3.265 \(\int \frac {(c e+d e x)^4}{(a+b \sin ^{-1}(c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=412 \[ \frac {\sqrt {\frac {\pi }{2}} e^4 \sin \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{3/2} d}-\frac {3 \sqrt {\frac {3 \pi }{2}} e^4 \sin \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}+\frac {\sqrt {\frac {5 \pi }{2}} e^4 \sin \left (\frac {5 a}{b}\right ) C\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}-\frac {\sqrt {\frac {\pi }{2}} e^4 \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{3/2} d}+\frac {3 \sqrt {\frac {3 \pi }{2}} e^4 \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}-\frac {\sqrt {\frac {5 \pi }{2}} e^4 \cos \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}-\frac {2 e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}} \]

[Out]

-1/4*e^4*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/b^(3/2)/d+1/4*
e^4*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*sin(a/b)*2^(1/2)*Pi^(1/2)/b^(3/2)/d+3/8*e^4*c
os(3*a/b)*FresnelS(6^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*6^(1/2)*Pi^(1/2)/b^(3/2)/d-3/8*e^4*Fres
nelC(6^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*sin(3*a/b)*6^(1/2)*Pi^(1/2)/b^(3/2)/d-1/8*e^4*cos(5*a
/b)*FresnelS(10^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*10^(1/2)*Pi^(1/2)/b^(3/2)/d+1/8*e^4*FresnelC
(10^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*sin(5*a/b)*10^(1/2)*Pi^(1/2)/b^(3/2)/d-2*e^4*(d*x+c)^4*(
1-(d*x+c)^2)^(1/2)/b/d/(a+b*arcsin(d*x+c))^(1/2)

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Rubi [A]  time = 0.87, antiderivative size = 412, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4805, 12, 4631, 3306, 3305, 3351, 3304, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} e^4 \sin \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{3/2} d}-\frac {3 \sqrt {\frac {3 \pi }{2}} e^4 \sin \left (\frac {3 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}+\frac {\sqrt {\frac {5 \pi }{2}} e^4 \sin \left (\frac {5 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}-\frac {\sqrt {\frac {\pi }{2}} e^4 \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{3/2} d}+\frac {3 \sqrt {\frac {3 \pi }{2}} e^4 \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}-\frac {\sqrt {\frac {5 \pi }{2}} e^4 \cos \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}-\frac {2 e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^4/(a + b*ArcSin[c + d*x])^(3/2),x]

[Out]

(-2*e^4*(c + d*x)^4*Sqrt[1 - (c + d*x)^2])/(b*d*Sqrt[a + b*ArcSin[c + d*x]]) - (e^4*Sqrt[Pi/2]*Cos[a/b]*Fresne
lS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(2*b^(3/2)*d) + (3*e^4*Sqrt[(3*Pi)/2]*Cos[(3*a)/b]*Fresn
elS[(Sqrt[6/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(4*b^(3/2)*d) - (e^4*Sqrt[(5*Pi)/2]*Cos[(5*a)/b]*Fresne
lS[(Sqrt[10/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(4*b^(3/2)*d) + (e^4*Sqrt[Pi/2]*FresnelC[(Sqrt[2/Pi]*Sq
rt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(2*b^(3/2)*d) - (3*e^4*Sqrt[(3*Pi)/2]*FresnelC[(Sqrt[6/Pi]*Sqrt[
a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(3*a)/b])/(4*b^(3/2)*d) + (e^4*Sqrt[(5*Pi)/2]*FresnelC[(Sqrt[10/Pi]*Sqrt[
a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(5*a)/b])/(4*b^(3/2)*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 3304

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

Rule 3305

Int[sin[(e_.) + (f_.)*(x_)]/Sqrt[(c_.) + (d_.)*(x_)], x_Symbol] :> Dist[2/d, Subst[Int[Sin[(f*x^2)/d], x], x,
Sqrt[c + d*x]], x] /; FreeQ[{c, d, e, f}, x] && ComplexFreeQ[f] && EqQ[d*e - c*f, 0]

