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

Optimal. Leaf size=398 \[ \frac {e^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{2 b c \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 e^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {e^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b c e^2 x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {4 b e^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {4 b^2 e^2 \left (1-c^2 x^2\right )}{c \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b^2 e^2 x \left (1-c^2 x^2\right )}{4 \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {b^2 e^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{4 c \sqrt {c d x+d} \sqrt {e-c e x}} \]

[Out]

-4*b^2*e^2*(-c^2*x^2+1)/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/4*b^2*e^2*x*(-c^2*x^2+1)/(c*d*x+d)^(1/2)/(-c*e*x+
e)^(1/2)+2*e^2*(-c^2*x^2+1)*(a+b*arcsin(c*x))^2/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/2*e^2*x*(-c^2*x^2+1)*(a+b
*arcsin(c*x))^2/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)-1/4*b^2*e^2*arcsin(c*x)*(-c^2*x^2+1)^(1/2)/c/(c*d*x+d)^(1/2)/
(-c*e*x+e)^(1/2)-4*b*e^2*x*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/2*b*c*e^2*x
^2*(a+b*arcsin(c*x))*(-c^2*x^2+1)^(1/2)/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)+1/2*e^2*(a+b*arcsin(c*x))^3*(-c^2*x^2
+1)^(1/2)/b/c/(c*d*x+d)^(1/2)/(-c*e*x+e)^(1/2)

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Rubi [A]  time = 0.58, antiderivative size = 398, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {4673, 4773, 3317, 3296, 2638, 3311, 32, 2635, 8} \[ \frac {e^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )^3}{2 b c \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {2 e^2 \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{c \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {e^2 x \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )^2}{2 \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b c e^2 x^2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{2 \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {4 b e^2 x \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {c d x+d} \sqrt {e-c e x}}-\frac {4 b^2 e^2 \left (1-c^2 x^2\right )}{c \sqrt {c d x+d} \sqrt {e-c e x}}+\frac {b^2 e^2 x \left (1-c^2 x^2\right )}{4 \sqrt {c d x+d} \sqrt {e-c e x}}-\frac {b^2 e^2 \sqrt {1-c^2 x^2} \sin ^{-1}(c x)}{4 c \sqrt {c d x+d} \sqrt {e-c e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b^2*e^2*(1 - c^2*x^2))/(c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (b^2*e^2*x*(1 - c^2*x^2))/(4*Sqrt[d + c*d*x]*
Sqrt[e - c*e*x]) - (b^2*e^2*Sqrt[1 - c^2*x^2]*ArcSin[c*x])/(4*c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) - (4*b*e^2*x*
Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x]))/(Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (b*c*e^2*x^2*Sqrt[1 - c^2*x^2]*(a +
b*ArcSin[c*x]))/(2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (2*e^2*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(c*Sqrt[d +
c*d*x]*Sqrt[e - c*e*x]) - (e^2*x*(1 - c^2*x^2)*(a + b*ArcSin[c*x])^2)/(2*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]) + (
e^2*Sqrt[1 - c^2*x^2]*(a + b*ArcSin[c*x])^3)/(2*b*c*Sqrt[d + c*d*x]*Sqrt[e - c*e*x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

Rule 4673

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :> D
ist[((d + e*x)^q*(f + g*x)^q)/(1 - c^2*x^2)^q, Int[(d + e*x)^(p - q)*(1 - c^2*x^2)^q*(a + b*ArcSin[c*x])^n, x]
, x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 - e^2, 0] && HalfIntegerQ[p, q]
&& GeQ[p - q, 0]

Rule 4773

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sin[x])^m, x], x, ArcSin[c*x]], x] /; FreeQ[{a,
b, c, d, e, f, g, n}, x] && EqQ[c^2*d + e, 0] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rubi steps

