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

Optimal. Leaf size=266 \[ \frac{2 b (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right ),\frac{1}{2}\right )}{3 d e^{5/2} \sqrt{(c+d x)^2+1}}-\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac{4 b \sqrt{(c+d x)^2+1}}{3 d e^2 \sqrt{e (c+d x)}}+\frac{4 b \sqrt{(c+d x)^2+1} \sqrt{e (c+d x)}}{3 d e^3 (c+d x+1)}-\frac{4 b (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{3 d e^{5/2} \sqrt{(c+d x)^2+1}} \]

[Out]

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

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Rubi [A]  time = 0.237651, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {5865, 5661, 325, 329, 305, 220, 1196} \[ -\frac{2 \left (a+b \sinh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}}-\frac{4 b \sqrt{(c+d x)^2+1}}{3 d e^2 \sqrt{e (c+d x)}}+\frac{4 b \sqrt{(c+d x)^2+1} \sqrt{e (c+d x)}}{3 d e^3 (c+d x+1)}+\frac{2 b (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{3 d e^{5/2} \sqrt{(c+d x)^2+1}}-\frac{4 b (c+d x+1) \sqrt{\frac{(c+d x)^2+1}{(c+d x+1)^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{e (c+d x)}}{\sqrt{e}}\right )|\frac{1}{2}\right )}{3 d e^{5/2} \sqrt{(c+d x)^2+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^(5/2),x]

[Out]

(-4*b*Sqrt[1 + (c + d*x)^2])/(3*d*e^2*Sqrt[e*(c + d*x)]) + (4*b*Sqrt[e*(c + d*x)]*Sqrt[1 + (c + d*x)^2])/(3*d*
e^3*(1 + c + d*x)) - (2*(a + b*ArcSinh[c + d*x]))/(3*d*e*(e*(c + d*x))^(3/2)) - (4*b*(1 + c + d*x)*Sqrt[(1 + (
c + d*x)^2)/(1 + c + d*x)^2]*EllipticE[2*ArcTan[Sqrt[e*(c + d*x)]/Sqrt[e]], 1/2])/(3*d*e^(5/2)*Sqrt[1 + (c + d
*x)^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)]/Sqrt
[e]], 1/2])/(3*d*e^(5/2)*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 305

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

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]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

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

Mathematica [C]  time = 0.0290358, size = 58, normalized size = 0.22 \[ -\frac{2 \left (2 b (c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-(c+d x)^2\right )+a+b \sinh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c + d*x])/(c*e + d*e*x)^(5/2),x]

[Out]

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

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Maple [C]  time = 0.011, size = 202, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{de} \left ( -1/3\,{\frac{a}{ \left ( dex+ce \right ) ^{3/2}}}+b \left ( -1/3\,{\frac{1}{ \left ( dex+ce \right ) ^{3/2}}{\it Arcsinh} \left ({\frac{dex+ce}{e}} \right ) }+2/3\,{\frac{1}{e} \left ( -{\frac{1}{\sqrt{dex+ce}}\sqrt{{\frac{ \left ( dex+ce \right ) ^{2}}{{e}^{2}}}+1}}+{\frac{i}{e}\sqrt{1-{\frac{i \left ( dex+ce \right ) }{e}}}\sqrt{1+{\frac{i \left ( dex+ce \right ) }{e}}} \left ({\it EllipticF} \left ( \sqrt{dex+ce}\sqrt{{\frac{i}{e}}},i \right ) -{\it EllipticE} \left ( \sqrt{dex+ce}\sqrt{{\frac{i}{e}}},i \right ) \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)^(5/2),x)

[Out]

2/d/e*(-1/3*a/(d*e*x+c*e)^(3/2)+b*(-1/3/(d*e*x+c*e)^(3/2)*arcsinh(1/e*(d*e*x+c*e))+2/3/e*(-(1/e^2*(d*e*x+c*e)^
2+1)^(1/2)/(d*e*x+c*e)^(1/2)+I/e/(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)-EllipticE((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)^(5/2),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^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, 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)^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \operatorname{asinh}{\left (c + d x \right )}}{\left (e \left (c + d x\right )\right )^{\frac{5}{2}}}\, dx \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)**(5/2),x)

[Out]

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

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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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(d*x + c) + a)/(d*e*x + c*e)^(5/2), x)