3.6.60 \(\int \frac {\sqrt {d+i c d x} (a+b \sinh ^{-1}(c x))}{(f-i c f x)^{3/2}} \, dx\) [560]

Optimal. Leaf size=180 \[ -\frac {2 i d^2 (1+i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {d^2 \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{3/2} \log (i+c x)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.26, antiderivative size = 180, 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 = {5796, 5844, 651, 5837, 12, 641, 31, 5783} \begin {gather*} -\frac {d^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 d^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 d^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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*d^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)) - (d^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*d^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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
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 651

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 5783

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

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 5837

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 5844

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 {d+i c d x} \left (a+b \sinh ^{-1}(c x)\right )}{(f-i c f x)^{3/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{3/2} \int \frac {(d+i c d 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 d^2-c d^2 x\right ) \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{3/2}}-\frac {d^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 d^2-c d^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 (d^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 d^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 {d^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 {d^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 d^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 {d^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 d^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 d^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 {d^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 d^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 d^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 {d^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 d^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 = 0.78, size = 285, normalized size = 1.58 \begin {gather*} \frac {\frac {4 a \sqrt {d+i c d x} \sqrt {f-i c f x}}{i+c x}-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 (-\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 )+4 \sinh ^{-1}(c x) \left (-i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+2 \left (4 \text {ArcTan}\left (\tanh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )+i \log \left (1+c^2 x^2\right )\right ) \left (i \cosh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )+\sinh \left (\frac {1}{2} \sinh ^{-1}(c x)\right )\right )\right )}{\sqrt {1+c^2 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 )}}{2 c f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((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]^2*(Cosh[ArcSinh[c*x]/
2] - I*Sinh[ArcSinh[c*x]/2])) + 4*ArcSinh[c*x]*((-I)*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/2]) + 2*(4*ArcTa
n[Tanh[ArcSinh[c*x]/2]] + I*Log[1 + c^2*x^2])*(I*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/2])))/(Sqrt[1 + c^2*
x^2]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/(2*c*f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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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))*(d+I*c*d*x)^(1/2)/(f-I*c*f*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*f^2*x^2 + 2*I*c*f^2*x - f^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {i d \left (c x - i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\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))*(d+I*c*d*x)**(1/2)/(f-I*c*f*x)**(3/2),x)

[Out]

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

[Out]

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

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

[Out]

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

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