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

Optimal. Leaf size=134 \[ -\frac {16 b^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};-(c+d x)^2\right )}{15 d e^3 \sqrt {e (c+d x)}}-\frac {8 b \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}} \]

[Out]

-2/5*(a+b*arcsinh(d*x+c))^2/d/e/(e*(d*x+c))^(5/2)-8/15*b*(a+b*arcsinh(d*x+c))*hypergeom([-3/4, 1/2],[1/4],-(d*
x+c)^2)/d/e^2/(e*(d*x+c))^(3/2)-16/15*b^2*HypergeometricPFQ([-1/4, -1/4, 1],[1/4, 3/4],-(d*x+c)^2)/d/e^3/(e*(d
*x+c))^(1/2)

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Rubi [A]  time = 0.23, 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} \[ -\frac {16 b^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};-(c+d x)^2\right )}{15 d e^3 \sqrt {e (c+d x)}}-\frac {8 b \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-(c+d x)^2\right ) \left (a+b \sinh ^{-1}(c+d x)\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSinh[c + d*x])^2)/(5*d*e*(e*(c + d*x))^(5/2)) - (8*b*(a + b*ArcSinh[c + d*x])*Hypergeometric2F1[
-3/4, 1/2, 1/4, -(c + d*x)^2])/(15*d*e^2*(e*(c + d*x))^(3/2)) - (16*b^2*HypergeometricPFQ[{-1/4, -1/4, 1}, {1/
4, 3/4}, -(c + d*x)^2])/(15*d*e^3*Sqrt[e*(c + d*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]

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)^{7/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \sinh ^{-1}(x)\right )^2}{(e x)^{7/2}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}}+\frac {(4 b) \operatorname {Subst}\left (\int \frac {a+b \sinh ^{-1}(x)}{(e x)^{5/2} \sqrt {1+x^2}} \, dx,x,c+d x\right )}{5 d e}\\ &=-\frac {2 \left (a+b \sinh ^{-1}(c+d x)\right )^2}{5 d e (e (c+d x))^{5/2}}-\frac {8 b \left (a+b \sinh ^{-1}(c+d x)\right ) \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-(c+d x)^2\right )}{15 d e^2 (e (c+d x))^{3/2}}-\frac {16 b^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};-(c+d x)^2\right )}{15 d e^3 \sqrt {e (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 110, normalized size = 0.82 \[ -\frac {2 \left (8 b^2 (c+d x)^2 \, _3F_2\left (-\frac {1}{4},-\frac {1}{4},1;\frac {1}{4},\frac {3}{4};-(c+d x)^2\right )+\left (a+b \sinh ^{-1}(c+d x)\right ) \left (3 \left (a+b \sinh ^{-1}(c+d x)\right )+4 b (c+d x) \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-(c+d x)^2\right )\right )\right )}{15 d e (e (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*((a + b*ArcSinh[c + d*x])*(3*(a + b*ArcSinh[c + d*x]) + 4*b*(c + d*x)*Hypergeometric2F1[-3/4, 1/2, 1/4, -(
c + d*x)^2]) + 8*b^2*(c + d*x)^2*HypergeometricPFQ[{-1/4, -1/4, 1}, {1/4, 3/4}, -(c + d*x)^2]))/(15*d*e*(e*(c
+ d*x))^(5/2))

