3.566 \(\int \frac {\sqrt {d+i c d x} (a+b \sinh ^{-1}(c x))}{(f-i c f x)^{5/2}} \, dx\)

Optimal. Leaf size=185 \[ -\frac {i d^3 (1+i c x)^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b d^3 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b d^3 \left (c^2 x^2+1\right )^{5/2} \log (c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

2/3*I*b*d^3*(c^2*x^2+1)^(5/2)/c/(I+c*x)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*I*d^3*(1+I*c*x)^3*(c^2*x^2+1)*
(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*b*d^3*(c^2*x^2+1)^(5/2)*ln(I+c*x)/c/(d+I*c*d*x)^(
5/2)/(f-I*c*f*x)^(5/2)

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Rubi [A]  time = 0.31, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {5712, 651, 5819, 12, 627, 43} \[ -\frac {i d^3 (1+i c x)^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b d^3 \left (c^2 x^2+1\right )^{5/2}}{3 c (c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b d^3 \left (c^2 x^2+1\right )^{5/2} \log (c x+i)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[d + I*c*d*x]*(a + b*ArcSinh[c*x]))/(f - I*c*f*x)^(5/2),x]

[Out]

(((2*I)/3)*b*d^3*(1 + c^2*x^2)^(5/2))/(c*(I + c*x)*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - ((I/3)*d^3*(1 +
I*c*x)^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) + (b*d^3*(1 + c^2*x^2
)^(5/2)*Log[I + c*x])/(3*c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

Rule 12

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 651

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a + c*x^2)^(p + 1))
/(2*c*d*(p + 1)), x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p
+ 2, 0]

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 {\sqrt {d+i c d x} \left (a+b \sinh ^{-1}(c x)\right )}{(f-i c f x)^{5/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(d+i c d x)^3 \left (a+b \sinh ^{-1}(c x)\right )}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {i d^3 (1+i c x)^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (b c \left (1+c^2 x^2\right )^{5/2}\right ) \int -\frac {i d^3 (1+i c x)^3}{3 c \left (1+c^2 x^2\right )^2} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {i d^3 (1+i c x)^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (i b d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {(1+i c x)^3}{\left (1+c^2 x^2\right )^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {i d^3 (1+i c x)^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (i b d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1+i c x}{(1-i c x)^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {i d^3 (1+i c x)^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (i b d^3 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {2}{(i+c x)^2}-\frac {i}{i+c x}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {2 i b d^3 \left (1+c^2 x^2\right )^{5/2}}{3 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 (1+i c x)^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b d^3 \left (1+c^2 x^2\right )^{5/2} \log (i+c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 131, normalized size = 0.71 \[ -\frac {i d \sqrt {f-i c f x} \left ((c x-i) \left (a c x-i a+b \sqrt {c^2 x^2+1}\right )-b (c x+i) \sqrt {c^2 x^2+1} \log (d (-1+i c x))+b (c x-i)^2 \sinh ^{-1}(c x)\right )}{3 c f^3 (c x+i)^2 \sqrt {d+i c d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[d + I*c*d*x]*(a + b*ArcSinh[c*x]))/(f - I*c*f*x)^(5/2),x]

[Out]

