\(\int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \, dx\) [591]

Optimal result

Integrand size = 37, antiderivative size = 59 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \, dx=\frac {\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {5796, 5783} \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \, dx=\frac {\sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[In]

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

[Out]

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

Rule 5783

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(1/(b*c*(n + 1)))*S
imp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]]*(a + b*ArcSinh[c*x])^(n + 1), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ
[e, c^2*d] && NeQ[n, -1]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {1+c^2 x^2} \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \\ & = \frac {\sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(168\) vs. \(2(59)=118\).

Time = 2.26 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.85 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \, dx=\frac {a b \sqrt {1+c^2 x^2} \text {arcsinh}(c x)^2}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {b^2 \sqrt {1+c^2 x^2} \text {arcsinh}(c x)^3}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {a^2 \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )}{c \sqrt {d} \sqrt {f}} \]

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.90 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \, dx=\frac {b^{2} \operatorname {arsinh}\left (c x\right )^{3}}{3 \, \sqrt {d f} c} + \frac {a b \operatorname {arsinh}\left (c x\right )^{2}}{\sqrt {d f} c} + \frac {a^{2} \operatorname {arsinh}\left (c x\right )}{\sqrt {d f} c} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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