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

Optimal. Leaf size=109 \[ \frac {16 b^2 c^2 \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{3 d^3}-\frac {8 b c \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d x}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}} \]

[Out]

-2/3*(a+b*arcsin(c*x))^2/d/(d*x)^(3/2)-8/3*b*c*(a+b*arcsin(c*x))*hypergeom([-1/4, 1/2],[3/4],c^2*x^2)/d^2/(d*x
)^(1/2)+16/3*b^2*c^2*HypergeometricPFQ([1/4, 1/4, 1],[3/4, 5/4],c^2*x^2)*(d*x)^(1/2)/d^3

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Rubi [A]  time = 0.15, 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} \[ \frac {16 b^2 c^2 \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{3 d^3}-\frac {8 b c \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{3 d^2 \sqrt {d x}}-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSin[c*x])^2)/(3*d*(d*x)^(3/2)) - (8*b*c*(a + b*ArcSin[c*x])*Hypergeometric2F1[-1/4, 1/2, 3/4, c^
2*x^2])/(3*d^2*Sqrt[d*x]) + (16*b^2*c^2*Sqrt[d*x]*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2])/(3*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 \frac {\left (a+b \sin ^{-1}(c x)\right )^2}{(d x)^{5/2}} \, dx &=-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}}+\frac {(4 b c) \int \frac {a+b \sin ^{-1}(c x)}{(d x)^{3/2} \sqrt {1-c^2 x^2}} \, dx}{3 d}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c x)\right )^2}{3 d (d x)^{3/2}}-\frac {8 b c \left (a+b \sin ^{-1}(c x)\right ) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};c^2 x^2\right )}{3 d^2 \sqrt {d x}}+\frac {16 b^2 c^2 \sqrt {d x} \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )}{3 d^3}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 87, normalized size = 0.80 \[ \frac {x \left (16 b^2 c^2 x^2 \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};c^2 x^2\right )-2 \left (a+b \sin ^{-1}(c x)\right ) \left (a+4 b c x \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};c^2 x^2\right )+b \sin ^{-1}(c x)\right )\right )}{3 (d x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-2*(a + b*ArcSin[c*x])*(a + b*ArcSin[c*x] + 4*b*c*x*Hypergeometric2F1[-1/4, 1/2, 3/4, c^2*x^2]) + 16*b^2*c
^2*x^2*HypergeometricPFQ[{1/4, 1/4, 1}, {3/4, 5/4}, c^2*x^2]))/(3*(d*x)^(5/2))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (c x\right )^{2} + 2 \, a b \arcsin \left (c x\right ) + a^{2}\right )} \sqrt {d x}}{d^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arcsin(c*x)^2 + 2*a*b*arcsin(c*x) + a^2)*sqrt(d*x)/(d^3*x^3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {{\left (3 \, a^{2} c^{2} \sqrt {d} {\left (\frac {2 \, \arctan \left (\sqrt {c} \sqrt {x}\right )}{\sqrt {c} d^{3}} - \frac {\log \left (\frac {c \sqrt {x} - \sqrt {c}}{c \sqrt {x} + \sqrt {c}}\right )}{\sqrt {c} d^{3}}\right )} - 12 \, a b c^{2} \sqrt {d} \int \frac {x^{\frac {5}{2}} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{c^{2} d^{3} x^{5} - d^{3} x^{3}}\,{d x} + 8 \, b^{2} c \sqrt {d} \int \frac {\sqrt {c x + 1} \sqrt {-c x + 1} x^{\frac {3}{2}} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{c^{2} d^{3} x^{5} - d^{3} x^{3}}\,{d x} - a^{2} \sqrt {d} {\left (\frac {6 \, c^{\frac {3}{2}} \arctan \left (\sqrt {c} \sqrt {x}\right )}{d^{3}} - \frac {3 \, c^{\frac {3}{2}} \log \left (\frac {c \sqrt {x} - \sqrt {c}}{c \sqrt {x} + \sqrt {c}}\right )}{d^{3}} - \frac {4}{d^{3} x^{\frac {3}{2}}}\right )} + 12 \, a b \sqrt {d} \int \frac {\sqrt {x} \arctan \left (\frac {c x}{\sqrt {c x + 1} \sqrt {-c x + 1}}\right )}{c^{2} d^{3} x^{5} - d^{3} x^{3}}\,{d x}\right )} d^{\frac {5}{2}} x^{\frac {3}{2}} + 4 \, b^{2} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )^{2}}{6 \, d^{\frac {5}{2}} x^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((3*a^2*c^2*sqrt(d)*(2*arctan(sqrt(c)*sqrt(x))/(sqrt(c)*d^3) - log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqr
t(c)))/(sqrt(c)*d^3)) - 36*a*b*c^2*sqrt(d)*integrate(1/3*x^(5/2)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c
^2*d^3*x^5 - d^3*x^3), x) + 24*b^2*c*sqrt(d)*integrate(1/3*sqrt(c*x + 1)*sqrt(-c*x + 1)*x^(3/2)*arctan(c*x/(sq
rt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*d^3*x^5 - d^3*x^3), x) - a^2*sqrt(d)*(6*c^(3/2)*arctan(sqrt(c)*sqrt(x))/d^3
- 3*c^(3/2)*log((c*sqrt(x) - sqrt(c))/(c*sqrt(x) + sqrt(c)))/d^3 - 4/(d^3*x^(3/2))) + 36*a*b*sqrt(d)*integrate
(1/3*sqrt(x)*arctan(c*x/(sqrt(c*x + 1)*sqrt(-c*x + 1)))/(c^2*d^3*x^5 - d^3*x^3), x))*d^(5/2)*x^(3/2) + 4*b^2*a
rctan2(c*x, sqrt(c*x + 1)*sqrt(-c*x + 1))^2)/(d^(5/2)*x^(3/2))

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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