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

Optimal. Leaf size=181 \[ -\frac {f^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i f^2 (1-i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b f^2 \left (c^2 x^2+1\right )^{3/2} \log (-c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.40, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {5712, 5833, 637, 5819, 12, 627, 31, 5675} \[ -\frac {f^2 \left (c^2 x^2+1\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i f^2 (1-i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b f^2 \left (c^2 x^2+1\right )^{3/2} \log (-c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 637

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(-(a*e) + c*d*x)/(a*c*Sqrt[a + c*x^2]),
x] /; FreeQ[{a, c, d, e}, x]

Rule 5675

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

Rule 5712

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 5819

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Wit
h[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcSinh[c*x], u, x] - Dist[b*c, Int[Dist[1/Sqrt[1 +
c^2*x^2], u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && ILtQ[p + 1/2, 0]
&& GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])

Rule 5833

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Int[ExpandIntegrand[(a + b*ArcSinh[c*x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; Fr
eeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IntegerQ[m] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 1.52, size = 283, normalized size = 1.56 \[ -\frac {-\frac {4 a \sqrt {d+i c d x} \sqrt {f-i c f x}}{c x-i}+2 a \sqrt {d} \sqrt {f} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )+\frac {b \sqrt {d+i c d x} \sqrt {f-i c f x} \left (2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right ) \left (\log \left (c^2 x^2+1\right )+4 i \tan ^{-1}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )+\sinh ^{-1}(c x)^2 \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+\sinh ^{-1}(c x) \left (-4 \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )-4 i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{\sqrt {c^2 x^2+1} \left (\cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+i \sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )}}{2 c d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((-4*a*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])/(-I + c*x) + 2*a*Sqrt[d]*Sqrt[f]*Log[c*d*f*x + Sqrt[d]*Sqrt[f
]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]] + (b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(ArcSinh[c*x]*((-4*I)*Cosh[Arc
Sinh[c*x]/2] - 4*Sinh[ArcSinh[c*x]/2]) + ArcSinh[c*x]^2*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) + 2*((
4*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + Log[1 + c^2*x^2])*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])))/(Sqrt[
1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])))/(c*d^2)

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} a}{c^{2} d^{2} x^{2} - 2 i \, c d^{2} x - d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(f-I*c*f*x)^(1/2)/(d+I*c*d*x)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {-i \, c f x + f} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (i \, c d x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(f-I*c*f*x)^(1/2)/(d+I*c*d*x)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right ) \sqrt {-i c f x +f}}{\left (i c d x +d \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a {\left (\frac {2 i \, \sqrt {c^{2} d f x^{2} + d f}}{i \, c^{2} d^{2} x + c d^{2}} - \frac {f \operatorname {arsinh}\left (c x\right )}{c d^{2} \sqrt {\frac {f}{d}}}\right )} + b \int \frac {\sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{{\left (i \, c d x + d\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))*(f-I*c*f*x)^(1/2)/(d+I*c*d*x)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}}}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))*(f-I*c*f*x)**(1/2)/(d+I*c*d*x)**(3/2),x)

[Out]

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

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