Rule 3306

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

Rule 3351

Int[Sin[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelS[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

Rule 3352

Int[Cos[(d_.)*((e_.) + (f_.)*(x_))^2], x_Symbol] :> Simp[(Sqrt[Pi/2]*FresnelC[Sqrt[2/Pi]*Rt[d, 2]*(e + f*x)])/
(f*Rt[d, 2]), x] /; FreeQ[{d, e, f}, x]

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 )^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\left (a+b \sin ^{-1}(x)\right )^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}+\frac {\left (2 e^4\right ) \operatorname {Subst}\left (\int \left (-\frac {\sin (x)}{8 \sqrt {a+b x}}+\frac {9 \sin (3 x)}{16 \sqrt {a+b x}}-\frac {5 \sin (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {e^4 \operatorname {Subst}\left (\int \frac {\sin (x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 b d}-\frac {\left (5 e^4\right ) \operatorname {Subst}\left (\int \frac {\sin (5 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b d}+\frac {\left (9 e^4\right ) \operatorname {Subst}\left (\int \frac {\sin (3 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {\left (e^4 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 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 )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 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 )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 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 )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{4 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 )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 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 )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 b d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {\left (e^4 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{2 b^2 d}+\frac {\left (9 e^4 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {3 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{4 b^2 d}-\frac {\left (5 e^4 \cos \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \sin \left (\frac {5 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{4 b^2 d}+\frac {\left (e^4 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{2 b^2 d}-\frac {\left (9 e^4 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {3 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{4 b^2 d}+\frac {\left (5 e^4 \sin \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \cos \left (\frac {5 x^2}{b}\right ) \, dx,x,\sqrt {a+b \sin ^{-1}(c+d x)}\right )}{4 b^2 d}\\ &=-\frac {2 e^4 (c+d x)^4 \sqrt {1-(c+d x)^2}}{b d \sqrt {a+b \sin ^{-1}(c+d x)}}-\frac {e^4 \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{2 b^{3/2} d}+\frac {3 e^4 \sqrt {\frac {3 \pi }{2}} \cos \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}-\frac {e^4 \sqrt {\frac {5 \pi }{2}} \cos \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 b^{3/2} d}+\frac {e^4 \sqrt {\frac {\pi }{2}} C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{2 b^{3/2} d}-\frac {3 e^4 \sqrt {\frac {3 \pi }{2}} C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{4 b^{3/2} d}+\frac {e^4 \sqrt {\frac {5 \pi }{2}} C\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {5 a}{b}\right )}{4 b^{3/2} d}\\ \end {align*}

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Mathematica [C]  time = 0.74, size = 572, normalized size = 1.39 \[ \frac {e^4 e^{-\frac {5 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \left (3 e^{\frac {5 i a}{b}+2 i \sin ^{-1}(c+d x)}-2 e^{\frac {5 i a}{b}+4 i \sin ^{-1}(c+d x)}-2 e^{\frac {5 i a}{b}+6 i \sin ^{-1}(c+d x)}+3 e^{\frac {5 i a}{b}+8 i \sin ^{-1}(c+d x)}-e^{\frac {5 i \left (a+2 b \sin ^{-1}(c+d x)\right )}{b}}+2 e^{\frac {4 i a}{b}+5 i \sin ^{-1}(c+d x)} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+2 e^{\frac {6 i a}{b}+5 i \sin ^{-1}(c+d x)} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-3 \sqrt {3} e^{\frac {2 i a}{b}+5 i \sin ^{-1}(c+d x)} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-3 \sqrt {3} e^{\frac {8 i a}{b}+5 i \sin ^{-1}(c+d x)} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+\sqrt {5} e^{5 i \sin ^{-1}(c+d x)} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {5 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+\sqrt {5} e^{\frac {5 i \left (2 a+b \sin ^{-1}(c+d x)\right )}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {5 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-e^{\frac {5 i a}{b}}\right )}{16 b d \sqrt {a+b \sin ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c*e + d*e*x)^4/(a + b*ArcSin[c + d*x])^(3/2),x]