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

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Mathematica [A]  time = 2.39, size = 358, normalized size = 0.90 \[ \frac {e \sqrt {c d x+d} \sqrt {e-c e x} \left (-4 \left (a^2 (c x-4) \sqrt {1-c^2 x^2}+8 a b c x+8 b^2 \sqrt {1-c^2 x^2}\right )-2 a b \cos \left (2 \sin ^{-1}(c x)\right )+b^2 \sin \left (2 \sin ^{-1}(c x)\right )\right )-12 a^2 \sqrt {d} e^{3/2} \sqrt {1-c^2 x^2} \tan ^{-1}\left (\frac {c x \sqrt {c d x+d} \sqrt {e-c e x}}{\sqrt {d} \sqrt {e} \left (c^2 x^2-1\right )}\right )+2 b e \sqrt {c d x+d} \sqrt {e-c e x} \sin ^{-1}(c x)^2 \left (6 a+8 b \sqrt {1-c^2 x^2}-b \sin \left (2 \sin ^{-1}(c x)\right )\right )-2 b e \sqrt {c d x+d} \sqrt {e-c e x} \sin ^{-1}(c x) \left (4 a (c x-4) \sqrt {1-c^2 x^2}+16 b c x+b \cos \left (2 \sin ^{-1}(c x)\right )\right )+4 b^2 e \sqrt {c d x+d} \sqrt {e-c e x} \sin ^{-1}(c x)^3}{8 c d \sqrt {1-c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b^2*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*ArcSin[c*x]^3 - 12*a^2*Sqrt[d]*e^(3/2)*Sqrt[1 - c^2*x^2]*ArcTan[(c*x*
Sqrt[d + c*d*x]*Sqrt[e - c*e*x])/(Sqrt[d]*Sqrt[e]*(-1 + c^2*x^2))] - 2*b*e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]*Arc
Sin[c*x]*(16*b*c*x + 4*a*(-4 + c*x)*Sqrt[1 - c^2*x^2] + b*Cos[2*ArcSin[c*x]]) + 2*b*e*Sqrt[d + c*d*x]*Sqrt[e -
c*e*x]*ArcSin[c*x]^2*(6*a + 8*b*Sqrt[1 - c^2*x^2] - b*Sin[2*ArcSin[c*x]]) + e*Sqrt[d + c*d*x]*Sqrt[e - c*e*x]
*(-4*(8*a*b*c*x + 8*b^2*Sqrt[1 - c^2*x^2] + a^2*(-4 + c*x)*Sqrt[1 - c^2*x^2]) - 2*a*b*Cos[2*ArcSin[c*x]] + b^2
*Sin[2*ArcSin[c*x]]))/(8*c*d*Sqrt[1 - c^2*x^2])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (a^{2} c e x - a^{2} e + {\left (b^{2} c e x - b^{2} e\right )} \arcsin \left (c x\right )^{2} + 2 \, {\left (a b c e x - a b e\right )} \arcsin \left (c x\right )\right )} \sqrt {-c e x + e}}{\sqrt {c d x + d}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(a^2*c*e*x - a^2*e + (b^2*c*e*x - b^2*e)*arcsin(c*x)^2 + 2*(a*b*c*e*x - a*b*e)*arcsin(c*x))*sqrt(-c*
e*x + e)/sqrt(c*d*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:Warning, integrat
ion of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep)]S
implification assuming t_nostep near 0Simplification assuming t_nostep near 0Simplification assuming t_nostep
near 0Simplification assuming t_nostep near 0Warning, integration of abs or sign assumes constant sign by inte
rvals (correct if the argument is real):Check [abs(t_nostep)]Warning, integration of abs or sign assumes const
ant sign by intervals (correct if the argument is real):Check [abs(t_nostep)]Simplification assuming t_nostep
near 0Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Simplification assuming t_
nostep near 0Simplification assuming t_nostep near 0Simplification assuming t_nostep near 0Simplification assu
ming t_nostep near 0Simplification assuming t_nostep near 0Warning, integration of abs or sign assumes constan
t sign by intervals (correct if the argument is real):Check [abs(t_nostep)]Warning, integration of abs or sign
assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep)]sym2poly/r2sym(cons
t gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (-c e x +e \right )^{\frac {3}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}}{\sqrt {c d x +d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{2} \, {\left (\frac {\sqrt {-c^{2} d e x^{2} + d e} e x}{d} - \frac {3 \, e^{2} \arcsin \left (c x\right )}{\sqrt {d e} c} - \frac {4 \, \sqrt {-c^{2} d e x^{2} + d e} e}{c d}\right )} a^{2} - \sqrt {d} \sqrt {e} \int \frac {{\left ({\left (b^{2} c e x - b^{2} e\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + 2 \, {\left (a b c e x - a b e\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )\right )} \sqrt {c x + 1} \sqrt {-c x + 1}}{c d x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(sqrt(-c^2*d*e*x^2 + d*e)*e*x/d - 3*e^2*arcsin(c*x)/(sqrt(d*e)*c) - 4*sqrt(-c^2*d*e*x^2 + d*e)*e/(c*d))*a
^2 - sqrt(d)*sqrt(e)*integrate(((b^2*c*e*x - b^2*e)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 2*(a*b*c*e*
x - a*b*e)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1)))*sqrt(c*x + 1)*sqrt(-c*x + 1)/(c*d*x + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- e \left (c x - 1\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asin}{\left (c x \right )}\right )^{2}}{\sqrt {d \left (c x + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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