3.3.42 \(\int \frac {(a+b \sinh ^{-1}(c+d x))^2}{(c e+d e x)^{5/2}} \, dx\) [242]

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

[Out]

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

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Rubi [A]
time = 0.16, 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 = {5859, 5776, 5817} \begin {gather*} \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}} \end {gather*}

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 5776

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 5817

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[((f*x
)^(m + 1)/(f*(m + 1)))*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])*Hypergeometric2F1[1/2, (1
+ m)/2, (3 + m)/2, (-c^2)*x^2], x] - Simp[b*c*((f*x)^(m + 2)/(f^2*(m + 1)*(m + 2)))*Simp[Sqrt[1 + c^2*x^2]/Sqr
t[d + e*x^2]]*HypergeometricPFQ[{1, 1 + m/2, 1 + m/2}, {3/2 + m/2, 2 + m/2}, (-c^2)*x^2], x] /; FreeQ[{a, b, c
, d, e, f, m}, x] && EqQ[e, c^2*d] &&  !IntegerQ[m]

Rule 5859

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)^{5/2}} \, dx &=\frac {\text {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) \text {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*}

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Mathematica [A]
time = 0.06, size = 106, normalized size = 0.79 \begin {gather*} -\frac {2 \left (\left (a+b \sinh ^{-1}(c+d x)\right )^2+4 b (c+d x) \left (\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 )-2 b (c+d x) \, _3F_2\left (\frac {1}{4},\frac {1}{4},1;\frac {3}{4},\frac {5}{4};-(c+d x)^2\right )\right )\right )}{3 d e (e (c+d x))^{3/2}} \end {gather*}

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 = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (d x +c \right )\right )^{2}}{\left (d e x +c e \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))^2/(d*e*x+c*e)^(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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*x + c)*e^(-5/2)/(d^3*x^3 + 3*c*d^2*x^2
+ 3*c^2*d*x + c^3), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c+d\,x\right )\right )}^2}{{\left (c\,e+d\,e\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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