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

Optimal. Leaf size=130 \[ \frac {16 b^2 (e (c+d x))^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};(c+d x)^2\right )}{315 d e^3}-\frac {8 b (e (c+d x))^{7/2} \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{35 d e^2}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e} \]

[Out]

2/5*(e*(d*x+c))^(5/2)*(a+b*arcsin(d*x+c))^2/d/e-8/35*b*(e*(d*x+c))^(7/2)*(a+b*arcsin(d*x+c))*hypergeom([1/2, 7
/4],[11/4],(d*x+c)^2)/d/e^2+16/315*b^2*(e*(d*x+c))^(9/2)*HypergeometricPFQ([1, 9/4, 9/4],[11/4, 13/4],(d*x+c)^
2)/d/e^3

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Rubi [A]  time = 0.21, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {4805, 4627, 4711} \[ \frac {16 b^2 (e (c+d x))^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};(c+d x)^2\right )}{315 d e^3}-\frac {8 b (e (c+d x))^{7/2} \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{35 d e^2}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(5/2)*(a + b*ArcSin[c + d*x])^2)/(5*d*e) - (8*b*(e*(c + d*x))^(7/2)*(a + b*ArcSin[c + d*x])*H
ypergeometric2F1[1/2, 7/4, 11/4, (c + d*x)^2])/(35*d*e^2) + (16*b^2*(e*(c + d*x))^(9/2)*HypergeometricPFQ[{1,
9/4, 9/4}, {11/4, 13/4}, (c + d*x)^2])/(315*d*e^3)

Rule 4627

Int[((a_.) + ArcSin[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcSi
n[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSin[c*x])^(n - 1))/Sqrt[1
- c^2*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

Rule 4711

Int[(((a_.) + ArcSin[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x)
^(m + 1)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, c^2*x^2])/(Sqrt[d]*f*(m + 1)), x] -
Simp[(b*c*(f*x)^(m + 2)*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, c^2*x^2])/(Sqrt[d]*f^2*
(m + 1)*(m + 2)), x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0] &&  !IntegerQ[m]

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 (c e+d e x)^{3/2} \left (a+b \sin ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int (e x)^{3/2} \left (a+b \sin ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {(e x)^{5/2} \left (a+b \sin ^{-1}(x)\right )}{\sqrt {1-x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=\frac {2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e}-\frac {8 b (e (c+d x))^{7/2} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};(c+d x)^2\right )}{35 d e^2}+\frac {16 b^2 (e (c+d x))^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};(c+d x)^2\right )}{315 d e^3}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 107, normalized size = 0.82 \[ \frac {2 (e (c+d x))^{5/2} \left (8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};(c+d x)^2\right )+9 \left (a+b \sin ^{-1}(c+d x)\right ) \left (7 \left (a+b \sin ^{-1}(c+d x)\right )-4 b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};(c+d x)^2\right )\right )\right )}{315 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(5/2)*(9*(a + b*ArcSin[c + d*x])*(7*(a + b*ArcSin[c + d*x]) - 4*b*(c + d*x)*Hypergeometric2F1
[1/2, 7/4, 11/4, (c + d*x)^2]) + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 9/4, 9/4}, {11/4, 13/4}, (c + d*x)^2]
))/(315*d*e)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} d e x + a^{2} c e + {\left (b^{2} d e x + b^{2} c e\right )} \arcsin \left (d x + c\right )^{2} + 2 \, {\left (a b d e x + a b c e\right )} \arcsin \left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{\frac {3}{2}} \left (a +b \arcsin \left (d x +c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(4*(b^2*d^2*e*x^2 + 2*b^2*c*d*e*x + b^2*c^2*e)*sqrt(d*x + c)*sqrt(e)*arctan2(d*x + c, sqrt(d*x + c + 1)*s
qrt(-d*x - c + 1))^2 + (100*a*b*d^3*e^(3/2)*integrate(1/5*sqrt(d*x + c)*x^3*arctan(d*x/(sqrt(d*x + c + 1)*sqrt
(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 300*a*b*c*d^2*
e^(3/2)*integrate(1/5*sqrt(d*x + c)*x^2*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c +
1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 300*a*b*c^2*d*e^(3/2)*integrate(1/5*sqrt(d*x + c)*
x*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c
*d*x + c^2 - 1), x) + 100*a*b*c^3*e^(3/2)*integrate(1/5*sqrt(d*x + c)*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x
- c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 5*a^2*c^3*e^(3/2)*(2
*arctan(sqrt(d*x + c)) - log(sqrt(d*x + c) + 1) + log(sqrt(d*x + c) - 1))/d + 40*b^2*d^2*e^(3/2)*integrate(1/5
*sqrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*x^2*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/
(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 80*b^2*c*d*e^(3/2)*integrate(1/5*s
qrt(d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*x*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqr
t(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) + 40*b^2*c^2*e^(3/2)*integrate(1/5*sqrt(
d*x + c + 1)*sqrt(d*x + c)*sqrt(-d*x - c + 1)*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x
+ c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) - 15*(2*(c + 1)*arctan(sqrt(d*x + c)) - (c - 1
)*log(sqrt(d*x + c) + 1) + (c - 1)*log(sqrt(d*x + c) - 1) - 4*sqrt(d*x + c))*a^2*c^2*e^(3/2)/d - 100*a*b*d*e^(
3/2)*integrate(1/5*sqrt(d*x + c)*x*arctan(d*x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sq
rt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2 - 1), x) - 100*a*b*c*e^(3/2)*integrate(1/5*sqrt(d*x + c)*arctan(d*
x/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)) + c/(sqrt(d*x + c + 1)*sqrt(-d*x - c + 1)))/(d^2*x^2 + 2*c*d*x + c^2
- 1), x) + 5*(6*(c^2 + 2*c + 1)*arctan(sqrt(d*x + c)) - 3*(c^2 - 2*c + 1)*log(sqrt(d*x + c) + 1) + 3*(c^2 - 2*
c + 1)*log(sqrt(d*x + c) - 1) + 4*(d*x + c)^(3/2) - 24*sqrt(d*x + c)*c)*a^2*c*e^(3/2)/d - 5*a^2*c*e^(3/2)*(2*a
rctan(sqrt(d*x + c)) - log(sqrt(d*x + c) + 1) + log(sqrt(d*x + c) - 1))/d + (4*(d*x + c)^(5/2) - 20*(d*x + c)^
(3/2)*c - 10*(c^3 + 3*c^2 + 3*c + 1)*arctan(sqrt(d*x + c)) + 5*(c^3 - 3*c^2 + 3*c - 1)*log(sqrt(d*x + c) + 1)
- 5*(c^3 - 3*c^2 + 3*c - 1)*log(sqrt(d*x + c) - 1) + 20*(3*c^2 + 1)*sqrt(d*x + c))*a^2*e^(3/2)/d + 5*(2*(c + 1
)*arctan(sqrt(d*x + c)) - (c - 1)*log(sqrt(d*x + c) + 1) + (c - 1)*log(sqrt(d*x + c) - 1) - 4*sqrt(d*x + c))*a
^2*e^(3/2)/d)*d)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,e+d\,e\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {asin}\left (c+d\,x\right )\right )}^2 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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