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

Optimal. Leaf size=145 \[ -\frac{2 b (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right ),\frac{1}{2}\right )}{15 d e^{7/2} \sqrt{(c+d x)^2+1}}-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}-\frac{4 b \sqrt{(c+d x)^2+1}}{15 d e^2 (e (c+d x))^{3/2}} \]

[Out]

(-4*b*Sqrt[1 + (c + d*x)^2])/(15*d*e^2*(e*(c + d*x))^(3/2)) - (2*(a + b*ArcSinh[c + d*x]))/(5*d*e*(e*(c + d*x)
)^(5/2)) - (2*b*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d*x)]/Sqr
t[e]], 1/2])/(15*d*e^(7/2)*Sqrt[1 + (c + d*x)^2])

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Rubi [A]  time = 0.135654, antiderivative size = 145, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {5865, 5661, 325, 329, 220} \[ -\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{5 d e (e (c+d x))^{5/2}}-\frac{4 b \sqrt{(c+d x)^2+1}}{15 d e^2 (e (c+d x))^{3/2}}-\frac{2 b (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{15 d e^{7/2} \sqrt{(c+d x)^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*b*Sqrt[1 + (c + d*x)^2])/(15*d*e^2*(e*(c + d*x))^(3/2)) - (2*(a + b*ArcSinh[c + d*x]))/(5*d*e*(e*(c + d*x)
)^(5/2)) - (2*b*(1 + c + d*x)*Sqrt[(1 + (c + d*x)^2)/(1 + c + d*x)^2]*EllipticF[2*ArcTan[Sqrt[e*(c + d*x)]/Sqr
t[e]], 1/2])/(15*d*e^(7/2)*Sqrt[1 + (c + d*x)^2])

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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

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

Mathematica [C]  time = 0.0356996, size = 61, normalized size = 0.42 \[ \frac{-4 b (c+d x) \text{Hypergeometric2F1}\left (-\frac{3}{4},\frac{1}{2},\frac{1}{4},-(c+d x)^2\right )-6 \left (a+b \sinh ^{-1}(c+d x)\right )}{15 d e (e (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(a + b*ArcSinh[c + d*x]) - 4*b*(c + d*x)*Hypergeometric2F1[-3/4, 1/2, 1/4, -(c + d*x)^2])/(15*d*e*(e*(c +
d*x))^(5/2))

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Maple [C]  time = 0.014, size = 176, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{de} \left ( -1/5\,{\frac{a}{ \left ( dex+ce \right ) ^{5/2}}}+b \left ( -1/5\,{\frac{1}{ \left ( dex+ce \right ) ^{5/2}}{\it Arcsinh} \left ({\frac{dex+ce}{e}} \right ) }+2/5\,{\frac{1}{e} \left ( -1/3\,{\frac{1}{ \left ( dex+ce \right ) ^{3/2}}\sqrt{{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}-1/3\,{\frac{1}{{e}^{2}}\sqrt{1-{\frac{i \left ( dex+ce \right ) }{e}}}\sqrt{1+{\frac{i \left ( dex+ce \right ) }{e}}}{\it EllipticF} \left ( \sqrt{dex+ce}\sqrt{{\frac{i}{e}}},i \right ){\frac{1}{\sqrt{{\frac{i}{e}}}}}{\frac{1}{\sqrt{{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d/e*(-1/5*a/(d*e*x+c*e)^(5/2)+b*(-1/5/(d*e*x+c*e)^(5/2)*arcsinh(1/e*(d*e*x+c*e))+2/5/e*(-1/3*(1/e^2*(d*e*x+c
*e)^2+1)^(1/2)/(d*e*x+c*e)^(3/2)-1/3/e^2/(I/e)^(1/2)*(1-I/e*(d*e*x+c*e))^(1/2)*(1+I/e*(d*e*x+c*e))^(1/2)/(1/e^
2*(d*e*x+c*e)^2+1)^(1/2)*EllipticF((d*e*x+c*e)^(1/2)*(I/e)^(1/2),I))))

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d e x + c e}{\left (b \operatorname{arsinh}\left (d x + c\right ) + a\right )}}{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*e*x + c*e)*(b*arcsinh(d*x + c) + a)/(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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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((a+b*asinh(d*x+c))/(d*e*x+c*e)**(7/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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