3.221 \(\int \frac {(c e+d e x)^4}{(a+b \sin ^{-1}(c+d x))^2} \, dx\)

Optimal. Leaf size=258 \[ \frac {e^4 \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b^2 d}-\frac {9 e^4 \sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}+\frac {5 e^4 \sin \left (\frac {5 a}{b}\right ) \text {Ci}\left (\frac {5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a+b \sin ^{-1}(c+d x)}{b}\right )}{8 b^2 d}+\frac {9 e^4 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac {5 e^4 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )}{16 b^2 d}-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

[Out]

-1/8*e^4*cos(a/b)*Si((a+b*arcsin(d*x+c))/b)/b^2/d+9/16*e^4*cos(3*a/b)*Si(3*(a+b*arcsin(d*x+c))/b)/b^2/d-5/16*e
^4*cos(5*a/b)*Si(5*(a+b*arcsin(d*x+c))/b)/b^2/d+1/8*e^4*Ci((a+b*arcsin(d*x+c))/b)*sin(a/b)/b^2/d-9/16*e^4*Ci(3
*(a+b*arcsin(d*x+c))/b)*sin(3*a/b)/b^2/d+5/16*e^4*Ci(5*(a+b*arcsin(d*x+c))/b)*sin(5*a/b)/b^2/d-e^4*(d*x+c)^4*(
1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))

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Rubi [A]  time = 0.37, antiderivative size = 254, normalized size of antiderivative = 0.98, number of steps used = 13, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4805, 12, 4631, 3303, 3299, 3302} \[ \frac {e^4 \sin \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^2 d}-\frac {9 e^4 \sin \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b^2 d}+\frac {5 e^4 \sin \left (\frac {5 a}{b}\right ) \text {CosIntegral}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^2 d}+\frac {9 e^4 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {5 e^4 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*ArcSin[c + d*x])^2,x]

[Out]

-((e^4*(c + d*x)^4*Sqrt[1 - (c + d*x)^2])/(b*d*(a + b*ArcSin[c + d*x]))) + (e^4*CosIntegral[a/b + ArcSin[c + d
*x]]*Sin[a/b])/(8*b^2*d) - (9*e^4*CosIntegral[(3*a)/b + 3*ArcSin[c + d*x]]*Sin[(3*a)/b])/(16*b^2*d) + (5*e^4*C
osIntegral[(5*a)/b + 5*ArcSin[c + d*x]]*Sin[(5*a)/b])/(16*b^2*d) - (e^4*Cos[a/b]*SinIntegral[a/b + ArcSin[c +
d*x]])/(8*b^2*d) + (9*e^4*Cos[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcSin[c + d*x]])/(16*b^2*d) - (5*e^4*Cos[(5*a)
/b]*SinIntegral[(5*a)/b + 5*ArcSin[c + d*x]])/(16*b^2*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 4631

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[1/(b*c^(m + 1)*(n + 1)), Subst[Int[ExpandTrigReduce[(a + b*x)^(n + 1)
, Sin[x]^(m - 1)*(m - (m + 1)*Sin[x]^2), x], x], x, ArcSin[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[m, 0] && G
eQ[n, -2] && LtQ[n, -1]

