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

Optimal. Leaf size=259 \[ \frac {2 i a b f x \sqrt {c^2 x^2+1}}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {f \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i f \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b^2 f \left (c^2 x^2+1\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 f x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[Out]

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

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Rubi [A]  time = 0.51, antiderivative size = 259, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {5712, 5821, 5675, 5717, 5653, 261} \[ \frac {2 i a b f x \sqrt {c^2 x^2+1}}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {f \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i f \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b^2 f \left (c^2 x^2+1\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 f x \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*a*b*f*x*Sqrt[1 + c^2*x^2])/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - ((2*I)*b^2*f*(1 + c^2*x^2))/(c*Sqrt[
d + I*c*d*x]*Sqrt[f - I*c*f*x]) + ((2*I)*b^2*f*x*Sqrt[1 + c^2*x^2]*ArcSinh[c*x])/(Sqrt[d + I*c*d*x]*Sqrt[f - I
*c*f*x]) - (I*f*(1 + c^2*x^2)*(a + b*ArcSinh[c*x])^2)/(c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (f*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 261

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

Rule 5653

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.), x_Symbol] :> Simp[x*(a + b*ArcSinh[c*x])^n, x] - Dist[b*c*n, In
t[(x*(a + b*ArcSinh[c*x])^(n - 1))/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c}, x] && GtQ[n, 0]

Rule 5675

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(a + b*ArcSinh[c*x]
)^(n + 1)/(b*c*Sqrt[d]*(n + 1)), x] /; FreeQ[{a, b, c, d, e, n}, x] && EqQ[e, c^2*d] && GtQ[d, 0] && NeQ[n, -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 5717

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*(x_)*((d_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)
^(p + 1)*(a + b*ArcSinh[c*x])^n)/(2*e*(p + 1)), x] - Dist[(b*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*c*(p +
1)*(1 + c^2*x^2)^FracPart[p]), Int[(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x] /; FreeQ[{a,
b, c, d, e, p}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && NeQ[p, -1]

Rule 5821

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol]
:> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g
}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n, 0] && ((EqQ[n, 1] && GtQ[p,
-1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] && LtQ[p, -2]))

Rubi steps

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

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Mathematica [A]  time = 1.17, size = 315, normalized size = 1.22 \[ \frac {-3 i \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a^2 \sqrt {c^2 x^2+1}-2 a b c x+2 b^2 \sqrt {c^2 x^2+1}\right )+3 a^2 \sqrt {d} \sqrt {f} \sqrt {c^2 x^2+1} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )+3 b \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2 \left (a-i b \sqrt {c^2 x^2+1}\right )+6 i b \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (b c x-a \sqrt {c^2 x^2+1}\right )+b^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^3}{3 c d \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {-i \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} - 2 i \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) - i \, \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} a^{2}}{c d x - i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} {\left (\frac {f \operatorname {arsinh}\left (c x\right )}{c d \sqrt {\frac {f}{d}}} - \frac {i \, \sqrt {c^{2} d f x^{2} + d f}}{c d}\right )} + \int \frac {\sqrt {-i \, c f x + f} b^{2} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2}}{\sqrt {i \, c d x + d}} + \frac {2 \, \sqrt {-i \, c f x + f} a b \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )}{\sqrt {i \, c d x + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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