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

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

[Out]

-2/5*(a+b*arcsinh(d*x+c))/d/e/(e*(d*x+c))^(5/2)-4/15*b*(1+(d*x+c)^2)^(1/2)/d/e^2/(e*(d*x+c))^(3/2)-2/15*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(si
n(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^(7/2)/(1+(d*x+c)^2)^
(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 145, normalized size of antiderivative = 1.00, 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 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 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 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)^{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*}

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Mathematica [C]  time = 0.04, size = 61, normalized size = 0.42 \[ \frac {-6 \left (a+b \sinh ^{-1}(c+d x)\right )-4 b (c+d x) \, _2F_1\left (-\frac {3}{4},\frac {1}{2};\frac {1}{4};-(c+d x)^2\right )}{15 d e (e (c+d x))^{5/2}} \]

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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fricas [F]  time = 0.59, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d e x + c e} {\left (b \operatorname {arsinh}\left (d x + c\right ) + a\right )}}{d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + 6 \, c^{2} d^{2} e^{4} x^{2} + 4 \, c^{3} d e^{4} x + c^{4} e^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 176, normalized size = 1.21 \[ \frac {-\frac {2 a}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+2 b \left (-\frac {\arcsinh \left (\frac {d e x +c e}{e}\right )}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+\frac {-\frac {2 \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{15 \left (d e x +c e \right )^{\frac {3}{2}}}-\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{15 e^{2} \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(d*x+c))/(d*e*x+c*e)^(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((d*e*x+c*e)/e)+2/5/e*(-1/3*((d*e*x+c*e)^2/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)/((d*e*x+c
*e)^2/e^2+1)^(1/2)*EllipticF((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 \[ \frac {1}{10} \, {\left (20 \, \sqrt {e} \int \frac {1}{5 \, {\left (d^{5} e^{4} x^{5} + 5 \, c d^{4} e^{4} x^{4} + c^{5} e^{4} + c^{3} e^{4} + {\left (10 \, c^{2} d^{3} e^{4} + d^{3} e^{4}\right )} x^{3} + {\left (10 \, c^{3} d^{2} e^{4} + 3 \, c d^{2} e^{4}\right )} x^{2} + {\left (5 \, c^{4} d e^{4} + 3 \, c^{2} d e^{4}\right )} x + {\left (d^{4} e^{4} x^{4} + 4 \, c d^{3} e^{4} x^{3} + c^{4} e^{4} + c^{2} e^{4} + {\left (6 \, c^{2} d^{2} e^{4} + d^{2} e^{4}\right )} x^{2} + 2 \, {\left (2 \, c^{3} d e^{4} + c d e^{4}\right )} x\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \sqrt {d x + c}}\,{d x} + \frac {\sqrt {e} {\left (\frac {i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} - i \, \sqrt {2} {\left (\log \left (\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right ) - \log \left (-\frac {1}{2} i \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {d x + c}\right )} + 1\right )\right )} + \sqrt {2} \log \left (d x + \sqrt {2} \sqrt {d x + c} + c + 1\right ) - \sqrt {2} \log \left (d x - \sqrt {2} \sqrt {d x + c} + c + 1\right )}{e^{4}} - \frac {8}{\sqrt {d x + c} e^{4}}\right )}}{d} - \frac {4 \, \sqrt {e} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )}{{\left (d^{3} e^{4} x^{2} + 2 \, c d^{2} e^{4} x + c^{2} d e^{4}\right )} \sqrt {d x + c}}\right )} b - \frac {2 \, a}{5 \, {\left (d e x + c e\right )}^{\frac {5}{2}} d e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(20*sqrt(e)*integrate(1/5/((d^5*e^4*x^5 + 5*c*d^4*e^4*x^4 + c^5*e^4 + c^3*e^4 + (10*c^2*d^3*e^4 + d^3*e^4
)*x^3 + (10*c^3*d^2*e^4 + 3*c*d^2*e^4)*x^2 + (5*c^4*d*e^4 + 3*c^2*d*e^4)*x + (d^4*e^4*x^4 + 4*c*d^3*e^4*x^3 +
c^4*e^4 + c^2*e^4 + (6*c^2*d^2*e^4 + d^2*e^4)*x^2 + 2*(2*c^3*d*e^4 + c*d*e^4)*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*sqrt(d*x + c)) + 1) - log(-1/2*
I*sqrt(2)*(sqrt(2) + 2*sqrt(d*x + c)) + 1)) - 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)) + sqrt(2)*log(d*x + sqrt(2)*sqrt(d*x + c) + c + 1) - sqrt
(2)*log(d*x - sqrt(2)*sqrt(d*x + c) + c + 1))/e^4 - 8/(sqrt(d*x + c)*e^4))/d - 4*sqrt(e)*log(d*x + c + sqrt(d^
2*x^2 + 2*c*d*x + c^2 + 1))/((d^3*e^4*x^2 + 2*c*d^2*e^4*x + c^2*d*e^4)*sqrt(d*x + c)))*b - 2/5*a/((d*e*x + c*e
)^(5/2)*d*e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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