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

Optimal. Leaf size=365 \[ \frac {\sqrt {\frac {\pi }{2}} e^4 \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {b} d}-\frac {\sqrt {\frac {3 \pi }{2}} e^4 \cos \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{10}} e^4 \cos \left (\frac {5 a}{b}\right ) C\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{2}} e^4 \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {b} d}-\frac {\sqrt {\frac {3 \pi }{2}} e^4 \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{10}} e^4 \sin \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d} \]

[Out]

1/80*e^4*cos(5*a/b)*FresnelC(10^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*10^(1/2)*Pi^(1/2)/d/b^(1/2)+
1/80*e^4*FresnelS(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)/d/b^(1/2)+
1/8*e^4*cos(a/b)*FresnelC(2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*2^(1/2)*Pi^(1/2)/d/b^(1/2)+1/8*e
^4*FresnelS(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)/d/b^(1/2)-1/16*e^4*c
os(3*a/b)*FresnelC(6^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*6^(1/2)*Pi^(1/2)/d/b^(1/2)-1/16*e^4*Fre
snelS(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)/d/b^(1/2)

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Rubi [A]  time = 0.92, antiderivative size = 365, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4805, 12, 4635, 4406, 3306, 3305, 3351, 3304, 3352} \[ \frac {\sqrt {\frac {\pi }{2}} e^4 \cos \left (\frac {a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {b} d}-\frac {\sqrt {\frac {3 \pi }{2}} e^4 \cos \left (\frac {3 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{10}} e^4 \cos \left (\frac {5 a}{b}\right ) \text {FresnelC}\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{2}} e^4 \sin \left (\frac {a}{b}\right ) S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {b} d}-\frac {\sqrt {\frac {3 \pi }{2}} e^4 \sin \left (\frac {3 a}{b}\right ) S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{10}} e^4 \sin \left (\frac {5 a}{b}\right ) S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^4*Sqrt[Pi/2]*Cos[a/b]*FresnelC[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(4*Sqrt[b]*d) - (e^4*Sqrt
[(3*Pi)/2]*Cos[(3*a)/b]*FresnelC[(Sqrt[6/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(8*Sqrt[b]*d) + (e^4*Sqrt[
Pi/10]*Cos[(5*a)/b]*FresnelC[(Sqrt[10/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(8*Sqrt[b]*d) + (e^4*Sqrt[Pi/
2]*FresnelS[(Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[a/b])/(4*Sqrt[b]*d) - (e^4*Sqrt[(3*Pi)/2]*Fr
esnelS[(Sqrt[6/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(3*a)/b])/(8*Sqrt[b]*d) + (e^4*Sqrt[Pi/10]*Fresne
lS[(Sqrt[10/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(5*a)/b])/(8*Sqrt[b]*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 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 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 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}{\sqrt {a+b \sin ^{-1}(c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {e^4 x^4}{\sqrt {a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {x^4}{\sqrt {a+b \sin ^{-1}(x)}} \, dx,x,c+d x\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^4(x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{8 \sqrt {a+b x}}-\frac {3 \cos (3 x)}{16 \sqrt {a+b x}}+\frac {\cos (5 x)}{16 \sqrt {a+b x}}\right ) \, dx,x,\sin ^{-1}(c+d x)\right )}{d}\\ &=\frac {e^4 \operatorname {Subst}\left (\int \frac {\cos (5 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}+\frac {e^4 \operatorname {Subst}\left (\int \frac {\cos (x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{8 d}-\frac {\left (3 e^4\right ) \operatorname {Subst}\left (\int \frac {\cos (3 x)}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}\\ &=\frac {\left (e^4 \cos \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 )}{8 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 )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}+\frac {\left (e^4 \cos \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 )}{16 d}+\frac {\left (e^4 \sin \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 )}{8 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 )}{\sqrt {a+b x}} \, dx,x,\sin ^{-1}(c+d x)\right )}{16 d}+\frac {\left (e^4 \sin \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 )}{16 d}\\ &=\frac {\left (e^4 \cos \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 )}{4 b d}-\frac {\left (3 e^4 \cos \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 )}{8 b d}+\frac {\left (e^4 \cos \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 )}{8 b d}+\frac {\left (e^4 \sin \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 )}{4 b d}-\frac {\left (3 e^4 \sin \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 )}{8 b d}+\frac {\left (e^4 \sin \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 )}{8 b d}\\ &=\frac {e^4 \sqrt {\frac {\pi }{2}} \cos \left (\frac {a}{b}\right ) C\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{4 \sqrt {b} d}-\frac {e^4 \sqrt {\frac {3 \pi }{2}} \cos \left (\frac {3 a}{b}\right ) C\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {e^4 \sqrt {\frac {\pi }{10}} \cos \left (\frac {5 a}{b}\right ) C\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {e^4 \sqrt {\frac {\pi }{2}} S\left (\frac {\sqrt {\frac {2}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {a}{b}\right )}{4 \sqrt {b} d}-\frac {e^4 \sqrt {\frac {3 \pi }{2}} S\left (\frac {\sqrt {\frac {6}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {3 a}{b}\right )}{8 \sqrt {b} d}+\frac {e^4 \sqrt {\frac {\pi }{10}} S\left (\frac {\sqrt {\frac {10}{\pi }} \sqrt {a+b \sin ^{-1}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {5 a}{b}\right )}{8 \sqrt {b} d}\\ \end {align*}

