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

Optimal. Leaf size=134 \[ \frac {16 b^2 (e (c+d x))^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};-(c+d x)^2\right )}{315 d e^3}-\frac {8 b (e (c+d x))^{7/2} \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{35 d e^2}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e} \]

[Out]

2/5*(e*(d*x+c))^(5/2)*(a+b*arcsinh(d*x+c))^2/d/e-8/35*b*(e*(d*x+c))^(7/2)*(a+b*arcsinh(d*x+c))*hypergeom([1/2,
7/4],[11/4],-(d*x+c)^2)/d/e^2+16/315*b^2*(e*(d*x+c))^(9/2)*HypergeometricPFQ([1, 9/4, 9/4],[11/4, 13/4],-(d*x
+c)^2)/d/e^3

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Rubi [A]  time = 0.22, antiderivative size = 134, 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 = {5865, 5661, 5762} \[ \frac {16 b^2 (e (c+d x))^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};-(c+d x)^2\right )}{315 d e^3}-\frac {8 b (e (c+d x))^{7/2} \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{35 d e^2}+\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(5/2)*(a + b*ArcSinh[c + d*x])^2)/(5*d*e) - (8*b*(e*(c + d*x))^(7/2)*(a + b*ArcSinh[c + d*x])
*Hypergeometric2F1[1/2, 7/4, 11/4, -(c + d*x)^2])/(35*d*e^2) + (16*b^2*(e*(c + d*x))^(9/2)*HypergeometricPFQ[{
1, 9/4, 9/4}, {11/4, 13/4}, -(c + d*x)^2])/(315*d*e^3)

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]

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
]

Rubi steps

\begin {align*} \int (c e+d e x)^{3/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2 \, dx &=\frac {\operatorname {Subst}\left (\int (e x)^{3/2} \left (a+b \sinh ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e}-\frac {(4 b) \operatorname {Subst}\left (\int \frac {(e x)^{5/2} \left (a+b \sinh ^{-1}(x)\right )}{\sqrt {1+x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=\frac {2 (e (c+d x))^{5/2} \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e}-\frac {8 b (e (c+d x))^{7/2} \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};-(c+d x)^2\right )}{35 d e^2}+\frac {16 b^2 (e (c+d x))^{9/2} \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};-(c+d x)^2\right )}{315 d e^3}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 110, normalized size = 0.82 \[ \frac {2 (e (c+d x))^{5/2} \left (8 b^2 (c+d x)^2 \, _3F_2\left (1,\frac {9}{4},\frac {9}{4};\frac {11}{4},\frac {13}{4};-(c+d x)^2\right )-36 b (c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{4};\frac {11}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )+63 \left (a+b \sinh ^{-1}(c+d x)\right )^2\right )}{315 d e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(e*(c + d*x))^(5/2)*(63*(a + b*ArcSinh[c + d*x])^2 - 36*b*(c + d*x)*(a + b*ArcSinh[c + d*x])*Hypergeometric
2F1[1/2, 7/4, 11/4, -(c + d*x)^2] + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{1, 9/4, 9/4}, {11/4, 13/4}, -(c + d*x
)^2]))/(315*d*e)

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} d e x + a^{2} c e + {\left (b^{2} d e x + b^{2} c e\right )} \operatorname {arsinh}\left (d x + c\right )^{2} + 2 \, {\left (a b d e x + a b c e\right )} \operatorname {arsinh}\left (d x + c\right )\right )} \sqrt {d e x + c e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (d e x +c e \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (d x +c \right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(d*e*x + c*e)^(5/2)*a^2/(d*e) + 2/5*(b^2*d^2*e^(3/2)*x^2 + 2*b^2*c*d*e^(3/2)*x + b^2*c^2*e^(3/2))*sqrt(d*x
+ c)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2/d + integrate(-2/5*((2*b^2*c^3*e^(3/2) - (5*a*b*d^3*e
^(3/2) - 2*b^2*d^3*e^(3/2))*x^3 - 5*(c^3*e^(3/2) + c*e^(3/2))*a*b - 3*(5*a*b*c*d^2*e^(3/2) - 2*b^2*c*d^2*e^(3/
2))*x^2 + (6*b^2*c^2*d*e^(3/2) - 5*(3*c^2*d*e^(3/2) + d*e^(3/2))*a*b)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*sqr
t(d*x + c) - ((5*a*b*d^4*e^(3/2) - 2*b^2*d^4*e^(3/2))*x^4 + 4*(5*a*b*c*d^3*e^(3/2) - 2*b^2*c*d^3*e^(3/2))*x^3
+ 5*(c^4*e^(3/2) + c^2*e^(3/2))*a*b - 2*(c^4*e^(3/2) + c^2*e^(3/2))*b^2 + (5*(6*c^2*d^2*e^(3/2) + d^2*e^(3/2))
*a*b - 2*(6*c^2*d^2*e^(3/2) + d^2*e^(3/2))*b^2)*x^2 + 2*(5*(2*c^3*d*e^(3/2) + c*d*e^(3/2))*a*b - 2*(2*c^3*d*e^
(3/2) + c*d*e^(3/2))*b^2)*x)*sqrt(d*x + c))*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^3*x^3 + 3*c*d^
2*x^2 + c^3 + (3*c^2*d + d)*x + (d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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