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 )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sin ^{-1}(x)\right )^2} \, dx,x,c+d x\right )}{d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {e^4 \operatorname {Subst}\left (\int \left (-\frac {\sin (x)}{8 (a+b x)}+\frac {9 \sin (3 x)}{16 (a+b x)}-\frac {5 \sin (5 x)}{16 (a+b x)}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {e^4 \operatorname {Subst}\left (\int \frac {\sin (x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b d}-\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {\sin (5 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}+\frac {\left (9 e^4\right ) \operatorname {Subst}\left (\int \frac {\sin (3 x)}{a+b x} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}-\frac {\left (e^4 \cos \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 )}{8 b d}+\frac {\left (9 e^4 \cos \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 )}{16 b d}-\frac {\left (5 e^4 \cos \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 )}{16 b d}+\frac {\left (e^4 \sin \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 )}{8 b d}-\frac {\left (9 e^4 \sin \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 )}{16 b d}+\frac {\left (5 e^4 \sin \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 )}{16 b d}\\ &=-\frac {e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \left (a+b \sin ^{-1}(c+d x)\right )}+\frac {e^4 \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right ) \sin \left (\frac {a}{b}\right )}{8 b^2 d}-\frac {9 e^4 \text {Ci}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right ) \sin \left (\frac {3 a}{b}\right )}{16 b^2 d}+\frac {5 e^4 \text {Ci}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right ) \sin \left (\frac {5 a}{b}\right )}{16 b^2 d}-\frac {e^4 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )}{8 b^2 d}+\frac {9 e^4 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \sin ^{-1}(c+d x)\right )}{16 b^2 d}-\frac {5 e^4 \cos \left (\frac {5 a}{b}\right ) \text {Si}\left (\frac {5 a}{b}+5 \sin ^{-1}(c+d x)\right )}{16 b^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.29, size = 283, normalized size = 1.10 \[ \frac {e^4 \left (16 \left (-3 \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )+\sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+3 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-\cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )+5 \left (10 \sin \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )-5 \sin \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )+\sin \left (\frac {5 a}{b}\right ) \text {Ci}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-10 \cos \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )+5 \cos \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )-\cos \left (\frac {5 a}{b}\right ) \text {Si}\left (5 \left (\frac {a}{b}+\sin ^{-1}(c+d x)\right )\right )\right )-\frac {16 b \sqrt {1-(c+d x)^2} (c+d x)^4}{a+b \sin ^{-1}(c+d x)}\right )}{16 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*ArcSin[c + d*x])^2,x]

[Out]

(e^4*((-16*b*(c + d*x)^4*Sqrt[1 - (c + d*x)^2])/(a + b*ArcSin[c + d*x]) + 16*(-3*CosIntegral[a/b + ArcSin[c +
d*x]]*Sin[a/b] + CosIntegral[3*(a/b + ArcSin[c + d*x])]*Sin[(3*a)/b] + 3*Cos[a/b]*SinIntegral[a/b + ArcSin[c +
d*x]] - Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c + d*x])]) + 5*(10*CosIntegral[a/b + ArcSin[c + d*x]]*Sin[a
/b] - 5*CosIntegral[3*(a/b + ArcSin[c + d*x])]*Sin[(3*a)/b] + CosIntegral[5*(a/b + ArcSin[c + d*x])]*Sin[(5*a)
/b] - 10*Cos[a/b]*SinIntegral[a/b + ArcSin[c + d*x]] + 5*Cos[(3*a)/b]*SinIntegral[3*(a/b + ArcSin[c + d*x])] -
Cos[(5*a)/b]*SinIntegral[5*(a/b + ArcSin[c + d*x])])))/(16*b^2*d)

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}{b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [B]  time = 0.47, size = 1374, normalized size = 5.33 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^2,x, algorithm="giac")

[Out]

5*b*arcsin(d*x + c)*cos(a/b)^4*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a
*b^2*d) - 5*b*arcsin(d*x + c)*cos(a/b)^5*e^4*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) +
a*b^2*d) + 5*a*cos(a/b)^4*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*
d) - 5*a*cos(a/b)^5*e^4*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 15/4*b*arc
sin(d*x + c)*cos(a/b)^2*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d)
- 9/4*b*arcsin(d*x + c)*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c
) + a*b^2*d) + 25/4*b*arcsin(d*x + c)*cos(a/b)^3*e^4*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^3*d*arcsin(d*x
+ c) + a*b^2*d) + 9/4*b*arcsin(d*x + c)*cos(a/b)^3*e^4*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^3*d*arcsin(
d*x + c) + a*b^2*d) - 15/4*a*cos(a/b)^2*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x
+ c) + a*b^2*d) - 9/4*a*cos(a/b)^2*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c
) + a*b^2*d) + 25/4*a*cos(a/b)^3*e^4*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d)
+ 9/4*a*cos(a/b)^3*e^4*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 5/16*b*arc
sin(d*x + c)*cos_integral(5*a/b + 5*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 9/16*b*a
rcsin(d*x + c)*cos_integral(3*a/b + 3*arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 1/8*b*
arcsin(d*x + c)*cos_integral(a/b + arcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 25/16*b*a
rcsin(d*x + c)*cos(a/b)*e^4*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 27/16*
b*arcsin(d*x + c)*cos(a/b)*e^4*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - 1/8
*b*arcsin(d*x + c)*cos(a/b)*e^4*sin_integral(a/b + arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) - ((d*x
+ c)^2 - 1)^2*sqrt(-(d*x + c)^2 + 1)*b*e^4/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 5/16*a*cos_integral(5*a/b + 5*a
rcsin(d*x + c))*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 9/16*a*cos_integral(3*a/b + 3*arcsin(d*x + c)
)*e^4*sin(a/b)/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 1/8*a*cos_integral(a/b + arcsin(d*x + c))*e^4*sin(a/b)/(b^3
*d*arcsin(d*x + c) + a*b^2*d) - 25/16*a*cos(a/b)*e^4*sin_integral(5*a/b + 5*arcsin(d*x + c))/(b^3*d*arcsin(d*x
+ c) + a*b^2*d) - 27/16*a*cos(a/b)*e^4*sin_integral(3*a/b + 3*arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2
*d) - 1/8*a*cos(a/b)*e^4*sin_integral(a/b + arcsin(d*x + c))/(b^3*d*arcsin(d*x + c) + a*b^2*d) + 2*(-(d*x + c)
^2 + 1)^(3/2)*b*e^4/(b^3*d*arcsin(d*x + c) + a*b^2*d) - sqrt(-(d*x + c)^2 + 1)*b*e^4/(b^3*d*arcsin(d*x + c) +
a*b^2*d)

