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

Optimal. Leaf size=126 \[ -\frac {16 b^2 (e (c+d x))^{3/2} \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};(c+d x)^2\right )}{3 d e^3}+\frac {8 b \sqrt {e (c+d x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 126, 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} \[ -\frac {16 b^2 (e (c+d x))^{3/2} \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};(c+d x)^2\right )}{3 d e^3}+\frac {8 b \sqrt {e (c+d x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};(c+d x)^2\right ) \left (a+b \sin ^{-1}(c+d x)\right )}{d e^2}-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSin[c + d*x])^2)/(d*e*Sqrt[e*(c + d*x)]) + (8*b*Sqrt[e*(c + d*x)]*(a + b*ArcSin[c + d*x])*Hyperg
eometric2F1[1/4, 1/2, 5/4, (c + d*x)^2])/(d*e^2) - (16*b^2*(e*(c + d*x))^(3/2)*HypergeometricPFQ[{3/4, 3/4, 1}
, {5/4, 7/4}, (c + d*x)^2])/(3*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 \frac {\left (a+b \sin ^{-1}(c+d x)\right )^2}{(c e+d e x)^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sin ^{-1}(x)\right )^2}{(e x)^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {a+b \sin ^{-1}(x)}{\sqrt {e x} \sqrt {1-x^2}} \, dx,x,c+d x\right )}{d e}\\ &=-\frac {2 \left (a+b \sin ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}}+\frac {8 b \sqrt {e (c+d x)} \left (a+b \sin ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};(c+d x)^2\right )}{d e^2}-\frac {16 b^2 (e (c+d x))^{3/2} \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};(c+d x)^2\right )}{3 d e^3}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 104, normalized size = 0.83 \[ -\frac {2 \left (8 b^2 (c+d x)^2 \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};(c+d x)^2\right )+3 \left (a+b \sin ^{-1}(c+d x)\right ) \left (a-4 b (c+d x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};(c+d x)^2\right )+b \sin ^{-1}(c+d x)\right )\right )}{3 d e \sqrt {e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*(a + b*ArcSin[c + d*x])*(a + b*ArcSin[c + d*x] - 4*b*(c + d*x)*Hypergeometric2F1[1/4, 1/2, 5/4, (c + d*
x)^2]) + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{3/4, 3/4, 1}, {5/4, 7/4}, (c + d*x)^2]))/(3*d*e*Sqrt[e*(c + d*x)
])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \arcsin \left (d x + c\right )^{2} + 2 \, a b \arcsin \left (d x + c\right ) + a^{2}\right )} \sqrt {d e x + c e}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.39, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsin \left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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