3.3.60 \(\int \frac {(c e+d e x)^3}{\sqrt {a+b \text {ArcSin}(c+d x)}} \, dx\) [260]

Optimal. Leaf size=233 \[ -\frac {e^3 \sqrt {\frac {\pi }{2}} \cos \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {e^3 \sqrt {\pi } \cos \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{4 \sqrt {b} d}-\frac {e^3 \sqrt {\pi } \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right ) \sin \left (\frac {2 a}{b}\right )}{4 \sqrt {b} d}+\frac {e^3 \sqrt {\frac {\pi }{2}} \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right ) \sin \left (\frac {4 a}{b}\right )}{8 \sqrt {b} d} \]

[Out]

-1/16*e^3*cos(4*a/b)*FresnelS(2*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)
+1/16*e^3*FresnelC(2*2^(1/2)/Pi^(1/2)*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2))*sin(4*a/b)*2^(1/2)*Pi^(1/2)/d/b^(1/2)
+1/4*e^3*cos(2*a/b)*FresnelS(2*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*Pi^(1/2)/d/b^(1/2)-1/4*e^3*FresnelC
(2*(a+b*arcsin(d*x+c))^(1/2)/b^(1/2)/Pi^(1/2))*sin(2*a/b)*Pi^(1/2)/d/b^(1/2)

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Rubi [A]
time = 0.32, antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4889, 12, 4731, 4491, 3387, 3386, 3432, 3385, 3433} \begin {gather*} -\frac {\sqrt {\pi } e^3 \sin \left (\frac {2 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {\pi } \sqrt {b}}\right )}{4 \sqrt {b} d}+\frac {\sqrt {\frac {\pi }{2}} e^3 \sin \left (\frac {4 a}{b}\right ) \text {FresnelC}\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}-\frac {\sqrt {\frac {\pi }{2}} e^3 \cos \left (\frac {4 a}{b}\right ) S\left (\frac {2 \sqrt {\frac {2}{\pi }} \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b}}\right )}{8 \sqrt {b} d}+\frac {\sqrt {\pi } e^3 \cos \left (\frac {2 a}{b}\right ) S\left (\frac {2 \sqrt {a+b \text {ArcSin}(c+d x)}}{\sqrt {b} \sqrt {\pi }}\right )}{4 \sqrt {b} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(e^3*Sqrt[Pi/2]*Cos[(4*a)/b]*FresnelS[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]])/(Sqrt[b]*d) +
(e^3*Sqrt[Pi]*Cos[(2*a)/b]*FresnelS[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])])/(4*Sqrt[b]*d) - (e^3*
Sqrt[Pi]*FresnelC[(2*Sqrt[a + b*ArcSin[c + d*x]])/(Sqrt[b]*Sqrt[Pi])]*Sin[(2*a)/b])/(4*Sqrt[b]*d) + (e^3*Sqrt[
Pi/2]*FresnelC[(2*Sqrt[2/Pi]*Sqrt[a + b*ArcSin[c + d*x]])/Sqrt[b]]*Sin[(4*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 3385

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 3386

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 3387

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 3432

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

Rule 3433

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

Rule 4491

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 4731

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

Rule 4889

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.09, size = 249, normalized size = 1.07 \begin {gather*} \frac {e^3 e^{-\frac {4 i a}{b}} \left (-2 \sqrt {2} e^{\frac {2 i a}{b}} \sqrt {-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},-\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}\right )-2 \sqrt {2} e^{\frac {6 i a}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},\frac {2 i (a+b \text {ArcSin}(c+d x))}{b}\right )+\sqrt {-\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},-\frac {4 i (a+b \text {ArcSin}(c+d x))}{b}\right )+e^{\frac {8 i a}{b}} \sqrt {\frac {i (a+b \text {ArcSin}(c+d x))}{b}} \text {Gamma}\left (\frac {1}{2},\frac {4 i (a+b \text {ArcSin}(c+d x))}{b}\right )\right )}{32 d \sqrt {a+b \text {ArcSin}(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*(-2*Sqrt[2]*E^(((2*I)*a)/b)*Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-2*I)*(a + b*ArcSin[c + d
*x]))/b] - 2*Sqrt[2]*E^(((6*I)*a)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((2*I)*(a + b*ArcSin[c + d
*x]))/b] + Sqrt[((-I)*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((-4*I)*(a + b*ArcSin[c + d*x]))/b] + E^(((8*I)*a
)/b)*Sqrt[(I*(a + b*ArcSin[c + d*x]))/b]*Gamma[1/2, ((4*I)*(a + b*ArcSin[c + d*x]))/b]))/(32*d*E^(((4*I)*a)/b)
*Sqrt[a + b*ArcSin[c + d*x]])