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maple [A]  time = 0.23, size = 397, normalized size = 1.54 \[ \frac {e^{4} \left (9 \arcsin \left (d x +c \right ) \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) b -9 \arcsin \left (d x +c \right ) \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) b -5 \arcsin \left (d x +c \right ) \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) b +5 \arcsin \left (d x +c \right ) \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) b -2 \arcsin \left (d x +c \right ) \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) b +2 \arcsin \left (d x +c \right ) \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) b +9 \Si \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \cos \left (\frac {3 a}{b}\right ) a -9 \Ci \left (3 \arcsin \left (d x +c \right )+\frac {3 a}{b}\right ) \sin \left (\frac {3 a}{b}\right ) a -5 \Si \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \cos \left (\frac {5 a}{b}\right ) a +5 \Ci \left (5 \arcsin \left (d x +c \right )+\frac {5 a}{b}\right ) \sin \left (\frac {5 a}{b}\right ) a -2 \Si \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \cos \left (\frac {a}{b}\right ) a +2 \Ci \left (\arcsin \left (d x +c \right )+\frac {a}{b}\right ) \sin \left (\frac {a}{b}\right ) a +3 \cos \left (3 \arcsin \left (d x +c \right )\right ) b -\cos \left (5 \arcsin \left (d x +c \right )\right ) b -2 \sqrt {1-\left (d x +c \right )^{2}}\, b \right )}{16 d \left (a +b \arcsin \left (d x +c \right )\right ) b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d*e^4*(9*arcsin(d*x+c)*Si(3*arcsin(d*x+c)+3*a/b)*cos(3*a/b)*b-9*arcsin(d*x+c)*Ci(3*arcsin(d*x+c)+3*a/b)*s
in(3*a/b)*b-5*arcsin(d*x+c)*Si(5*arcsin(d*x+c)+5*a/b)*cos(5*a/b)*b+5*arcsin(d*x+c)*Ci(5*arcsin(d*x+c)+5*a/b)*s
in(5*a/b)*b-2*arcsin(d*x+c)*Si(arcsin(d*x+c)+a/b)*cos(a/b)*b+2*arcsin(d*x+c)*Ci(arcsin(d*x+c)+a/b)*sin(a/b)*b+
9*Si(3*arcsin(d*x+c)+3*a/b)*cos(3*a/b)*a-9*Ci(3*arcsin(d*x+c)+3*a/b)*sin(3*a/b)*a-5*Si(5*arcsin(d*x+c)+5*a/b)*
cos(5*a/b)*a+5*Ci(5*arcsin(d*x+c)+5*a/b)*sin(5*a/b)*a-2*Si(arcsin(d*x+c)+a/b)*cos(a/b)*a+2*Ci(arcsin(d*x+c)+a/
b)*sin(a/b)*a+3*cos(3*arcsin(d*x+c))*b-cos(5*arcsin(d*x+c))*b-2*(1-(d*x+c)^2)^(1/2)*b)/(a+b*arcsin(d*x+c))/b^2

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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))^2,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^2,x)

[Out]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asin(d*x+c))**2,x)

[Out]

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

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