\(\int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx\) [556]

Optimal result

Integrand size = 35, antiderivative size = 111 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\frac {f (i+c x) \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b f \left (1+c^2 x^2\right )^{3/2} \log (i-c x)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[Out]

f*(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*f*(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)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {5796, 651, 5837, 12, 641, 31} \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\frac {f (c x+i) \left (c^2 x^2+1\right ) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b f \left (c^2 x^2+1\right )^{3/2} \log (-c x+i)}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]

[In]

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

[Out]

(f*(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*f*(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*} \text {integral}& = \frac {\left (1+c^2 x^2\right )^{3/2} \int \frac {(f-i c f x) (a+b \text {arcsinh}(c x))}{\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 {f (i+c x) \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{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 {f (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 {f (i+c x) \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {\left (b f \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 {f (i+c x) \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {\left (b f \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 {f (i+c x) \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b f \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*}

Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.02 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left (a \sqrt {1+c^2 x^2}+b \sqrt {1+c^2 x^2} \text {arcsinh}(c x)+b (i-c x) \log (d+i c d x)\right )}{c d^2 f (-i+c x) \sqrt {1+c^2 x^2}} \]

[In]

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

[Out]

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

Maple [F]

\[\int \frac {a +b \,\operatorname {arcsinh}\left (c x \right )}{\left (i c d x +d \right )^{\frac {3}{2}} \sqrt {-i c f x +f}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 443 vs. \(2 (89) = 178\).

Time = 0.31 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.99 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\frac {2 \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (c^{2} d^{2} f x - i \, c d^{2} f\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f}} \log \left (-\frac {{\left (i \, b c^{6} x^{2} + 2 \, b c^{5} x - 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} - {\left (i \, c^{9} d^{2} f x^{4} + 2 \, c^{8} d^{2} f x^{3} + i \, c^{7} d^{2} f x^{2} + 2 \, c^{6} d^{2} f x\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f}}}{8 \, {\left (b c^{3} x^{3} - i \, b c^{2} x^{2} + b c x - i \, b\right )}}\right ) - {\left (c^{2} d^{2} f x - i \, c d^{2} f\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f}} \log \left (-\frac {{\left (i \, b c^{6} x^{2} + 2 \, b c^{5} x - 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} - {\left (-i \, c^{9} d^{2} f x^{4} - 2 \, c^{8} d^{2} f x^{3} - i \, c^{7} d^{2} f x^{2} - 2 \, c^{6} d^{2} f x\right )} \sqrt {\frac {b^{2}}{c^{2} d^{3} f}}}{8 \, {\left (b c^{3} x^{3} - i \, b c^{2} x^{2} + b c x - i \, b\right )}}\right ) + 2 \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} a}{2 \, {\left (c^{2} d^{2} f x - i \, c d^{2} f\right )}} \]

[In]

integrate((a+b*arcsinh(c*x))/(d+I*c*d*x)^(3/2)/(f-I*c*f*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^2*f*x - I*c*d^2*f)*sqrt(b^
2/(c^2*d^3*f))*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^2*f*x^4 + 2*c^8*d^2*f*x^3 + I*c^7*d^2*f*x^2 + 2*c^6*d^2*f*x)*sqrt(b^2/(c^2*d^3*f)))/(b*c^3*
x^3 - I*b*c^2*x^2 + b*c*x - I*b)) - (c^2*d^2*f*x - I*c*d^2*f)*sqrt(b^2/(c^2*d^3*f))*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^2*f*x^4 - 2*c^8*d^2*f
*x^3 - I*c^7*d^2*f*x^2 - 2*c^6*d^2*f*x)*sqrt(b^2/(c^2*d^3*f)))/(b*c^3*x^3 - I*b*c^2*x^2 + b*c*x - I*b)) + 2*sq
rt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a)/(c^2*d^2*f*x - I*c*d^2*f)

Sympy [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\left (i d \left (c x - i\right )\right )^{\frac {3}{2}} \sqrt {- i f \left (c x + i\right )}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.88 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\frac {i \, \sqrt {c^{2} d f x^{2} + d f} b \operatorname {arsinh}\left (c x\right )}{i \, c^{2} d^{2} f x + c d^{2} f} + \frac {i \, \sqrt {c^{2} d f x^{2} + d f} a}{i \, c^{2} d^{2} f x + c d^{2} f} - \frac {b \log \left (i \, c x + 1\right )}{c d^{\frac {3}{2}} \sqrt {f}} \]

[In]

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

Giac [F]

\[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (i \, c d x + d\right )}^{\frac {3}{2}} \sqrt {-i \, c f x + f}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}}} \,d x \]

[In]

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

[Out]

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