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

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

[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)

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Rubi [A]  time = 0.916338, antiderivative size = 365, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, 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 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]

Rule 12

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

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 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 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 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 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 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 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]

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*}

Mathematica [C]  time = 0.245863, size = 370, normalized size = 1.01 \[ \frac{i e^4 e^{-\frac{5 i a}{b}} \left (-10 e^{\frac{4 i a}{b}} \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+10 e^{\frac{6 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+5 \sqrt{3} e^{\frac{2 i a}{b}} \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-5 \sqrt{3} e^{\frac{8 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{3 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )-\sqrt{5} \sqrt{-\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},-\frac{5 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )+\sqrt{5} e^{\frac{10 i a}{b}} \sqrt{\frac{i \left (a+b \sin ^{-1}(c+d x)\right )}{b}} \text{Gamma}\left (\frac{1}{2},\frac{5 i \left (a+b \sin ^{-1}(c+d x)\right )}{b}\right )\right )}{160 d \sqrt{a+b \sin ^{-1}(c+d x)}} \]

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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Maple [A]  time = 0.086, size = 263, normalized size = 0.7 \begin{align*}{\frac{{e}^{4}\sqrt{\pi }\sqrt{2}}{80\,d}\sqrt{{b}^{-1}} \left ( \sqrt{5}\cos \left ( 5\,{\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{5}}{\sqrt{\pi }b}\sqrt{a+b\arcsin \left ( dx+c \right ) }{\frac{1}{\sqrt{{b}^{-1}}}}} \right ) +\sqrt{5}\sin \left ( 5\,{\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{5}}{\sqrt{\pi }b}\sqrt{a+b\arcsin \left ( dx+c \right ) }{\frac{1}{\sqrt{{b}^{-1}}}}} \right ) -5\,\sqrt{3}\cos \left ( 3\,{\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{3}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) -5\,\sqrt{3}\sin \left ( 3\,{\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{3}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) +10\,\cos \left ({\frac{a}{b}} \right ){\it FresnelC} \left ({\frac{\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) +10\,\sin \left ({\frac{a}{b}} \right ){\it FresnelS} \left ({\frac{\sqrt{2}\sqrt{a+b\arcsin \left ( dx+c \right ) }}{\sqrt{\pi }\sqrt{{b}^{-1}}b}} \right ) \right ) } \end{align*}

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^(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)-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)+10*cos(a/b)*Fresnel
C(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., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d e x + c e\right )}^{4}}{\sqrt{b \arcsin \left (d x + c\right ) + a}}\,{d x} \end{align*}

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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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

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: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

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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Giac [A]  time = 2.50202, size = 694, normalized size = 1.9 \begin{align*} \text{result too large to display} \end{align*}

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)