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Mathematica [C]  time = 0.27, size = 370, normalized size = 1.01 \[ \frac {i e^4 e^{-\frac {5 i a}{b}} \left (-10 e^{\frac {4 i a}{b}} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+10 e^{\frac {6 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+5 \sqrt {3} e^{\frac {2 i a}{b}} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-5 \sqrt {3} e^{\frac {8 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-\sqrt {5} \sqrt {-\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},-\frac {5 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+\sqrt {5} e^{\frac {10 i a}{b}} \sqrt {\frac {i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \Gamma \left (\frac {1}{2},\frac {5 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )}{160 d \sqrt {a+b \sin ^{-1}(c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/160)*e^4*(-10*E^(((4*I)*a)/b)*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-I)*(a + b*ArcSin[c + d*
x]))/b] + 10*E^(((6*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, (I*(a + b*ArcSin[c + d*x]))/b] + 5
*Sqrt[3]*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-3*I)*(a + b*ArcSin[c + d*x]))/b]
- 5*Sqrt[3]*E^(((8*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((3*I)*(a + b*ArcSin[c + d*x]))/b]
- Sqrt[5]*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-5*I)*(a + b*ArcSin[c + d*x]))/b] + Sqrt[5]*E^(
((10*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((5*I)*(a + b*ArcSin[c + d*x]))/b]))/(d*E^(((5*I)
*a)/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))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 3.16, size = 514, normalized size = 1.41 \[ -\frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {\sqrt {10} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b} i}{2 \, {\left | b \right |}} - \frac {\sqrt {10} \sqrt {b \arcsin \left (d x + c\right ) + a}}{2 \, \sqrt {b}}\right ) e^{\left (\frac {5 \, a i}{b} + 4\right )}}{16 \, {\left (\frac {\sqrt {10} b^{\frac {3}{2}} i}{{\left | b \right |}} + \sqrt {10} \sqrt {b}\right )} d} + \frac {\sqrt {6} \sqrt {\pi } \operatorname {erf}\left (-\frac {\sqrt {6} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b} i}{2 \, {\left | b \right |}} - \frac {\sqrt {6} \sqrt {b \arcsin \left (d x + c\right ) + a}}{2 \, \sqrt {b}}\right ) e^{\left (\frac {3 \, a i}{b} + 4\right )}}{32 \, \sqrt {b} d {\left (\frac {b i}{{\left | b \right |}} + 1\right )}} - \frac {\sqrt {\pi } \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a} i}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (\frac {a i}{b} + 4\right )}}{8 \, {\left (\frac {\sqrt {2} b i}{\sqrt {{\left | b \right |}}} + \sqrt {2} \sqrt {{\left | b \right |}}\right )} d} + \frac {\sqrt {\pi } \operatorname {erf}\left (\frac {\sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a} i}{2 \, \sqrt {{\left | b \right |}}} - \frac {\sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {{\left | b \right |}}}{2 \, b}\right ) e^{\left (-\frac {a i}{b} + 4\right )}}{8 \, {\left (\frac {\sqrt {2} b i}{\sqrt {{\left | b \right |}}} - \sqrt {2} \sqrt {{\left | b \right |}}\right )} d} - \frac {\sqrt {6} \sqrt {\pi } \operatorname {erf}\left (\frac {\sqrt {6} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b} i}{2 \, {\left | b \right |}} - \frac {\sqrt {6} \sqrt {b \arcsin \left (d x + c\right ) + a}}{2 \, \sqrt {b}}\right ) e^{\left (-\frac {3 \, a i}{b} + 4\right )}}{32 \, \sqrt {b} d {\left (\frac {b i}{{\left | b \right |}} - 1\right )}} + \frac {\sqrt {\pi } \operatorname {erf}\left (\frac {\sqrt {10} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b} i}{2 \, {\left | b \right |}} - \frac {\sqrt {10} \sqrt {b \arcsin \left (d x + c\right ) + a}}{2 \, \sqrt {b}}\right ) e^{\left (-\frac {5 \, a i}{b} + 4\right )}}{16 \, {\left (\frac {\sqrt {10} b^{\frac {3}{2}} i}{{\left | b \right |}} - \sqrt {10} \sqrt {b}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/16*sqrt(pi)*erf(-1/2*sqrt(10)*sqrt(b*arcsin(d*x + c) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(10)*sqrt(b*arcsin(d*x
+ c) + a)/sqrt(b))*e^(5*a*i/b + 4)/((sqrt(10)*b^(3/2)*i/abs(b) + sqrt(10)*sqrt(b))*d) + 1/32*sqrt(6)*sqrt(pi)
*erf(-1/2*sqrt(6)*sqrt(b*arcsin(d*x + c) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(6)*sqrt(b*arcsin(d*x + c) + a)/sqrt(
b))*e^(3*a*i/b + 4)/(sqrt(b)*d*(b*i/abs(b) + 1)) - 1/8*sqrt(pi)*erf(-1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*i
/sqrt(abs(b)) - 1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(a*i/b + 4)/((sqrt(2)*b*i/sqrt(abs(b
)) + sqrt(2)*sqrt(abs(b)))*d) + 1/8*sqrt(pi)*erf(1/2*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*i/sqrt(abs(b)) - 1/2*
sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*sqrt(abs(b))/b)*e^(-a*i/b + 4)/((sqrt(2)*b*i/sqrt(abs(b)) - sqrt(2)*sqrt(a
bs(b)))*d) - 1/32*sqrt(6)*sqrt(pi)*erf(1/2*sqrt(6)*sqrt(b*arcsin(d*x + c) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(6)*
sqrt(b*arcsin(d*x + c) + a)/sqrt(b))*e^(-3*a*i/b + 4)/(sqrt(b)*d*(b*i/abs(b) - 1)) + 1/16*sqrt(pi)*erf(1/2*sqr
t(10)*sqrt(b*arcsin(d*x + c) + a)*sqrt(b)*i/abs(b) - 1/2*sqrt(10)*sqrt(b*arcsin(d*x + c) + a)/sqrt(b))*e^(-5*a
*i/b + 4)/((sqrt(10)*b^(3/2)*i/abs(b) - sqrt(10)*sqrt(b))*d)

