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

Optimal. Leaf size=112 \[ -\frac {d (-c x+i) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b d \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

-d*(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)-b*d*(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 = 0.25, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {5712, 637, 5819, 12, 627, 31} \[ -\frac {d (-c x+i) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b d \left (c^2 x^2+1\right )^{3/2} \log (c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*(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))) - (b*d*(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 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])

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {d+i c d x} (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) \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 {d (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 {\left (b c \left (1+c^2 x^2\right )^{3/2}\right ) \int \frac {d (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 {d (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 {\left (b d \left (1+c^2 x^2\right )^{3/2}\right ) \int \frac {i-c x}{1+c^2 x^2} \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=-\frac {d (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 {\left (b d \left (1+c^2 x^2\right )^{3/2}\right ) \int \frac {1}{-i-c x} \, dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\\ &=-\frac {d (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 {b d \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 = 0.41, size = 94, normalized size = 0.84 \[ \frac {\sqrt {f-i c f x} \left (i a c x+a-i b \sqrt {c^2 x^2+1} \log (d (-1+i c x))+(b+i b c x) \sinh ^{-1}(c x)\right )}{c f^2 (c x+i) \sqrt {d+i c d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.65, size = 445, normalized size = 3.97 \[ \frac {2 \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - {\left (c^{2} d f^{2} x + i \, c d f^{2}\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{3}}} \log \left (\frac {{\left (2 i \, b c^{6} x^{2} - 4 \, b c^{5} x - 4 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + 2 \, {\left (i \, c^{9} d f^{2} x^{4} - 2 \, c^{8} d f^{2} x^{3} + i \, c^{7} d f^{2} x^{2} - 2 \, c^{6} d f^{2} x\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{3}}}}{16 \, b c^{3} x^{3} + 16 i \, b c^{2} x^{2} + 16 \, b c x + 16 i \, b}\right ) + {\left (c^{2} d f^{2} x + i \, c d f^{2}\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{3}}} \log \left (\frac {{\left (2 i \, b c^{6} x^{2} - 4 \, b c^{5} x - 4 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + 2 \, {\left (-i \, c^{9} d f^{2} x^{4} + 2 \, c^{8} d f^{2} x^{3} - i \, c^{7} d f^{2} x^{2} + 2 \, c^{6} d f^{2} x\right )} \sqrt {\frac {b^{2}}{c^{2} d f^{3}}}}{16 \, b c^{3} x^{3} + 16 i \, b c^{2} x^{2} + 16 \, b c x + 16 i \, b}\right ) + 2 \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} a}{2 \, {\left (c^{2} d f^{2} x + i \, c d f^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*log(c*x + sqrt(c^2*x^2 + 1)) - (c^2*d*f^2*x + I*c*d*f^2)*sqrt(b^
2/(c^2*d*f^3))*log(((2*I*b*c^6*x^2 - 4*b*c^5*x - 4*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x
+ f) + 2*(I*c^9*d*f^2*x^4 - 2*c^8*d*f^2*x^3 + I*c^7*d*f^2*x^2 - 2*c^6*d*f^2*x)*sqrt(b^2/(c^2*d*f^3)))/(16*b*c^
3*x^3 + 16*I*b*c^2*x^2 + 16*b*c*x + 16*I*b)) + (c^2*d*f^2*x + I*c*d*f^2)*sqrt(b^2/(c^2*d*f^3))*log(((2*I*b*c^6
*x^2 - 4*b*c^5*x - 4*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) + 2*(-I*c^9*d*f^2*x^4 + 2
*c^8*d*f^2*x^3 - I*c^7*d*f^2*x^2 + 2*c^6*d*f^2*x)*sqrt(b^2/(c^2*d*f^3)))/(16*b*c^3*x^3 + 16*I*b*c^2*x^2 + 16*b
*c*x + 16*I*b)) + 2*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a)/(c^2*d*f^2*x + I*c*d*f^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.65, size = 98, normalized size = 0.88 \[ -\frac {i \, \sqrt {c^{2} d f x^{2} + d f} b \operatorname {arsinh}\left (c x\right )}{-i \, c^{2} d f^{2} x + c d f^{2}} - \frac {i \, \sqrt {c^{2} d f x^{2} + d f} a}{-i \, c^{2} d f^{2} x + c d f^{2}} - \frac {b \log \left (i \, c x - 1\right )}{c \sqrt {d} f^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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