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

Optimal. Leaf size=130 \[ -\frac {16 b^2 (e (c+d x))^{3/2} \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};-(c+d x)^2\right )}{3 d e^3}+\frac {8 b \sqrt {e (c+d x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}} \]

[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*arcsinh(d*x+c)
)^2/d/e/(e*(d*x+c))^(1/2)+8*b*(a+b*arcsinh(d*x+c))*hypergeom([1/4, 1/2],[5/4],-(d*x+c)^2)*(e*(d*x+c))^(1/2)/d/
e^2

________________________________________________________________________________________

Rubi [A]  time = 0.22, antiderivative size = 130, 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} \[ -\frac {16 b^2 (e (c+d x))^{3/2} \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};-(c+d x)^2\right )}{3 d e^3}+\frac {8 b \sqrt {e (c+d x)} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{d e^2}-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSinh[c + d*x])^2)/(d*e*Sqrt[e*(c + d*x)]) + (8*b*Sqrt[e*(c + d*x)]*(a + b*ArcSinh[c + d*x])*Hype
rgeometric2F1[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 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 \frac {\left (a+b \sinh ^{-1}(c+d x)\right )^2}{(c e+d e x)^{3/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{(e x)^{3/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{\sqrt {e x} \sqrt {1+x^2}} \, dx,x,c+d x\right )}{d e}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{d e \sqrt {e (c+d x)}}+\frac {8 b \sqrt {e (c+d x)} \left (a+b \sinh ^{-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*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 109, normalized size = 0.84 \[ \frac {2 \left (-4 b (c+d x) \left (2 b (c+d x) \, _3F_2\left (\frac {3}{4},\frac {3}{4},1;\frac {5}{4},\frac {7}{4};-(c+d x)^2\right )-3 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )\right )-3 \left (a+b \sinh ^{-1}(c+d x)\right )^2\right )}{3 d e \sqrt {e (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arsinh}\left (d x + c\right ) + a^{2}\right )} \sqrt {d e x + c e}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {2 \, \sqrt {d x + c} b^{2} \sqrt {e} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )^{2}}{d^{2} e^{2} x + c d e^{2}} - \frac {2 \, a^{2}}{\sqrt {d e x + c e} d e} + \int \frac {2 \, {\left ({\left (2 \, b^{2} c^{2} + {\left (c^{2} + 1\right )} a b + {\left (a b d^{2} + 2 \, b^{2} d^{2}\right )} x^{2} + 2 \, {\left (a b c d + 2 \, b^{2} c d\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1} \sqrt {d x + c} + {\left ({\left (a b d^{3} + 2 \, b^{2} d^{3}\right )} x^{3} + {\left (c^{3} + c\right )} a b + 2 \, {\left (c^{3} + c\right )} b^{2} + 3 \, {\left (a b c d^{2} + 2 \, b^{2} c d^{2}\right )} x^{2} + {\left ({\left (3 \, c^{2} d + d\right )} a b + 2 \, {\left (3 \, c^{2} d + d\right )} b^{2}\right )} x\right )} \sqrt {d x + c}\right )} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{d^{5} e^{\frac {3}{2}} x^{5} + 5 \, c d^{4} e^{\frac {3}{2}} x^{4} + c^{5} e^{\frac {3}{2}} + c^{3} e^{\frac {3}{2}} + {\left (10 \, c^{2} d^{3} e^{\frac {3}{2}} + d^{3} e^{\frac {3}{2}}\right )} x^{3} + {\left (10 \, c^{3} d^{2} e^{\frac {3}{2}} + 3 \, c d^{2} e^{\frac {3}{2}}\right )} x^{2} + {\left (5 \, c^{4} d e^{\frac {3}{2}} + 3 \, c^{2} d e^{\frac {3}{2}}\right )} x + {\left (d^{4} e^{\frac {3}{2}} x^{4} + 4 \, c d^{3} e^{\frac {3}{2}} x^{3} + c^{4} e^{\frac {3}{2}} + c^{2} e^{\frac {3}{2}} + {\left (6 \, c^{2} d^{2} e^{\frac {3}{2}} + d^{2} e^{\frac {3}{2}}\right )} x^{2} + 2 \, {\left (2 \, c^{3} d e^{\frac {3}{2}} + c d e^{\frac {3}{2}}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*sqrt(d*x + c)*b^2*sqrt(e)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))^2/(d^2*e^2*x + c*d*e^2) - 2*a^2/
(sqrt(d*e*x + c*e)*d*e) + integrate(2*((2*b^2*c^2 + (c^2 + 1)*a*b + (a*b*d^2 + 2*b^2*d^2)*x^2 + 2*(a*b*c*d + 2
*b^2*c*d)*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)*sqrt(d*x + c) + ((a*b*d^3 + 2*b^2*d^3)*x^3 + (c^3 + c)*a*b + 2*
(c^3 + c)*b^2 + 3*(a*b*c*d^2 + 2*b^2*c*d^2)*x^2 + ((3*c^2*d + d)*a*b + 2*(3*c^2*d + d)*b^2)*x)*sqrt(d*x + c))*
log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/(d^5*e^(3/2)*x^5 + 5*c*d^4*e^(3/2)*x^4 + c^5*e^(3/2) + c^3*e^
(3/2) + (10*c^2*d^3*e^(3/2) + d^3*e^(3/2))*x^3 + (10*c^3*d^2*e^(3/2) + 3*c*d^2*e^(3/2))*x^2 + (5*c^4*d*e^(3/2)
+ 3*c^2*d*e^(3/2))*x + (d^4*e^(3/2)*x^4 + 4*c*d^3*e^(3/2)*x^3 + c^4*e^(3/2) + c^2*e^(3/2) + (6*c^2*d^2*e^(3/2
) + d^2*e^(3/2))*x^2 + 2*(2*c^3*d*e^(3/2) + c*d*e^(3/2))*x)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________