 e^4 (c+d x)^5}{2 b^2 d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {e^4 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{16 b^3 d}+\frac {27 e^4 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{32 b^3 d}-\frac {25 e^4 \sin \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{32 b^3 d}\\ \end {align*}

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Mathematica [A]  time = 1.60, size = 317, normalized size = 0.98 \[ \frac {e^4 \left (-\frac {16 b^2 \sqrt {1-(c+d x)^2} (c+d x)^4}{\left (a+b \sin ^{-1}(c+d x)\right )^2}+48 \left (\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-\cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )-25 \left (2 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\cos \left (\frac {5 a}{b}\right ) \text {Ci}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+2 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )+\frac {16 b \left (5 (c+d x)^5-4 (c+d x)^3\right )}{a+b \sin ^{-1}(c+d x)}\right )}{32 b^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*ArcSin[c + d*x])^3,x]

[Out]

(e^4*((-16*b^2*(c + d*x)^4*Sqrt[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x])^2 + (16*b*(-4*(c + d*x)^3 + 5*(c + d
*x)^5))/(a + b*ArcSin[c + d*x]) + 48*(Cos[a/b]*CosIntegral[a/b + ArcSin[c + d*x]] - Cos[(3*a)/b]*CosIntegral[3
*(a/b + ArcSin[c + d*x])] + Sin[a/b]*SinIntegral[a/b + ArcSin[c + d*x]] - Sin[(3*a)/b]*SinIntegral[3*(a/b + Ar
cSin[c + d*x])]) - 25*(2*Cos[a/b]*CosIntegral[a/b + ArcSin[c + d*x]] - 3*Cos[(3*a)/b]*CosIntegral[3*(a/b + Arc
Sin[c + d*x])] + Cos[(5*a)/b]*CosIntegral[5*(a/b + ArcSin[c + d*x])] + 2*Sin[a/b]*SinIntegral[a/b + ArcSin[c +
d*x]] - 3*Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c + d*x])] + Sin[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c +
d*x])])))/(32*b^3*d)

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b^{3} \arcsin \left (d x + c\right )^{3} + 3 \, a b^{2} \arcsin \left (d x + c\right )^{2} + 3 \, a^{2} b \arcsin \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^3,x, algorithm="fricas")

[Out]

integral((d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4)/(b^3*arcsin(d*x + c)^3
+ 3*a*b^2*arcsin(d*x + c)^2 + 3*a^2*b*arcsin(d*x + c) + a^3), x)

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giac [B]  time = 0.72, size = 3135, normalized size = 9.74 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^3,x, algorithm="giac")

[Out]