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fricas [F]  time = 0.73, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \operatorname {arsinh}\left (d x + c\right )^{2} + 2 \, a b \operatorname {arsinh}\left (d x + c\right ) + a^{2}\right )} \sqrt {d e x + c e}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^(7/2),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^4*e^4*x^4 + 4*c*d^3*e^4*
x^3 + 6*c^2*d^2*e^4*x^2 + 4*c^3*d*e^4*x + c^4*e^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {7}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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}}{5 \, {\left (d^{4} e^{4} x^{3} + 3 \, c d^{3} e^{4} x^{2} + 3 \, c^{2} d^{2} e^{4} x + c^{3} d e^{4}\right )}} - \frac {2 \, a^{2}}{5 \, {\left (d e x + c e\right )}^{\frac {5}{2}} d e} + \int \frac {2 \, {\left ({\left (2 \, b^{2} c^{2} + 5 \, {\left (c^{2} + 1\right )} a b + {\left (5 \, a b d^{2} + 2 \, b^{2} d^{2}\right )} x^{2} + 2 \, {\left (5 \, 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 (5 \, a b d^{3} + 2 \, b^{2} d^{3}\right )} x^{3} + 5 \, {\left (c^{3} + c\right )} a b + 2 \, {\left (c^{3} + c\right )} b^{2} + 3 \, {\left (5 \, a b c d^{2} + 2 \, b^{2} c d^{2}\right )} x^{2} + {\left (5 \, {\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 )}{5 \, {\left (d^{7} e^{\frac {7}{2}} x^{7} + 7 \, c d^{6} e^{\frac {7}{2}} x^{6} + c^{7} e^{\frac {7}{2}} + c^{5} e^{\frac {7}{2}} + {\left (21 \, c^{2} d^{5} e^{\frac {7}{2}} + d^{5} e^{\frac {7}{2}}\right )} x^{5} + 5 \, {\left (7 \, c^{3} d^{4} e^{\frac {7}{2}} + c d^{4} e^{\frac {7}{2}}\right )} x^{4} + 5 \, {\left (7 \, c^{4} d^{3} e^{\frac {7}{2}} + 2 \, c^{2} d^{3} e^{\frac {7}{2}}\right )} x^{3} + {\left (21 \, c^{5} d^{2} e^{\frac {7}{2}} + 10 \, c^{3} d^{2} e^{\frac {7}{2}}\right )} x^{2} + {\left (7 \, c^{6} d e^{\frac {7}{2}} + 5 \, c^{4} d e^{\frac {7}{2}}\right )} x + {\left (d^{6} e^{\frac {7}{2}} x^{6} + 6 \, c d^{5} e^{\frac {7}{2}} x^{5} + c^{6} e^{\frac {7}{2}} + c^{4} e^{\frac {7}{2}} + {\left (15 \, c^{2} d^{4} e^{\frac {7}{2}} + d^{4} e^{\frac {7}{2}}\right )} x^{4} + 4 \, {\left (5 \, c^{3} d^{3} e^{\frac {7}{2}} + c d^{3} e^{\frac {7}{2}}\right )} x^{3} + 3 \, {\left (5 \, c^{4} d^{2} e^{\frac {7}{2}} + 2 \, c^{2} d^{2} e^{\frac {7}{2}}\right )} x^{2} + 2 \, {\left (3 \, c^{5} d e^{\frac {7}{2}} + 2 \, c^{3} d e^{\frac {7}{2}}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}}\,{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)^(7/2),x, algorithm="maxima")

[Out]

-2/5*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^4*e^4*x^3 + 3*c*d^3*e^4*x
^2 + 3*c^2*d^2*e^4*x + c^3*d*e^4) - 2/5*a^2/((d*e*x + c*e)^(5/2)*d*e) + integrate(2/5*((2*b^2*c^2 + 5*(c^2 + 1
)*a*b + (5*a*b*d^2 + 2*b^2*d^2)*x^2 + 2*(5*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) + ((5*a*b*d^3 + 2*b^2*d^3)*x^3 + 5*(c^3 + c)*a*b + 2*(c^3 + c)*b^2 + 3*(5*a*b*c*d^2 + 2*b^2*c*d^2)*x^2 +
(5*(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^7*e^(7/2)*x^7 + 7*c*d^6*e^(7/2)*x^6 + c^7*e^(7/2) + c^5*e^(7/2) + (21*c^2*d^5*e^(7/2) + d^5*e^(7/2))*x^5 +
5*(7*c^3*d^4*e^(7/2) + c*d^4*e^(7/2))*x^4 + 5*(7*c^4*d^3*e^(7/2) + 2*c^2*d^3*e^(7/2))*x^3 + (21*c^5*d^2*e^(7/
2) + 10*c^3*d^2*e^(7/2))*x^2 + (7*c^6*d*e^(7/2) + 5*c^4*d*e^(7/2))*x + (d^6*e^(7/2)*x^6 + 6*c*d^5*e^(7/2)*x^5
+ c^6*e^(7/2) + c^4*e^(7/2) + (15*c^2*d^4*e^(7/2) + d^4*e^(7/2))*x^4 + 4*(5*c^3*d^3*e^(7/2) + c*d^3*e^(7/2))*x
^3 + 3*(5*c^4*d^2*e^(7/2) + 2*c^2*d^2*e^(7/2))*x^2 + 2*(3*c^5*d*e^(7/2) + 2*c^3*d*e^(7/2))*x)*sqrt(d^2*x^2 + 2
*c*d*x + c^2 + 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [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*asinh(d*x+c))**2/(d*e*x+c*e)**(7/2),x)

[Out]

Timed out

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