3.209 \(\int (d x)^{5/2} (a+b \sin ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=109 \[ \frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3}-\frac {8 b c (d x)^{9/2} \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{63 d^2}+\frac {2 (d x)^{7/2} \left (a+b \sin ^{-1}(c x)\right )^2}{7 d} \]

[Out]

2/7*(d*x)^(7/2)*(a+b*arcsin(c*x))^2/d-8/63*b*c*(d*x)^(9/2)*(a+b*arcsin(c*x))*hypergeom([1/2, 9/4],[13/4],c^2*x
^2)/d^2+16/693*b^2*c^2*(d*x)^(11/2)*HypergeometricPFQ([1, 11/4, 11/4],[13/4, 15/4],c^2*x^2)/d^3

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Rubi [A]  time = 0.14, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4627, 4711} \[ \frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3}-\frac {8 b c (d x)^{9/2} \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{63 d^2}+\frac {2 (d x)^{7/2} \left (a+b \sin ^{-1}(c x)\right )^2}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[(d*x)^(5/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(2*(d*x)^(7/2)*(a + b*ArcSin[c*x])^2)/(7*d) - (8*b*c*(d*x)^(9/2)*(a + b*ArcSin[c*x])*Hypergeometric2F1[1/2, 9/
4, 13/4, c^2*x^2])/(63*d^2) + (16*b^2*c^2*(d*x)^(11/2)*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, c^2*x^
2])/(693*d^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]

Rubi steps

\begin {align*} \int (d x)^{5/2} \left (a+b \sin ^{-1}(c x)\right )^2 \, dx &=\frac {2 (d x)^{7/2} \left (a+b \sin ^{-1}(c x)\right )^2}{7 d}-\frac {(4 b c) \int \frac {(d x)^{7/2} \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {1-c^2 x^2}} \, dx}{7 d}\\ &=\frac {2 (d x)^{7/2} \left (a+b \sin ^{-1}(c x)\right )^2}{7 d}-\frac {8 b c (d x)^{9/2} \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};c^2 x^2\right )}{63 d^2}+\frac {16 b^2 c^2 (d x)^{11/2} \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )}{693 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 90, normalized size = 0.83 \[ \frac {2}{693} x (d x)^{5/2} \left (8 b^2 c^2 x^2 \, _3F_2\left (1,\frac {11}{4},\frac {11}{4};\frac {13}{4},\frac {15}{4};c^2 x^2\right )+11 \left (a+b \sin ^{-1}(c x)\right ) \left (9 \left (a+b \sin ^{-1}(c x)\right )-4 b c x \, _2F_1\left (\frac {1}{2},\frac {9}{4};\frac {13}{4};c^2 x^2\right )\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(d*x)^(5/2)*(a + b*ArcSin[c*x])^2,x]

[Out]

(2*x*(d*x)^(5/2)*(11*(a + b*ArcSin[c*x])*(9*(a + b*ArcSin[c*x]) - 4*b*c*x*Hypergeometric2F1[1/2, 9/4, 13/4, c^
2*x^2]) + 8*b^2*c^2*x^2*HypergeometricPFQ[{1, 11/4, 11/4}, {13/4, 15/4}, c^2*x^2]))/693

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="fricas")

[Out]

integral((b^2*d^2*x^2*arcsin(c*x)^2 + 2*a*b*d^2*x^2*arcsin(c*x) + a^2*d^2*x^2)*sqrt(d*x), 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((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="giac")

[Out]

Exception raised: RuntimeError >> An error occurred running a Giac command:INPUT:sage2OUTPUT:sym2poly/r2sym(co
nst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [F]  time = 0.18, size = 0, normalized size = 0.00 \[ \int \left (d x \right )^{\frac {5}{2}} \left (a +b \arcsin \left (c x \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x)

[Out]

int((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2}{7} \, b^{2} d^{\frac {5}{2}} x^{\frac {7}{2}} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2} + \frac {1}{42} \, a^{2} c^{2} d^{\frac {5}{2}} {\left (\frac {4 \, {\left (3 \, c^{2} x^{\frac {7}{2}} + 7 \, x^{\frac {3}{2}}\right )}}{c^{4}} + \frac {42 \, \arctan \left (\sqrt {c} \sqrt {x}\right )}{c^{\frac {11}{2}}} + \frac {21 \, \log \left (\frac {c \sqrt {x} - \sqrt {c}}{c \sqrt {x} + \sqrt {c}}\right )}{c^{\frac {11}{2}}}\right )} + 14 \, a b c^{2} d^{\frac {5}{2}} \int \frac {x^{\frac {9}{2}} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{7 \, {\left (c^{2} x^{2} - 1\right )}}\,{d x} + 4 \, b^{2} c d^{\frac {5}{2}} \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{\frac {7}{2}} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{7 \, {\left (c^{2} x^{2} - 1\right )}}\,{d x} - \frac {1}{6} \, a^{2} d^{\frac {5}{2}} {\left (\frac {4 \, x^{\frac {3}{2}}}{c^{2}} + \frac {6 \, \arctan \left (\sqrt {c} \sqrt {x}\right )}{c^{\frac {7}{2}}} + \frac {3 \, \log \left (\frac {c \sqrt {x} - \sqrt {c}}{c \sqrt {x} + \sqrt {c}}\right )}{c^{\frac {7}{2}}}\right )} - 14 \, a b d^{\frac {5}{2}} \int \frac {x^{\frac {5}{2}} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{7 \, {\left (c^{2} x^{2} - 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)^(5/2)*(a+b*arcsin(c*x))^2,x, algorithm="maxima")

[Out]

2/7*b^2*d^(5/2)*x^(7/2)*arctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2 + 1/42*a^2*c^2*d^(5/2)*(4*(3*c^2*x^(7/2)
+ 7*x^(3/2))/c^4 + 42*arctan(sqrt(c)*sqrt(x))/c^(11/2) + 21*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/c
^(11/2)) + 14*a*b*c^2*d^(5/2)*integrate(1/7*x^(9/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*x^2 - 1),
x) + 4*b^2*c*d^(5/2)*integrate(1/7*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(7/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x +
1)))/(c^2*x^2 - 1), x) - 1/6*a^2*d^(5/2)*(4*x^(3/2)/c^2 + 6*arctan(sqrt(c)*sqrt(x))/c^(7/2) + 3*log((c*sqrt(x)
- sqrt(c))/(c*sqrt(x) + sqrt(c)))/c^(7/2)) - 14*a*b*d^(5/2)*integrate(1/7*x^(5/2)*arctan(c*x/(sqrt(c*x + 1)*s
qrt(-c*x + 1)))/(c^2*x^2 - 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}^2\,{\left (d\,x\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(c*x))^2*(d*x)^(5/2),x)

[Out]

int((a + b*asin(c*x))^2*(d*x)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x)**(5/2)*(a+b*asin(c*x))**2,x)

[Out]

Timed out

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