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

Optimal. Leaf size=130 \[ \frac{16 b^2 (e (c+d x))^{9/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{9}{4},\frac{9}{4}\right \},\left \{\frac{11}{4},\frac{13}{4}\right \},(c+d x)^2\right )}{315 d e^3}-\frac{8 b (e (c+d x))^{7/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{7}{4},\frac{11}{4},(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{35 d e^2}+\frac{2 (e (c+d x))^{5/2} \left (a+b \sin ^{-1}(c+d x)\right )^2}{5 d e} \]

[Out]

(2*(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)

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Rubi [A]  time = 0.205796, antiderivative size = 130, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, 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 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 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]

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*}

Mathematica [A]  time = 0.107328, size = 107, normalized size = 0.82 \[ \frac{2 (e (c+d x))^{5/2} \left (8 b^2 (c+d x)^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{9}{4},\frac{9}{4}\right \},\left \{\frac{11}{4},\frac{13}{4}\right \},(c+d x)^2\right )+9 \left (a+b \sin ^{-1}(c+d x)\right ) \left (7 \left (a+b \sin ^{-1}(c+d x)\right )-4 b (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{7}{4},\frac{11}{4},(c+d x)^2\right )\right )\right )}{315 d e} \]

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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Maple [F]  time = 0.293, size = 0, normalized size = 0. \begin{align*} \int \left ( dex+ce \right ) ^{{\frac{3}{2}}} \left ( a+b\arcsin \left ( dx+c \right ) \right ) ^{2}\, dx \end{align*}

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

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]

Exception raised: ValueError

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

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

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{\frac{3}{2}}{\left (b \arcsin \left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

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)