3.6.61 \(\int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {d+i c d x} (f-i c f x)^{3/2}} \, dx\) [561]

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

[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(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 = 0.16, 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 = {5796, 651, 5837, 12, 641, 31} \begin {gather*} -\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}} \end {gather*}

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 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 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])

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 = 0.23, size = 94, normalized size = 0.84 \begin {gather*} \frac {\sqrt {f-i c f x} \left (a+i a c x+(b+i b c x) \sinh ^{-1}(c x)-i b \sqrt {1+c^2 x^2} \log (d (-1+i c x))\right )}{c f^2 (i+c x) \sqrt {d+i c d x}} \end {gather*}

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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Maple [F]
time = 180.00, 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 = 0.49, size = 98, normalized size = 0.88 \begin {gather*} -\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}}} \end {gather*}

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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (88) = 176\).
time = 0.43, size = 443, normalized size = 3.96 \begin {gather*} \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 (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} - {\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}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\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 (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} - {\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}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\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 )}} \end {gather*}

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(-1/8*((-I*b*c^6*x^2 + 2*b*c^5*x + 2*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*
f*x + f) - (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)))/(b*c^3
*x^3 + I*b*c^2*x^2 + b*c*x + I*b)) + (c^2*d*f^2*x + I*c*d*f^2)*sqrt(b^2/(c^2*d*f^3))*log(-1/8*((-I*b*c^6*x^2 +
2*b*c^5*x + 2*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) - (-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)))/(b*c^3*x^3 + I*b*c^2*x^2 + b*c*x + 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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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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))/(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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \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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