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Maple [A]
time = 0.40, size = 211, normalized size = 0.91

method	result	size
default	\(\frac {e^{3} \sqrt {\pi }\, \sqrt {2}\, \sqrt {-\frac {1}{b}}\, \left (2 \cos \left (\frac {2 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {-\frac {2}{b}}\, b +2 \sin \left (\frac {2 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {2}{b}}\, b}\right ) \sqrt {-\frac {1}{b}}\, \sqrt {-\frac {2}{b}}\, b +\cos \left (\frac {4 a}{b}\right ) \mathrm {S}\left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )+\sin \left (\frac {4 a}{b}\right ) \FresnelC \left (\frac {2 \sqrt {2}\, \sqrt {a +b \arcsin \left (d x +c \right )}}{\sqrt {\pi }\, \sqrt {-\frac {1}{b}}\, b}\right )\right )}{16 d}\)	\(211\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16/d*e^3*Pi^(1/2)*2^(1/2)*(-1/b)^(1/2)*(2*cos(2*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d*
x+c))^(1/2)/b)*(-1/b)^(1/2)*(-2/b)^(1/2)*b+2*sin(2*a/b)*FresnelC(2*2^(1/2)/Pi^(1/2)/(-2/b)^(1/2)*(a+b*arcsin(d
*x+c))^(1/2)/b)*(-1/b)^(1/2)*(-2/b)^(1/2)*b+cos(4*a/b)*FresnelS(2*2^(1/2)/Pi^(1/2)/(-1/b)^(1/2)*(a+b*arcsin(d*
x+c))^(1/2)/b)+sin(4*a/b)*FresnelC(2*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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{3} \left (\int \frac {c^{3}}{\sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}}\, dx + \int \frac {d^{3} x^{3}}{\sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}}\, dx + \int \frac {3 c d^{2} x^{2}}{\sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}}\, dx + \int \frac {3 c^{2} d x}{\sqrt {a + b \operatorname {asin}{\left (c + d x \right )}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**3/(a+b*asin(d*x+c))**(1/2),x)

[Out]

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

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Giac [C] Result contains complex when optimal does not.
time = 0.82, size = 318, normalized size = 1.36 \begin {gather*} \frac {i \, \sqrt {\pi } e^{3} \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {4 i \, a}{b}\right )}}{16 \, {\left (\sqrt {2} \sqrt {b} + \frac {i \, \sqrt {2} b^{\frac {3}{2}}}{{\left | b \right |}}\right )} d} + \frac {i \, \sqrt {\pi } e^{3} \operatorname {erf}\left (-\frac {\sqrt {b \arcsin \left (d x + c\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {2 i \, a}{b}\right )}}{8 \, d {\left (\sqrt {b} - \frac {i \, b^{\frac {3}{2}}}{{\left | b \right |}}\right )}} - \frac {i \, \sqrt {\pi } e^{3} \operatorname {erf}\left (-\frac {\sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a}}{\sqrt {b}} + \frac {i \, \sqrt {2} \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (-\frac {4 i \, a}{b}\right )}}{16 \, {\left (\sqrt {2} \sqrt {b} - \frac {i \, \sqrt {2} b^{\frac {3}{2}}}{{\left | b \right |}}\right )} d} - \frac {i \, \sqrt {\pi } e^{3} \operatorname {erf}\left (-\frac {\sqrt {b \arcsin \left (d x + c\right ) + a}}{\sqrt {b}} - \frac {i \, \sqrt {b \arcsin \left (d x + c\right ) + a} \sqrt {b}}{{\left | b \right |}}\right ) e^{\left (\frac {2 i \, a}{b}\right )}}{8 \, \sqrt {b} d {\left (\frac {i \, b}{{\left | b \right |}} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)^3/(a+b*arcsin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/16*I*sqrt(pi)*e^3*erf(-sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(b) - I*sqrt(2)*sqrt(b*arcsin(d*x + c) + a)*s
qrt(b)/abs(b))*e^(4*I*a/b)/((sqrt(2)*sqrt(b) + I*sqrt(2)*b^(3/2)/abs(b))*d) + 1/8*I*sqrt(pi)*e^3*erf(-sqrt(b*a
rcsin(d*x + c) + a)/sqrt(b) + I*sqrt(b*arcsin(d*x + c) + a)*sqrt(b)/abs(b))*e^(-2*I*a/b)/(d*(sqrt(b) - I*b^(3/
2)/abs(b))) - 1/16*I*sqrt(pi)*e^3*erf(-sqrt(2)*sqrt(b*arcsin(d*x + c) + a)/sqrt(b) + I*sqrt(2)*sqrt(b*arcsin(d
*x + c) + a)*sqrt(b)/abs(b))*e^(-4*I*a/b)/((sqrt(2)*sqrt(b) - I*sqrt(2)*b^(3/2)/abs(b))*d) - 1/8*I*sqrt(pi)*e^
3*erf(-sqrt(b*arcsin(d*x + c) + a)/sqrt(b) - I*sqrt(b*arcsin(d*x + c) + a)*sqrt(b)/abs(b))*e^(2*I*a/b)/(sqrt(b
)*d*(I*b/abs(b) + 1))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c\,e+d\,e\,x\right )}^3}{\sqrt {a+b\,\mathrm {asin}\left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*e + d*e*x)^3/(a + b*asin(c + d*x))^(1/2),x)

[Out]

int((c*e + d*e*x)^3/(a + b*asin(c + d*x))^(1/2), x)

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