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

Optimal. Leaf size=381 \[ \frac {5 f^3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {c^2 x^2+1}}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {11 i b f^3 x \sqrt {c^2 x^2+1}}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i b c^2 f^3 x^3 \sqrt {c^2 x^2+1}}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[Out]

-11/3*I*f^3*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-3/2*f^3*x*(c^2*x^2+1)*(a+b*ar
csinh(c*x))/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+1/3*I*c*f^3*x^2*(c^2*x^2+1)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/
2)/(f-I*c*f*x)^(1/2)+11/3*I*b*f^3*x*(c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+3/4*b*c*f^3*x^2*(c^2
*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-1/9*I*b*c^2*f^3*x^3*(c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I
*c*f*x)^(1/2)+5/4*f^3*(a+b*arcsinh(c*x))^2*(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 = 0.63, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5712, 5821, 5675, 5717, 8, 5758, 30} \[ \frac {5 f^3 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i b c^2 f^3 x^3 \sqrt {c^2 x^2+1}}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {c^2 x^2+1}}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {11 i b f^3 x \sqrt {c^2 x^2+1}}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((11*I)/3)*b*f^3*x*Sqrt[1 + c^2*x^2])/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (3*b*c*f^3*x^2*Sqrt[1 + c^2*x^2
])/(4*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - ((I/9)*b*c^2*f^3*x^3*Sqrt[1 + c^2*x^2])/(Sqrt[d + I*c*d*x]*Sqrt[f
- I*c*f*x]) - (((11*I)/3)*f^3*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) - (
3*f^3*x*(1 + c^2*x^2)*(a + b*ArcSinh[c*x]))/(2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + ((I/3)*c*f^3*x^2*(1 + c^
2*x^2)*(a + b*ArcSinh[c*x]))/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (5*f^3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c
*x])^2)/(4*b*c*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 5758

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

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 {(f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+i c d x}} \, dx &=\frac {\sqrt {1+c^2 x^2} \int \frac {(f-i c f x)^3 \left (a+b \sinh ^{-1}(c x)\right )}{\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^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}-\frac {3 i c f^3 x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}-\frac {3 c^2 f^3 x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}+\frac {i c^3 f^3 x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}}\right ) \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=\frac {\left (f^3 \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (3 i c f^3 \sqrt {1+c^2 x^2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (3 c^2 f^3 \sqrt {1+c^2 x^2}\right ) \int \frac {x^2 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (i c^3 f^3 \sqrt {1+c^2 x^2}\right ) \int \frac {x^3 \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=-\frac {3 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{2 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (3 f^3 \sqrt {1+c^2 x^2}\right ) \int \frac {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (3 i b f^3 \sqrt {1+c^2 x^2}\right ) \int 1 \, dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (2 i c f^3 \sqrt {1+c^2 x^2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {1+c^2 x^2}} \, dx}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (3 b c f^3 \sqrt {1+c^2 x^2}\right ) \int x \, dx}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {\left (i b c^2 f^3 \sqrt {1+c^2 x^2}\right ) \int x^2 \, dx}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=\frac {3 i b f^3 x \sqrt {1+c^2 x^2}}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2}}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i b c^2 f^3 x^3 \sqrt {1+c^2 x^2}}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {5 f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {\left (2 i b f^3 \sqrt {1+c^2 x^2}\right ) \int 1 \, dx}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ &=\frac {11 i b f^3 x \sqrt {1+c^2 x^2}}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2}}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i b c^2 f^3 x^3 \sqrt {1+c^2 x^2}}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {5 f^3 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}\\ \end {align*}

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Mathematica [A]  time = 1.71, size = 465, normalized size = 1.22 \[ \frac {180 a \sqrt {d} f^{5/2} \sqrt {c^2 x^2+1} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )+24 i a c^2 f^2 x^2 \sqrt {c^2 x^2+1} \sqrt {d+i c d x} \sqrt {f-i c f x}-108 a c f^2 x \sqrt {c^2 x^2+1} \sqrt {d+i c d x} \sqrt {f-i c f x}-264 i a f^2 \sqrt {c^2 x^2+1} \sqrt {d+i c d x} \sqrt {f-i c f x}-8 i b c^3 f^2 x^3 \sqrt {d+i c d x} \sqrt {f-i c f x}-6 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (9 (2 c x+5 i) \sqrt {c^2 x^2+1}-i \cosh \left (3 \sinh ^{-1}(c x)\right )\right )+264 i b c f^2 x \sqrt {d+i c d x} \sqrt {f-i c f x}+90 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2+27 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )}{72 c d \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((264*I)*b*c*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (8*I)*b*c^3*f^2*x^3*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*
x] - (264*I)*a*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - 108*a*c*f^2*x*Sqrt[d + I*c*d*x]*Sqr
t[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + (24*I)*a*c^2*f^2*x^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2]
+ 90*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2 + 27*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*C
osh[2*ArcSinh[c*x]] - 6*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]*(9*(5*I + 2*c*x)*Sqrt[1 + c^2*x
^2] - I*Cosh[3*ArcSinh[c*x]]) + 180*a*Sqrt[d]*f^(5/2)*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]])/(72*c*d*Sqrt[1 + c^2*x^2])

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (i \, b c^{2} f^{2} x^{2} - 2 \, b c f^{2} x - i \, b f^{2}\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (i \, a c^{2} f^{2} x^{2} - 2 \, a c f^{2} x - i \, a f^{2}\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{c d x - i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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