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

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fricas [B]  time = 0.69, size = 565, normalized size = 3.05 \[ -\frac {24 \, \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x + 3 \, {\left (4 \, b c^{2} x^{2} - 8 i \, b c x - 4 \, b\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (3 \, c^{4} f^{3} x^{3} + 3 i \, c^{3} f^{3} x^{2} + 3 \, c^{2} f^{3} x + 3 i \, c f^{3}\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}} \log \left (\frac {3 \, {\left (2 i \, b c^{6} x^{2} - 4 \, b c^{5} x - 4 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + 2 \, {\left (3 i \, c^{9} f^{3} x^{4} - 6 \, c^{8} f^{3} x^{3} + 3 i \, c^{7} f^{3} x^{2} - 6 \, c^{6} f^{3} x\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}}}{3 \, {\left (16 \, b c^{3} x^{3} + 16 i \, b c^{2} x^{2} + 16 \, b c x + 16 i \, b\right )}}\right ) + 2 \, {\left (3 \, c^{4} f^{3} x^{3} + 3 i \, c^{3} f^{3} x^{2} + 3 \, c^{2} f^{3} x + 3 i \, c f^{3}\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}} \log \left (\frac {3 \, {\left (2 i \, b c^{6} x^{2} - 4 \, b c^{5} x - 4 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + 2 \, {\left (-3 i \, c^{9} f^{3} x^{4} + 6 \, c^{8} f^{3} x^{3} - 3 i \, c^{7} f^{3} x^{2} + 6 \, c^{6} f^{3} x\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}}}{3 \, {\left (16 \, b c^{3} x^{3} + 16 i \, b c^{2} x^{2} + 16 \, b c x + 16 i \, b\right )}}\right ) + 3 \, {\left (4 \, a c^{2} x^{2} - 8 i \, a c x - 4 \, a\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{3 \, {\left (12 \, c^{4} f^{3} x^{3} + 12 i \, c^{3} f^{3} x^{2} + 12 \, c^{2} f^{3} x + 12 i \, c f^{3}\right )}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(24*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x + 3*(4*b*c^2*x^2 - 8*I*b*c*x - 4*b)*sqrt
(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) - 2*(3*c^4*f^3*x^3 + 3*I*c^3*f^3*x^2 + 3*c^2*f^3
*x + 3*I*c*f^3)*sqrt(b^2*d/(c^2*f^5))*log(1/3*(3*(2*I*b*c^6*x^2 - 4*b*c^5*x - 4*I*b*c^4)*sqrt(c^2*x^2 + 1)*sqr
t(I*c*d*x + d)*sqrt(-I*c*f*x + f) + 2*(3*I*c^9*f^3*x^4 - 6*c^8*f^3*x^3 + 3*I*c^7*f^3*x^2 - 6*c^6*f^3*x)*sqrt(b
^2*d/(c^2*f^5)))/(16*b*c^3*x^3 + 16*I*b*c^2*x^2 + 16*b*c*x + 16*I*b)) + 2*(3*c^4*f^3*x^3 + 3*I*c^3*f^3*x^2 + 3
*c^2*f^3*x + 3*I*c*f^3)*sqrt(b^2*d/(c^2*f^5))*log(1/3*(3*(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*(-3*I*c^9*f^3*x^4 + 6*c^8*f^3*x^3 - 3*I*c^7*f^3*x^2 + 6*c^6*f^3*
x)*sqrt(b^2*d/(c^2*f^5)))/(16*b*c^3*x^3 + 16*I*b*c^2*x^2 + 16*b*c*x + 16*I*b)) + 3*(4*a*c^2*x^2 - 8*I*a*c*x -
4*a)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f))/(12*c^4*f^3*x^3 + 12*I*c^3*f^3*x^2 + 12*c^2*f^3*x + 12*I*c*f^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}}\,{d x} \]

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)^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*c*d*x + d)*(b*arcsinh(c*x) + a)/(-I*c*f*x + f)^(5/2), x)

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maple [F]  time = 0.34, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arcsinh \left (c x \right )\right ) \sqrt {i c d x +d}}{\left (-i c f x +f \right )^{\frac {5}{2}}}\, dx \]

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)^(5/2),x)

[Out]

int((a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(5/2),x)

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maxima [A]  time = 0.63, size = 220, normalized size = 1.19 \[ -\frac {1}{3} \, b c {\left (\frac {6 \, \sqrt {d}}{3 i \, c^{3} f^{\frac {5}{2}} x - 3 \, c^{2} f^{\frac {5}{2}}} - \frac {\sqrt {d} \log \left (c x + i\right )}{c^{2} f^{\frac {5}{2}}}\right )} + b {\left (\frac {2 i \, \sqrt {c^{2} d f x^{2} + d f}}{3 \, c^{3} f^{3} x^{2} + 6 i \, c^{2} f^{3} x - 3 \, c f^{3}} + \frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} f^{3} x + 3 \, c f^{3}}\right )} \operatorname {arsinh}\left (c x\right ) + a {\left (\frac {2 i \, \sqrt {c^{2} d f x^{2} + d f}}{3 \, c^{3} f^{3} x^{2} + 6 i \, c^{2} f^{3} x - 3 \, c f^{3}} + \frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} f^{3} x + 3 \, c f^{3}}\right )} \]

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)^(5/2),x, algorithm="maxima")

[Out]

-1/3*b*c*(6*sqrt(d)/(3*I*c^3*f^(5/2)*x - 3*c^2*f^(5/2)) - sqrt(d)*log(c*x + I)/(c^2*f^(5/2))) + b*(2*I*sqrt(c^
2*d*f*x^2 + d*f)/(3*c^3*f^3*x^2 + 6*I*c^2*f^3*x - 3*c*f^3) + I*sqrt(c^2*d*f*x^2 + d*f)/(-3*I*c^2*f^3*x + 3*c*f
^3))*arcsinh(c*x) + a*(2*I*sqrt(c^2*d*f*x^2 + d*f)/(3*c^3*f^3*x^2 + 6*I*c^2*f^3*x - 3*c*f^3) + I*sqrt(c^2*d*f*
x^2 + d*f)/(-3*I*c^2*f^3*x + 3*c*f^3))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

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)^(5/2),x)

[Out]

int(((a + b*asinh(c*x))*(d + c*d*x*1i)^(1/2))/(f - c*f*x*1i)^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(5/2),x)

[Out]

Timed out

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