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

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

[Out]

-2/3*(a+b*arcsinh(d*x+c))/d/e/(e*(d*x+c))^(3/2)-4/3*b*(1+(d*x+c)^2)^(1/2)/d/e^2/(e*(d*x+c))^(1/2)+4/3*b*(e*(d*
x+c))^(1/2)*(1+(d*x+c)^2)^(1/2)/d/e^3/(d*x+c+1)-4/3*b*(d*x+c+1)*(cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))^2)^(
1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*EllipticE(sin(2*arctan((e*(d*x+c))^(1/2)/e^(1/2))),1/2*2^(1/2))*
((1+(d*x+c)^2)/(d*x+c+1)^2)^(1/2)/d/e^(5/2)/(1+(d*x+c)^2)^(1/2)+2/3*b*(d*x+c+1)*(cos(2*arctan((e*(d*x+c))^(1/2
)/e^(1/2)))^2)^(1/2)/cos(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)))*EllipticF(sin(2*arctan((e*(d*x+c))^(1/2)/e^(1/2)
)),1/2*2^(1/2))*((1+(d*x+c)^2)/(d*x+c+1)^2)^(1/2)/d/e^(5/2)/(1+(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 266, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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 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 {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*}

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Mathematica [C]  time = 0.03, size = 58, normalized size = 0.22 \[ -\frac {2 \left (a+2 b (c+d x) \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-(c+d x)^2\right )+b \sinh ^{-1}(c+d x)\right )}{3 d e (e (c+d x))^{3/2}} \]

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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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d e x + c e} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}}{d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + 3 \, c^{2} d e^{3} x + c^{3} e^{3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \operatorname {arsinh}\left (d x + c\right ) + a}{{\left (d e x + c e\right )}^{\frac {5}{2}}}\,{d x} \]

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)

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maple [C]  time = 0.01, size = 202, normalized size = 0.76 \[ \frac {-\frac {2 a}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+2 b \left (-\frac {\arcsinh \left (\frac {d e x +c e}{e}\right )}{3 \left (d e x +c e \right )^{\frac {3}{2}}}+\frac {-\frac {2 \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{3 \sqrt {d e x +c e}}+\frac {2 i \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \left (\EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )-\EllipticE \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )\right )}{3 e \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e} \]

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((d*e*x+c*e)/e)+2/3/e*(-((d*e*x+c*e)^2/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)/((d*e*x+c*e)^2/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]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, {\left (12 \, \sqrt {e} \int \frac {1}{3 \, {\left (d^{4} e^{3} x^{4} + 4 \, c d^{3} e^{3} x^{3} + c^{4} e^{3} + c^{2} e^{3} + {\left (6 \, c^{2} d^{2} e^{3} + d^{2} e^{3}\right )} x^{2} + 2 \, {\left (2 \, c^{3} d e^{3} + c d e^{3}\right )} x + {\left (d^{3} e^{3} x^{3} + 3 \, c d^{2} e^{3} x^{2} + c^{3} e^{3} + c e^{3} + {\left (3 \, c^{2} d e^{3} + d e^{3}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \sqrt {d x + c}}\,{d x} - \frac {\sqrt {e} {\left (\frac {i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right )\right )}}{e^{3}} - \frac {i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right )\right )}}{e^{3}} - \frac {\sqrt {2} \log \left (d x + \sqrt {2} \sqrt {d x + c} + c + 1\right )}{e^{3}} + \frac {\sqrt {2} \log \left (d x - \sqrt {2} \sqrt {d x + c} + c + 1\right )}{e^{3}}\right )}}{d} - \frac {4 \, \sqrt {e} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{{\left (d^{2} e^{3} x + c d e^{3}\right )} \sqrt {d x + c}}\right )} b - \frac {2 \, a}{3 \, {\left (d e x + c e\right )}^{\frac {3}{2}} d e} \]

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]

1/6*(12*sqrt(e)*integrate(1/3/((d^4*e^3*x^4 + 4*c*d^3*e^3*x^3 + c^4*e^3 + c^2*e^3 + (6*c^2*d^2*e^3 + d^2*e^3)*
x^2 + 2*(2*c^3*d*e^3 + c*d*e^3)*x + (d^3*e^3*x^3 + 3*c*d^2*e^3*x^2 + c^3*e^3 + c*e^3 + (3*c^2*d*e^3 + d*e^3)*x
)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*sqrt(d*x + c)), x) - sqrt(e)*(I*sqrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2) + 2*s
qrt(d*x + c)) + 1) - log(-1/2*I*sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1))/e^3 - I*sqrt(2)*(log(1/2*I*sqrt(2)*(
sqrt(2) - 2*sqrt(d*x + c)) + 1) - log(-1/2*I*sqrt(2)*(sqrt(2) - 2*sqrt(d*x + c)) + 1))/e^3 - sqrt(2)*log(d*x +
sqrt(2)*sqrt(d*x + c) + c + 1)/e^3 + sqrt(2)*log(d*x - sqrt(2)*sqrt(d*x + c) + c + 1)/e^3)/d - 4*sqrt(e)*log(
d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/((d^2*e^3*x + c*d*e^3)*sqrt(d*x + c)))*b - 2/3*a/((d*e*x + c*e)^(
3/2)*d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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