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

Optimal. Leaf size=295 \[ \frac {f^2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 (1-i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b f^2 \left (c^2 x^2+1\right )^{5/2}}{3 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b f^2 \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i b f^2 \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.34, antiderivative size = 295, 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, 653, 191, 5819, 627, 44, 203, 260} \[ \frac {f^2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i f^2 (1-i c x) \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i b f^2 \left (c^2 x^2+1\right )^{5/2}}{3 c (-c x+i) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b f^2 \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i b f^2 \left (c^2 x^2+1\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/3)*b*f^2*(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)) + (((2*I)/3)*f^2*(1 -
I*c*x)*(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)) + (f^2*x*(1 + c^2*x^2)^
2*(a + b*ArcSinh[c*x]))/(3*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - ((I/3)*b*f^2*(1 + c^2*x^2)^(5/2)*ArcTan[
c*x])/(c*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)) - (b*f^2*(1 + c^2*x^2)^(5/2)*Log[1 + c^2*x^2])/(6*c*(d + I*c
*d*x)^(5/2)*(f - I*c*f*x)^(5/2))

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

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

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 653

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

Rubi steps

\begin {align*} \int \frac {a+b \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} \sqrt {f-i c f x}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(f-i c f x)^2 \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 {2 i f^2 (1-i c x) \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 {f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (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 \left (\frac {2 i f^2 (1-i c x)}{3 c \left (1+c^2 x^2\right )^2}+\frac {f^2 x}{3 \left (1+c^2 x^2\right )}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {2 i f^2 (1-i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1-i c x}{\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 {\left (b c f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {2 i f^2 (1-i c x) \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{(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 {2 i f^2 (1-i c x) \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 {f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (-\frac {1}{2 (-i+c x)^2}+\frac {1}{2 \left (1+c^2 x^2\right )}\right ) \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {i b f^2 \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 {2 i f^2 (1-i c x) \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 {f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (i b f^2 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {1}{1+c^2 x^2} \, dx}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {i b f^2 \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 {2 i f^2 (1-i c x) \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 {f^2 x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i b f^2 \left (1+c^2 x^2\right )^{5/2} \tan ^{-1}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b f^2 \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 143, normalized size = 0.48 \[ \frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left ((c x-2 i) \left (a \sqrt {c^2 x^2+1}+b c x-i b\right )+b (c x-2 i) \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)-b (c x-i)^2 \log (d+i c d x)\right )}{3 c d^3 f (c x-i)^2 \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.75, size = 593, normalized size = 2.01 \[ -\frac {12 \, \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x - 3 \, {\left (4 \, b c^{2} x^{2} - 4 i \, b c x + 8 \, b\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - 2 \, {\left (3 \, c^{4} d^{3} f x^{3} - 3 i \, c^{3} d^{3} f x^{2} + 3 \, c^{2} d^{3} f x - 3 i \, c d^{3} f\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f}} \log \left (\frac {3 \, {\left (-2 i \, b c^{6} x^{2} - 4 \, b c^{5} x + 4 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + 2 \, {\left (3 i \, c^{9} d^{3} f x^{4} + 6 \, c^{8} d^{3} f x^{3} + 3 i \, c^{7} d^{3} f x^{2} + 6 \, c^{6} d^{3} f x\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f}}}{3 \, {\left (16 \, b c^{3} x^{3} - 16 i \, b c^{2} x^{2} + 16 \, b c x - 16 i \, b\right )}}\right ) + 2 \, {\left (3 \, c^{4} d^{3} f x^{3} - 3 i \, c^{3} d^{3} f x^{2} + 3 \, c^{2} d^{3} f x - 3 i \, c d^{3} f\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f}} \log \left (\frac {3 \, {\left (-2 i \, b c^{6} x^{2} - 4 \, b c^{5} x + 4 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + 2 \, {\left (-3 i \, c^{9} d^{3} f x^{4} - 6 \, c^{8} d^{3} f x^{3} - 3 i \, c^{7} d^{3} f x^{2} - 6 \, c^{6} d^{3} f x\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f}}}{3 \, {\left (16 \, b c^{3} x^{3} - 16 i \, b c^{2} x^{2} + 16 \, b c x - 16 i \, b\right )}}\right ) - 3 \, {\left (4 \, a c^{2} x^{2} - 4 i \, a c x + 8 \, a\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{3 \, {\left (12 \, c^{4} d^{3} f x^{3} - 12 i \, c^{3} d^{3} f x^{2} + 12 \, c^{2} d^{3} f x - 12 i \, c d^{3} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arcsinh \left (c x \right )}{\left (i c d x +d \right )^{\frac {5}{2}} \sqrt {-i c f x +f}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.71, size = 233, normalized size = 0.79 \[ \frac {1}{3} \, b c {\left (\frac {3}{3 i \, c^{3} d^{\frac {5}{2}} \sqrt {f} x + 3 \, c^{2} d^{\frac {5}{2}} \sqrt {f}} - \frac {\log \left (c x - i\right )}{c^{2} d^{\frac {5}{2}} \sqrt {f}}\right )} + b {\left (-\frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{3 \, c^{3} d^{3} f x^{2} - 6 i \, c^{2} d^{3} f x - 3 \, c d^{3} f} + \frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{3 i \, c^{2} d^{3} f x + 3 \, c d^{3} f}\right )} \operatorname {arsinh}\left (c x\right ) + a {\left (-\frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{3 \, c^{3} d^{3} f x^{2} - 6 i \, c^{2} d^{3} f x - 3 \, c d^{3} f} + \frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{3 i \, c^{2} d^{3} f x + 3 \, c d^{3} f}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*asinh(c*x))/(d+I*c*d*x)**(5/2)/(f-I*c*f*x)**(1/2),x)

[Out]

Integral((a + b*asinh(c*x))/((I*d*(c*x - I))**(5/2)*sqrt(-I*f*(c*x + I))), x)

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