-25/2*b^2*arcsin(d*x + c)^2*cos(a/b)^5*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 +
2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 25/2*b^2*arcsin(d*x + c)^2*cos(a/b)^4*e^4*sin(a/b)*sin_integral(5*a/b
+ 5*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 25*a*b*arcsin(d*x +
c)*cos(a/b)^5*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c)
+ a^2*b^3*d) - 25*a*b*arcsin(d*x + c)*cos(a/b)^4*e^4*sin(a/b)*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^5*d*
arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 125/8*b^2*arcsin(d*x + c)^2*cos(a/b)^3*cos_integr
al(5*a/b + 5*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 25/2*a^2
*cos(a/b)^5*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) +
a^2*b^3*d) + 27/8*b^2*arcsin(d*x + c)^2*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4/(b^5*d*arcsin(
d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 75/8*b^2*arcsin(d*x + c)^2*cos(a/b)^2*e^4*sin(a/b)*sin_i
ntegral(5*a/b + 5*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 25/2*a^
2*cos(a/b)^4*e^4*sin(a/b)*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(
d*x + c) + a^2*b^3*d) + 27/8*b^2*arcsin(d*x + c)^2*cos(a/b)^2*e^4*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x +
c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 125/4*a*b*arcsin(d*x + c)*cos(a/b)^3*
cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d)
+ 27/4*a*b*arcsin(d*x + c)*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2
*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 75/4*a*b*arcsin(d*x + c)*cos(a/b)^2*e^4*sin(a/b)*sin_integral(5*a/b +
5*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 27/4*a*b*arcsin(d*x + c
)*cos(a/b)^2*e^4*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(
d*x + c) + a^2*b^3*d) + 5/2*((d*x + c)^2 - 1)^2*(d*x + c)*b^2*arcsin(d*x + c)*e^4/(b^5*d*arcsin(d*x + c)^2 + 2
*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 125/32*b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(5*a/b + 5*arcsin(d*
x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 125/8*a^2*cos(a/b)^3*cos_integ
ral(5*a/b + 5*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 81/32*b
^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d
*arcsin(d*x + c) + a^2*b^3*d) + 27/8*a^2*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4/(b^5*d*arcsin(
d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/16*b^2*arcsin(d*x + c)^2*cos(a/b)*cos_integral(a/b + a
rcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 25/32*b^2*arcsin(d*x +
c)^2*e^4*sin(a/b)*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c
) + a^2*b^3*d) + 75/8*a^2*cos(a/b)^2*e^4*sin(a/b)*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^5*d*arcsin(d*x +
c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 27/32*b^2*arcsin(d*x + c)^2*e^4*sin(a/b)*sin_integral(3*a/b +
3*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 27/8*a^2*cos(a/b)^2*e^4
*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b
^3*d) - 1/16*b^2*arcsin(d*x + c)^2*e^4*sin(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 +
2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 5/2*((d*x + c)^2 - 1)^2*(d*x + c)*a*b*e^4/(b^5*d*arcsin(d*x + c)^2 +
2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 3*((d*x + c)^2 - 1)*(d*x + c)*b^2*arcsin(d*x + c)*e^4/(b^5*d*arcsin(
d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 125/16*a*b*arcsin(d*x + c)*cos(a/b)*cos_integral(5*a/b +
5*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 81/16*a*b*arcsin(d
*x + c)*cos(a/b)*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x +
c) + a^2*b^3*d) - 1/8*a*b*arcsin(d*x + c)*cos(a/b)*cos_integral(a/b + arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x
+ c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 25/16*a*b*arcsin(d*x + c)*e^4*sin(a/b)*sin_integral(5*a/b +
5*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 27/16*a*b*arcsin(d*x +
c)*e^4*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) +
a^2*b^3*d) - 1/8*a*b*arcsin(d*x + c)*e^4*sin(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^
2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/2*((d*x + c)^2 - 1)^2*sqrt(-(d*x + c)^2 + 1)*b^2*e^4/(b^5*d*arc
sin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 3*((d*x + c)^2 - 1)*(d*x + c)*a*b*e^4/(b^5*d*arcsin(
d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*(d*x + c)*b^2*arcsin(d*x + c)*e^4/(b^5*d*arcsin(d*x
+ c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 125/32*a^2*cos(a/b)*cos_integral(5*a/b + 5*arcsin(d*x + c))*
e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 81/32*a^2*cos(a/b)*cos_integral(3*a/b
+ 3*arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/16*a^2*cos(a/b)
*cos_integral(a/b + arcsin(d*x + c))*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 2
5/32*a^2*e^4*sin(a/b)*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x
+ c) + a^2*b^3*d) - 27/32*a^2*e^4*sin(a/b)*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^5*d*arcsin(d*x + c)^2 +
2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) - 1/16*a^2*e^4*sin(a/b)*sin_integral(a/b + arcsin(d*x + c))/(b^5*d*arcs
in(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + (-(d*x + c)^2 + 1)^(3/2)*b^2*e^4/(b^5*d*arcsin(d*x +
c)^2 + 2*a*b^4*d*arcsin(d*x + c) + a^2*b^3*d) + 1/2*(d*x + c)*a*b*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arc
sin(d*x + c) + a^2*b^3*d) - 1/2*sqrt(-(d*x + c)^2 + 1)*b^2*e^4/(b^5*d*arcsin(d*x + c)^2 + 2*a*b^4*d*arcsin(d*x
+ c) + a^2*b^3*d)

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maple [B]  time = 0.26, size = 720, normalized size = 2.24 \[ \frac {e^{4} \left (27 \arcsin \left (d x +c \right )^{2} \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b^{2}-25 \arcsin \left (d x +c \right )^{2} \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b^{2}-25 \arcsin \left (d x +c \right )^{2} \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b^{2}-2 \arcsin \left (d x +c \right )^{2} \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b^{2}-2 \arcsin \left (d x +c \right )^{2} \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b^{2}+27 \arcsin \left (d x +c \right )^{2} \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b^{2}-4 \arcsin \left (d x +c \right ) \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a b -4 \arcsin \left (d x +c \right ) \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a b +54 \arcsin \left (d x +c \right ) \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a b +54 \arcsin \left (d x +c \right ) \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a b -50 \arcsin \left (d x +c \right ) \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a b -9 \sin \left (3 \arcsin \left (d x +c \right )\right ) \arcsin \left (d x +c \right ) b^{2}-2 \sqrt {1-\left (d x +c \right )^{2}}\, b^{2}+3 \cos \left (3 \arcsin \left (d x +c \right )\right ) b^{2}-\cos \left (5 \arcsin \left (d x +c \right )\right ) b^{2}-50 \arcsin \left (d x +c \right ) \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a b -25 \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a^{2}-25 \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a^{2}-2 \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a^{2}-2 \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a^{2}+27 \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a^{2}+27 \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a^{2}+5 \arcsin \left (d x +c \right ) \sin \left (5 \arcsin \left (d x +c \right )\right ) b^{2}+5 \sin \left (5 \arcsin \left (d x +c \right )\right ) a b +2 \arcsin \left (d x +c \right ) \left (d x +c \right ) b^{2}+2 \left (d x +c \right ) a b -9 \sin \left (3 \arcsin \left (d x +c \right )\right ) a b \right )}{32 d \left (a +b \arcsin \left (d x +c \right )\right )^{2} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^3,x)