[Out]

(e^4*(-E^(((5*I)*a)/b) + 3*E^(((5*I)*a)/b + (2*I)*ArcSin[c + d*x]) - 2*E^(((5*I)*a)/b + (4*I)*ArcSin[c + d*x])
- 2*E^(((5*I)*a)/b + (6*I)*ArcSin[c + d*x]) + 3*E^(((5*I)*a)/b + (8*I)*ArcSin[c + d*x]) - E^(((5*I)*(a + 2*b*
ArcSin[c + d*x]))/b) + 2*E^(((4*I)*a)/b + (5*I)*ArcSin[c + d*x])*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[
1/2, ((-I)*(a + b*ArcSin[c + d*x]))/b] + 2*E^(((6*I)*a)/b + (5*I)*ArcSin[c + d*x])*Sqrt[(I*(a + b*ArcSin[c + d
*x]))/b]*Gamma[1/2, (I*(a + b*ArcSin[c + d*x]))/b] - 3*Sqrt[3]*E^(((2*I)*a)/b + (5*I)*ArcSin[c + d*x])*Sqrt[((
-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-3*I)*(a + b*ArcSin[c + d*x]))/b] - 3*Sqrt[3]*E^(((8*I)*a)/b + (5
*I)*ArcSin[c + d*x])*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((3*I)*(a + b*ArcSin[c + d*x]))/b] + Sqrt[
5]*E^((5*I)*ArcSin[c + d*x])*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-5*I)*(a + b*ArcSin[c + d*x])
)/b] + Sqrt[5]*E^(((5*I)*(2*a + b*ArcSin[c + d*x]))/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((5*I)*(
a + b*ArcSin[c + d*x]))/b]))/(16*b*d*E^(((5*I)*(a + b*ArcSin[c + d*x]))/b)*Sqrt[a + b*ArcSin[c + d*x]])

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.41, size = 478, normalized size = 1.16 \[ -\frac {e^{4} \left (\sqrt {5}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {5 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-\sqrt {5}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {5 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-3 \sqrt {3}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+3 \sqrt {3}\, \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+2 \sqrt {2}\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \cos \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\pi }-2 \sqrt {2}\, \sqrt {\frac {1}{b}}\, \sqrt {a +b \arcsin \left (d x +c \right )}\, \sin \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right ) \sqrt {\pi }+2 \cos \left (\frac {a +b \arcsin \left (d x +c \right )}{b}-\frac {a}{b}\right )-3 \cos \left (\frac {3 a +3 b \arcsin \left (d x +c \right )}{b}-\frac {3 a}{b}\right )+\cos \left (\frac {5 a +5 b \arcsin \left (d x +c \right )}{b}-\frac {5 a}{b}\right )\right )}{8 d b \sqrt {a +b \arcsin \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8/d*e^4/b*(5^(1/2)*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(5*a/b)*FresnelS(2^(1/2)/Pi^(1
/2)*5^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)-5^(1/2)*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^
(1/2)*sin(5*a/b)*FresnelC(2^(1/2)/Pi^(1/2)*5^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)-3*3^(1/2)*(1/b)^(1
/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*cos(3*a/b)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*a
rcsin(d*x+c))^(1/2)/b)+3*3^(1/2)*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(a+b*arcsin(d*x+c))^(1/2)*sin(3*a/b)*FresnelC(2^
(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)+2*2^(1/2)*(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2
)*cos(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*Pi^(1/2)-2*2^(1/2)*(1/b)^(1/2)*(
a+b*arcsin(d*x+c))^(1/2)*sin(a/b)*FresnelC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)*Pi^(1/2)+
2*cos((a+b*arcsin(d*x+c))/b-a/b)-3*cos(3*(a+b*arcsin(d*x+c))/b-3*a/b)+cos(5*(a+b*arcsin(d*x+c))/b-5*a/b))/(a+b
*arcsin(d*x+c))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (c\,e+d\,e\,x\right )}^4}{{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^(3/2),x)

[Out]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asin(d*x+c))**(3/2),x)

[Out]

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

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