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

Optimal. Leaf size=134 \[ \frac{16 b^2 \sqrt{e (c+d x)} \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{4},1\right \},\left \{\frac{3}{4},\frac{5}{4}\right \},-(c+d x)^2\right )}{3 d e^3}-\frac{8 b \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^2 \sqrt{e (c+d x)}}-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}} \]

[Out]

(-2*(a + b*ArcSinh[c + d*x])^2)/(3*d*e*(e*(c + d*x))^(3/2)) - (8*b*(a + b*ArcSinh[c + d*x])*Hypergeometric2F1[
-1/4, 1/2, 3/4, -(c + d*x)^2])/(3*d*e^2*Sqrt[e*(c + d*x)]) + (16*b^2*Sqrt[e*(c + d*x)]*HypergeometricPFQ[{1/4,
1/4, 1}, {3/4, 5/4}, -(c + d*x)^2])/(3*d*e^3)

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Rubi [A]  time = 0.226133, antiderivative size = 134, 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 = {5865, 5661, 5762} \[ \frac{16 b^2 \sqrt{e (c+d x)} \, _3F_2\left (\frac{1}{4},\frac{1}{4},1;\frac{3}{4},\frac{5}{4};-(c+d x)^2\right )}{3 d e^3}-\frac{8 b \, _2F_1\left (-\frac{1}{4},\frac{1}{2};\frac{3}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e^2 \sqrt{e (c+d x)}}-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{3 d e (e (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSinh[c + d*x])^2)/(3*d*e*(e*(c + d*x))^(3/2)) - (8*b*(a + b*ArcSinh[c + d*x])*Hypergeometric2F1[
-1/4, 1/2, 3/4, -(c + d*x)^2])/(3*d*e^2*Sqrt[e*(c + d*x)]) + (16*b^2*Sqrt[e*(c + d*x)]*HypergeometricPFQ[{1/4,
1/4, 1}, {3/4, 5/4}, -(c + d*x)^2])/(3*d*e^3)

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 5661

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcS
inh[c*x])^n)/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcSinh[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 5762

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x
)^(m + 1)*(a + b*ArcSinh[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[e, c^2*d] && GtQ[d, 0] &&  !IntegerQ[m]

Rubi steps

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

Mathematica [A]  time = 0.0752128, size = 106, normalized size = 0.79 \[ -\frac{2 \left (4 b (c+d x) \left (\text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )-2 b (c+d x) \text{HypergeometricPFQ}\left (\left \{\frac{1}{4},\frac{1}{4},1\right \},\left \{\frac{3}{4},\frac{5}{4}\right \},-(c+d x)^2\right )\right )+\left (a+b \sinh ^{-1}(c+d x)\right )^2\right )}{3 d e (e (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.247, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( dx+c \right ) \right ) ^{2} \left ( dex+ce \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^(5/2),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((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^(5/2),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 (\frac{{\left (b^{2} \operatorname{arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname{arsinh}\left (d x + c\right ) + a^{2}\right )} \sqrt{d e x + c e}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{asinh}{\left (c + d x \right )}\right )^{2}}{\left (e \left (c + d x\right )\right )^{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*asinh(c + d*x))**2/(e*(c + d*x))**(5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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