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maple [A]  time = 0.48, size = 263, normalized size = 0.72 \[ \frac {e^{4} \sqrt {\frac {1}{b}}\, \sqrt {\pi }\, \sqrt {2}\, \left (-5 \sqrt {3}\, \cos \left (\frac {3 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )-5 \sqrt {3}\, \sin \left (\frac {3 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {3}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+\sqrt {5}\, \cos \left (\frac {5 a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+\sqrt {5}\, \sin \left (\frac {5 a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {5}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+10 \cos \left (\frac {a}{b}\right ) \FresnelC \left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )+10 \sin \left (\frac {a}{b}\right ) \mathrm {S}\left (\frac {\sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {\frac {1}{b}}\, b}\right )\right )}{80 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^(1/2),x)

[Out]

1/80/d*e^4*(1/b)^(1/2)*Pi^(1/2)*2^(1/2)*(-5*3^(1/2)*cos(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)-5*3^(1/2)*sin(3*a/b)*FresnelS(2^(1/2)/Pi^(1/2)*3^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x
+c))^(1/2)/b)+5^(1/2)*cos(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)+5^
(1/2)*sin(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)+10*cos(a/b)*Fresne
lC(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b)+10*sin(a/b)*FresnelS(2^(1/2)/Pi^(1/2)/(1/b)^(1/2)
*(a+b*arcsin(d*x+c))^(1/2)/b))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^4/(a+b*arcsin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^(1/2),x)

[Out]

int((c*e + d*e*x)^4/(a + b*asin(c + d*x))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**4/(a+b*asin(d*x+c))**(1/2),x)

[Out]

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

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