3.236 \(\int (c e+d e x)^{7/2} (a+b \sinh ^{-1}(c+d x))^2 \, dx\)

Optimal. Leaf size=134 \[ \frac{16 b^2 (e (c+d x))^{13/2} \text{HypergeometricPFQ}\left (\left \{1,\frac{13}{4},\frac{13}{4}\right \},\left \{\frac{15}{4},\frac{17}{4}\right \},-(c+d x)^2\right )}{1287 d e^3}-\frac{8 b (e (c+d x))^{11/2} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{11}{4},\frac{15}{4},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{99 d e^2}+\frac{2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d e} \]

[Out]

(2*(e*(c + d*x))^(9/2)*(a + b*ArcSinh[c + d*x])^2)/(9*d*e) - (8*b*(e*(c + d*x))^(11/2)*(a + b*ArcSinh[c + d*x]
)*Hypergeometric2F1[1/2, 11/4, 15/4, -(c + d*x)^2])/(99*d*e^2) + (16*b^2*(e*(c + d*x))^(13/2)*HypergeometricPF
Q[{1, 13/4, 13/4}, {15/4, 17/4}, -(c + d*x)^2])/(1287*d*e^3)

________________________________________________________________________________________

Rubi [A]  time = 0.212018, 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} \[ \frac{16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac{13}{4},\frac{13}{4};\frac{15}{4},\frac{17}{4};-(c+d x)^2\right )}{1287 d e^3}-\frac{8 b (e (c+d x))^{11/2} \, _2F_1\left (\frac{1}{2},\frac{11}{4};\frac{15}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{99 d e^2}+\frac{2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d e} \]

Antiderivative was successfully verified.

[In]

Int[(c*e + d*e*x)^(7/2)*(a + b*ArcSinh[c + d*x])^2,x]

[Out]

(2*(e*(c + d*x))^(9/2)*(a + b*ArcSinh[c + d*x])^2)/(9*d*e) - (8*b*(e*(c + d*x))^(11/2)*(a + b*ArcSinh[c + d*x]
)*Hypergeometric2F1[1/2, 11/4, 15/4, -(c + d*x)^2])/(99*d*e^2) + (16*b^2*(e*(c + d*x))^(13/2)*HypergeometricPF
Q[{1, 13/4, 13/4}, {15/4, 17/4}, -(c + d*x)^2])/(1287*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 (c e+d e x)^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int (e x)^{7/2} \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac{2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d e}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{(e x)^{9/2} \left (a+b \sinh ^{-1}(x)\right )}{\sqrt{1+x^2}} \, dx,x,c+d x\right )}{9 d e}\\ &=\frac{2 (e (c+d x))^{9/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{9 d e}-\frac{8 b (e (c+d x))^{11/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac{1}{2},\frac{11}{4};\frac{15}{4};-(c+d x)^2\right )}{99 d e^2}+\frac{16 b^2 (e (c+d x))^{13/2} \, _3F_2\left (1,\frac{13}{4},\frac{13}{4};\frac{15}{4},\frac{17}{4};-(c+d x)^2\right )}{1287 d e^3}\\ \end{align*}

Mathematica [A]  time = 0.13285, size = 110, normalized size = 0.82 \[ \frac{2 (e (c+d x))^{9/2} \left (8 b^2 (c+d x)^2 \text{HypergeometricPFQ}\left (\left \{1,\frac{13}{4},\frac{13}{4}\right \},\left \{\frac{15}{4},\frac{17}{4}\right \},-(c+d x)^2\right )-52 b (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{11}{4},\frac{15}{4},-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )+143 \left (a+b \sinh ^{-1}(c+d x)\right )^2\right )}{1287 d e} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*e + d*e*x)^(7/2)*(a + b*ArcSinh[c + d*x])^2,x]

[Out]

(2*(e*(c + d*x))^(9/2)*(143*(a + b*ArcSinh[c + d*x])^2 - 52*b*(c + d*x)*(a + b*ArcSinh[c + d*x])*Hypergeometri
c2F1[1/2, 11/4, 15/4, -(c + d*x)^2] + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 13/4, 13/4}, {15/4, 17/4}, -(c +
d*x)^2]))/(1287*d*e)

________________________________________________________________________________________

Maple [F]  time = 0.275, size = 0, normalized size = 0. \begin{align*} \int \left ( dex+ce \right ) ^{{\frac{7}{2}}} \left ( a+b{\it Arcsinh} \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)^(7/2)*(a+b*arcsinh(d*x+c))^2,x)

[Out]

int((d*e*x+c*e)^(7/2)*(a+b*arcsinh(d*x+c))^2,x)

________________________________________________________________________________________

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)^(7/2)*(a+b*arcsinh(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a^{2} d^{3} e^{3} x^{3} + 3 \, a^{2} c d^{2} e^{3} x^{2} + 3 \, a^{2} c^{2} d e^{3} x + a^{2} c^{3} e^{3} +{\left (b^{2} d^{3} e^{3} x^{3} + 3 \, b^{2} c d^{2} e^{3} x^{2} + 3 \, b^{2} c^{2} d e^{3} x + b^{2} c^{3} e^{3}\right )} \operatorname{arsinh}\left (d x + c\right )^{2} + 2 \,{\left (a b d^{3} e^{3} x^{3} + 3 \, a b c d^{2} e^{3} x^{2} + 3 \, a b c^{2} d e^{3} x + a b c^{3} e^{3}\right )} \operatorname{arsinh}\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)^(7/2)*(a+b*arcsinh(d*x+c))^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*e*x+c*e)**(7/2)*(a+b*asinh(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d e x + c e\right )}^{\frac{7}{2}}{\left (b \operatorname{arsinh}\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)^(7/2)*(a+b*arcsinh(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((d*e*x + c*e)^(7/2)*(b*arcsinh(d*x + c) + a)^2, x)