3.6.98 \(\int \frac {(a+b \sinh ^{-1}(c x))^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx\) [598]

Optimal. Leaf size=224 \[ \frac {x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b^2 \left (1+c^2 x^2\right )^{3/2} \text {PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

x*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))^2/
c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-2*b*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)
/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-b^2*(c^2*x^2+1)^(3/2)*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c/(d+I*c*d*
x)^(3/2)/(f-I*c*f*x)^(3/2)

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Rubi [A]
time = 0.28, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {5796, 5787, 5797, 3799, 2221, 2317, 2438} \begin {gather*} \frac {\left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b \left (c^2 x^2+1\right )^{3/2} \log \left (e^{2 \sinh ^{-1}(c x)}+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b^2 \left (c^2 x^2+1\right )^{3/2} \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcSinh[c*x])^2/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)),x]

[Out]

(x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) + ((1 + c^2*x^2)^(3/2)*(a +
b*ArcSinh[c*x])^2)/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (2*b*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]
)*Log[1 + E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)) - (b^2*(1 + c^2*x^2)^(3/2)*PolyLog[
2, -E^(2*ArcSinh[c*x])])/(c*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((c + d*x)^(m
+ 1)/(d*(m + 1))), x] + Dist[2*I, Int[(c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*fz*x)))), x]
, x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 5787

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh
[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Dist[b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]], Int[x*((a + b*ArcS
inh[c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5796

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_) + (g_.)*(x_))^(q_), x_Symbol] :>
Dist[(d + e*x)^q*((f + g*x)^q/(1 + c^2*x^2)^q), Int[(d + e*x)^(p - q)*(1 + c^2*x^2)^q*(a + b*ArcSinh[c*x])^n,
x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^2 + e^2, 0] && HalfIntegerQ[p,
q] && GeQ[p - q, 0]

Rule 5797

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/e, Subst[Int[(
a + b*x)^n*Tanh[x], x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{3/2} \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{3/2}} \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=\frac {x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {\left (2 b c \left (1+c^2 x^2\right )^{3/2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{1+c^2 x^2} \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=\frac {x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {\left (2 b \left (1+c^2 x^2\right )^{3/2}\right ) \text {Subst}\left (\int (a+b x) \tanh (x) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=\frac {x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {\left (4 b \left (1+c^2 x^2\right )^{3/2}\right ) \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1+e^{2 x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=\frac {x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (2 b^2 \left (1+c^2 x^2\right )^{3/2}\right ) \text {Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=\frac {x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (b^2 \left (1+c^2 x^2\right )^{3/2}\right ) \text {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=\frac {x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {\left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b^2 \left (1+c^2 x^2\right )^{3/2} \text {Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(488\) vs. \(2(224)=448\).
time = 0.82, size = 488, normalized size = 2.18 \begin {gather*} \frac {a^2 c x+2 a b c x \sinh ^{-1}(c x)-2 i b^2 \pi \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+b^2 c x \sinh ^{-1}(c x)^2-b^2 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)^2+i b^2 \pi \sqrt {1+c^2 x^2} \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-2 b^2 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \log \left (1-i e^{-\sinh ^{-1}(c x)}\right )-i b^2 \pi \sqrt {1+c^2 x^2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )-2 b^2 \sqrt {1+c^2 x^2} \sinh ^{-1}(c x) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )+4 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (1+e^{\sinh ^{-1}(c x)}\right )-a b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )+i b^2 \pi \sqrt {1+c^2 x^2} \log \left (-\cos \left (\frac {1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )-4 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )-i b^2 \pi \sqrt {1+c^2 x^2} \log \left (\sin \left (\frac {1}{4} \left (\pi +2 i \sinh ^{-1}(c x)\right )\right )\right )+2 b^2 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )+2 b^2 \sqrt {1+c^2 x^2} \text {PolyLog}\left (2,i e^{-\sinh ^{-1}(c x)}\right )}{c d f \sqrt {d+i c d x} \sqrt {f-i c f x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcSinh[c*x])^2/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)),x]

[Out]

(a^2*c*x + 2*a*b*c*x*ArcSinh[c*x] - (2*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + b^2*c*x*ArcSinh[c*x]^2 - b^2
*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 + I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 - I/E^ArcSinh[c*x]] - 2*b^2*Sqrt[1 + c^2*
x^2]*ArcSinh[c*x]*Log[1 - I/E^ArcSinh[c*x]] - I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 + I/E^ArcSinh[c*x]] - 2*b^2*Sqr
t[1 + c^2*x^2]*ArcSinh[c*x]*Log[1 + I/E^ArcSinh[c*x]] + (4*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[1 + E^ArcSinh[c*x]]
- a*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] + I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]]
- (4*I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[Cosh[ArcSinh[c*x]/2]] - I*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[Sin[(Pi + (2*I)*Ar
cSinh[c*x])/4]] + 2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, (-I)/E^ArcSinh[c*x]] + 2*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2,
I/E^ArcSinh[c*x]])/(c*d*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (i c d x +d \right )^{\frac {3}{2}} \left (-i c f x +f \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2),x)

[Out]

int((a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2),x)

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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(c*x))^2/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2),x, algorithm="maxima")

[Out]

b^2*integrate(log(c*x + sqrt(c^2*x^2 + 1))^2/((I*c*d*x + d)^(3/2)*(-I*c*f*x + f)^(3/2)), x) + 2*a*b*x*arcsinh(
c*x)/(sqrt(c^2*d*f*x^2 + d*f)*d*f) + a^2*x/(sqrt(c^2*d*f*x^2 + d*f)*d*f) - a*b*sqrt(1/(d*f))*log(x^2 + 1/c^2)/
(c*d*f)

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Fricas [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(c*x))^2/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2),x, algorithm="fricas")

[Out]

(sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b^2*x*log(c*x + sqrt(c^2*x^2 + 1))^2 + (c^2*d^2*f^2*x^2 + d^2*f^2)*integ
ral((sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a^2 - 2*(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b^2*
c*x - sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a*b)*log(c*x + sqrt(c^2*x^2 + 1)))/(c^4*d^2*f^2*x^4 + 2*c^2*d^2*f^2
*x^2 + d^2*f^2), x))/(c^2*d^2*f^2*x^2 + d^2*f^2)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (i d \left (c x - i\right )\right )^{\frac {3}{2}} \left (- i f \left (c x + i\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))**2/(d+I*c*d*x)**(3/2)/(f-I*c*f*x)**(3/2),x)

[Out]

Integral((a + b*asinh(c*x))**2/((I*d*(c*x - I))**(3/2)*(-I*f*(c*x + I))**(3/2)), 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(c*x))^2/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)^2/((I*c*d*x + d)^(3/2)*(-I*c*f*x + f)^(3/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(c*x))^2/((d + c*d*x*1i)^(3/2)*(f - c*f*x*1i)^(3/2)),x)

[Out]

int((a + b*asinh(c*x))^2/((d + c*d*x*1i)^(3/2)*(f - c*f*x*1i)^(3/2)), x)

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