[Out]

1/32/d*e^4*(27*arcsin(d*x+c)^2*Ci(3*arcsin(d*x+c)+3*a/b)*cos(3*a/b)*b^2-25*arcsin(d*x+c)^2*Si(5*arcsin(d*x+c)+
5*a/b)*sin(5*a/b)*b^2-25*arcsin(d*x+c)^2*Ci(5*arcsin(d*x+c)+5*a/b)*cos(5*a/b)*b^2-2*arcsin(d*x+c)^2*Si(arcsin(
d*x+c)+a/b)*sin(a/b)*b^2-2*arcsin(d*x+c)^2*Ci(arcsin(d*x+c)+a/b)*cos(a/b)*b^2+27*arcsin(d*x+c)^2*Si(3*arcsin(d
*x+c)+3*a/b)*sin(3*a/b)*b^2-9*sin(3*arcsin(d*x+c))*arcsin(d*x+c)*b^2-50*arcsin(d*x+c)*Ci(5*arcsin(d*x+c)+5*a/b
)*cos(5*a/b)*a*b-4*arcsin(d*x+c)*Si(arcsin(d*x+c)+a/b)*sin(a/b)*a*b-4*arcsin(d*x+c)*Ci(arcsin(d*x+c)+a/b)*cos(
a/b)*a*b+54*arcsin(d*x+c)*Si(3*arcsin(d*x+c)+3*a/b)*sin(3*a/b)*a*b+54*arcsin(d*x+c)*Ci(3*arcsin(d*x+c)+3*a/b)*
cos(3*a/b)*a*b-50*arcsin(d*x+c)*Si(5*arcsin(d*x+c)+5*a/b)*sin(5*a/b)*a*b-2*(1-(d*x+c)^2)^(1/2)*b^2+3*cos(3*arc
sin(d*x+c))*b^2-cos(5*arcsin(d*x+c))*b^2+5*arcsin(d*x+c)*sin(5*arcsin(d*x+c))*b^2-25*Si(5*arcsin(d*x+c)+5*a/b)
*sin(5*a/b)*a^2-25*Ci(5*arcsin(d*x+c)+5*a/b)*cos(5*a/b)*a^2+5*sin(5*arcsin(d*x+c))*a*b+2*arcsin(d*x+c)*(d*x+c)
*b^2-2*Si(arcsin(d*x+c)+a/b)*sin(a/b)*a^2-2*Ci(arcsin(d*x+c)+a/b)*cos(a/b)*a^2+2*(d*x+c)*a*b+27*Si(3*arcsin(d*
x+c)+3*a/b)*sin(3*a/b)*a^2+27*Ci(3*arcsin(d*x+c)+3*a/b)*cos(3*a/b)*a^2-9*sin(3*arcsin(d*x+c))*a*b)/(a+b*arcsin
(d*x+c))^2/b^3

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^3,x)

[Out]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ e^{4} \left (\int \frac {c^{4}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {d^{4} x^{4}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {4 c d^{3} x^{3}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {6 c^{2} d^{2} x^{2}}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx + \int \frac {4 c^{3} d x}{a^{3} + 3 a^{2} b \operatorname {asin}{\left (c + d x \right )} + 3 a b^{2} \operatorname {asin}^{2}{\left (c + d x \right )} + b^{3} \operatorname {asin}^{3}{\left (c + d x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asin(d*x+c))**3,x)

[Out]

e**4*(Integral(c**4/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x) +
Integral(d**4*x**4/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x) + I
ntegral(4*c*d**3*x**3/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3), x)
+ Integral(6*c**2*d**2*x**2/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3
), x) + Integral(4*c**3*d*x/(a**3 + 3*a**2*b*asin(c + d*x) + 3*a*b**2*asin(c + d*x)**2 + b**3*asin(c + d*x)**3
), x))

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