3.3.35 \(\int \frac {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^{7/2}} \, dx\) [235]

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

[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 = 0.09, 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 = {5859, 5776, 331, 335, 226} \begin {gather*} -\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 \text {ArcTan}\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}} \end {gather*}

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 331

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 335

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 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 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 {a+b \sinh ^{-1}(c+d x)}{(c e+d e x)^{7/2}} \, dx &=\frac {\text {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) \text {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) \text {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) \text {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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.03, size = 61, normalized size = 0.42 \begin {gather*} \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}} \end {gather*}

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] Result contains complex when optimal does not.
time = 3.06, size = 176, normalized size = 1.21

method	result	size
derivativedivides	\(\frac {-\frac {2 a}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+2 b \left (-\frac {\arcsinh \left (\frac {d e x +c e}{e}\right )}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+\frac {-\frac {2 \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{15 \left (d e x +c e \right )^{\frac {3}{2}}}-\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{15 e^{2} \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e}\)	\(176\)
default	\(\frac {-\frac {2 a}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+2 b \left (-\frac {\arcsinh \left (\frac {d e x +c e}{e}\right )}{5 \left (d e x +c e \right )^{\frac {5}{2}}}+\frac {-\frac {2 \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}{15 \left (d e x +c e \right )^{\frac {3}{2}}}-\frac {2 \sqrt {1-\frac {i \left (d e x +c e \right )}{e}}\, \sqrt {1+\frac {i \left (d e x +c e \right )}{e}}\, \EllipticF \left (\sqrt {d e x +c e}\, \sqrt {\frac {i}{e}}, i\right )}{15 e^{2} \sqrt {\frac {i}{e}}\, \sqrt {\frac {\left (d e x +c e \right )^{2}}{e^{2}}+1}}}{e}\right )}{d e}\)	\(176\)

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,method=_RETURNVERBOSE)

[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 \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))/(d*e*x+c*e)^(7/2),x, algorithm="maxima")

[Out]

1/10*(20*e^(1/2)*integrate(1/5/((d^5*x^5*e^4 + 5*c*d^4*x^4*e^4 + (10*c^2*d^3 + d^3)*x^3*e^4 + (10*c^3*d^2 + 3*
c*d^2)*x^2*e^4 + (5*c^4*d + 3*c^2*d)*x*e^4 + (c^5 + c^3)*e^4 + (d^4*x^4*e^4 + 4*c*d^3*x^3*e^4 + (6*c^2*d^2 + d
^2)*x^2*e^4 + 2*(2*c^3*d + c*d)*x*e^4 + (c^4 + c^2)*e^4)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*sqrt(d*x + c)), x)
- 4*e^(1/2)*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))/((d^3*x^2*e^4 + 2*c*d^2*x*e^4 + c^2*d*e^4)*sqrt(
d*x + c)) + ((I*sqrt(2)*(log(1/2*I*sqrt(2)*(sqrt(2)*e^(1/2) + 2*e^(1/2*log(d*x + c) + 1/2))*e^(-1/2) + 1) - lo
g(-1/2*I*sqrt(2)*(sqrt(2)*e^(1/2) + 2*e^(1/2*log(d*x + c) + 1/2))*e^(-1/2) + 1))*e^(-1/2) - I*sqrt(2)*(log(1/2
*I*sqrt(2)*(sqrt(2)*e^(1/2) - 2*e^(1/2*log(d*x + c) + 1/2))*e^(-1/2) + 1) - log(-1/2*I*sqrt(2)*(sqrt(2)*e^(1/2
) - 2*e^(1/2*log(d*x + c) + 1/2))*e^(-1/2) + 1))*e^(-1/2) + sqrt(2)*e^(-1/2)*log(sqrt(2)*e^(1/2*log(d*x + c) +
1) + e + e^(log(d*x + c) + 1)) - sqrt(2)*e^(-1/2)*log(-sqrt(2)*e^(1/2*log(d*x + c) + 1) + e + e^(log(d*x + c)
+ 1)))*e^(-3) - 8*e^(-1/2*log(d*x + c) - 7/2))/d)*b - 2/5*a*e^(-1)/((d*x*e + c*e)^(5/2)*d)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 383, normalized size = 2.64 \begin {gather*} -\frac {2 \, {\left (3 \, \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )} b d^{2} \log \left (d x + c + \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right ) + 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3}\right )} \sqrt {d^{3} \cosh \left (1\right ) + d^{3} \sinh \left (1\right )} {\rm weierstrassPInverse}\left (-\frac {4}{d^{2}}, 0, \frac {d x + c}{d}\right ) + {\left (3 \, a d^{2} + 2 \, {\left (b d^{3} x + b c d^{2}\right )} \sqrt {d^{2} x^{2} + 2 \, c d x + c^{2} + 1}\right )} \sqrt {{\left (d x + c\right )} \cosh \left (1\right ) + {\left (d x + c\right )} \sinh \left (1\right )}\right )}}{15 \, {\left ({\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )} \cosh \left (1\right )^{4} + 4 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )} \cosh \left (1\right )^{3} \sinh \left (1\right ) + 6 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )} \cosh \left (1\right )^{2} \sinh \left (1\right )^{2} + 4 \, {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )} \cosh \left (1\right ) \sinh \left (1\right )^{3} + {\left (d^{6} x^{3} + 3 \, c d^{5} x^{2} + 3 \, c^{2} d^{4} x + c^{3} d^{3}\right )} \sinh \left (1\right )^{4}\right )}} \end {gather*}

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]

-2/15*(3*sqrt((d*x + c)*cosh(1) + (d*x + c)*sinh(1))*b*d^2*log(d*x + c + sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1)) +
2*(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3)*sqrt(d^3*cosh(1) + d^3*sinh(1))*weierstrassPInverse(-4/d^2
, 0, (d*x + c)/d) + (3*a*d^2 + 2*(b*d^3*x + b*c*d^2)*sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1))*sqrt((d*x + c)*cosh(1)
+ (d*x + c)*sinh(1)))/((d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*cosh(1)^4 + 4*(d^6*x^3 + 3*c*d^5*x^2 +
3*c^2*d^4*x + c^3*d^3)*cosh(1)^3*sinh(1) + 6*(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*cosh(1)^2*sinh(1
)^2 + 4*(d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4*x + c^3*d^3)*cosh(1)*sinh(1)^3 + (d^6*x^3 + 3*c*d^5*x^2 + 3*c^2*d^4
*x + c^3*d^3)*sinh(1)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asinh}{\left (c + d x \right )}}{\left (e \left (c + d x\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

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]

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

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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