∫ a+b sinh−1(cx)___ (d+icdx)5∕2(f−icfx)5∕2 dx

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

Optimal. Leaf size=203 \[ \frac {2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b \left (c^2 x^2+1\right )^{3/2}}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

[Out]

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

^(5/2)/(f-I*c*f*x)^(5/2)+2/3*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*b*(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)

________________________________________________________________________________________

Rubi [A] time = 0.24, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {5712, 5690, 5687, 260, 261} \[ \frac {2 x \left (c^2 x^2+1\right )^2 \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b \left (c^2 x^2+1\right )^{3/2}}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b \left (c^2 x^2+1\right )^{5/2} \log \left (c^2 x^2+1\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

))

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 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 5687

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[(x*(a + b*ArcSinh

[c*x])^n)/(d*Sqrt[d + e*x^2]), x] - Dist[(b*c*n*Sqrt[1 + c^2*x^2])/(d*Sqrt[d + e*x^2]), Int[(x*(a + b*ArcSinh[

c*x])^(n - 1))/(1 + c^2*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]

Rule 5690

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x^2)^(p

+ 1)*(a + b*ArcSinh[c*x])^n)/(2*d*(p + 1)), x] + (Dist[(2*p + 3)/(2*d*(p + 1)), Int[(d + e*x^2)^(p + 1)*(a +

b*ArcSinh[c*x])^n, x], x] + Dist[(b*c*n*d^IntPart[p]*(d + e*x^2)^FracPart[p])/(2*(p + 1)*(1 + c^2*x^2)^FracPar

t[p]), Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ

[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]

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]

Rubi steps

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

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Mathematica [A] time = 0.57, size = 193, normalized size = 0.95 \[ \frac {i \sqrt {f-i c f x} \left (4 a c^3 x^3+6 a c x-2 b c^2 x^2 \sqrt {c^2 x^2+1} \log (d+i c d x)-2 b \left (c^2 x^2+1\right )^{3/2} \log (d (-1+i c x))-2 b \sqrt {c^2 x^2+1} \log (d+i c d x)+b \sqrt {c^2 x^2+1}+2 b c x \left (2 c^2 x^2+3\right ) \sinh ^{-1}(c x)\right )}{6 c d^2 f^3 (c x-i) (c x+i)^2 \sqrt {d+i c d x}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

((I/6)*Sqrt[f - I*c*f*x]*(6*a*c*x + 4*a*c^3*x^3 + b*Sqrt[1 + c^2*x^2] + 2*b*c*x*(3 + 2*c^2*x^2)*ArcSinh[c*x] -

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

^2*x^2]*Log[d + I*c*d*x]))/(c*d^2*f^3*(-I + c*x)*(I + c*x)^2*Sqrt[d + I*c*d*x])

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fricas [F] time = 0.84, size = 0, normalized size = 0.00 \[ -\frac {\sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x^{2} - 2 \, {\left (2 \, b c^{2} x^{3} + 3 \, b x\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 (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} \log \left (\frac {\sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d^{2} f^{2} x^{2} \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} + b c^{2} x^{4} + b x^{2}}{b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b}\right ) + {\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} \log \left (-\frac {\sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d^{2} f^{2} x^{2} \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} - b c^{2} x^{4} - b x^{2}}{b c^{4} x^{4} + 2 \, b c^{2} x^{2} + b}\right ) + 2 \, {\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} \log \left (\frac {\sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d^{2} f^{2} x \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} + b c^{2} x^{3} + b x}{b c^{2} x^{2} + b}\right ) - 2 \, {\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )} \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} \log \left (-\frac {\sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} c d^{2} f^{2} x \sqrt {\frac {b^{2}}{c^{2} d^{5} f^{5}}} - b c^{2} x^{3} - b x}{b c^{2} x^{2} + b}\right ) - 2 \, {\left (2 \, a c^{2} x^{3} + 3 \, a x\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} - 6 \, {\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )} {\rm integral}\left (-\frac {2 \, \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x}{3 \, {\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\right )}}, x\right )}{6 \, {\left (c^{4} d^{3} f^{3} x^{4} + 2 \, c^{2} d^{3} f^{3} x^{2} + d^{3} f^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x^2 - 2*(2*b*c^2*x^3 + 3*b*x)*sqrt(I*c*d*x +

d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) - (c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(b^2/

(c^2*d^5*f^5))*log((sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x^2*sqrt(b^2/(c^2*d^5*f^5

)) + b*c^2*x^4 + b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) + (c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(

b^2/(c^2*d^5*f^5))*log(-(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x^2*sqrt(b^2/(c^2*d^

5*f^5)) - b*c^2*x^4 - b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) + 2*(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3

)*sqrt(b^2/(c^2*d^5*f^5))*log((sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x*sqrt(b^2/(c^

2*d^5*f^5)) + b*c^2*x^3 + b*x)/(b*c^2*x^2 + b)) - 2*(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(b^2/(

c^2*d^5*f^5))*log(-(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x*sqrt(b^2/(c^2*d^5*f^5))

- b*c^2*x^3 - b*x)/(b*c^2*x^2 + b)) - 2*(2*a*c^2*x^3 + 3*a*x)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) - 6*(c^4*d

^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*integral(-2/3*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)

*b*c*x/(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3), x))/(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn

ing, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was

done assuming [c,d,t_nostep]=[-62,5,7]Warning, choosing root of [1,0,%%%{-8,[2,4,2]%%%}+%%%{-6,[0,0,0]%%%},%%%

{-32,[2,4,2]%%%}+%%%{-8,[0,0,0]%%%},%%%{16,[4,8,4]%%%}+%%%{-24,[2,4,2]%%%}+%%%{-3,[0,0,0]%%%}] at parameters v

alues [26,-89,63]Warning, need to choose a branch for the root of a polynomial with parameters. This might be

wrong.The choice was done assuming [c,d,t_nostep]=[50,45,-24]Warning, choosing root of [1,0,%%%{-8,[2,4,2]%%%}

+%%%{-6,[0,0,0]%%%},%%%{-32,[2,4,2]%%%}+%%%{-8,[0,0,0]%%%},%%%{16,[4,8,4]%%%}+%%%{-24,[2,4,2]%%%}+%%%{-3,[0,0,

0]%%%}] at parameters values [-64,-30,70]Warning, need to choose a branch for the root of a polynomial with pa

rameters. This might be wrong.The choice was done assuming [c,d,t_nostep]=[-43,-99,-71]schur row 3 -2.20626e-0

8Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice

was done assuming [c,d,t_nostep]=[29,40,-81]Warning, need to choose a branch for the root of a polynomial wit

h parameters. This might be wrong.The choice was done assuming [c,d,t_nostep]=[-69,45,-8]schur row 1 4.86814e-

07Francis algorithm not precise enough for[1.0,0.0,-6.053520512e+13,-2.4214082048e+14,9.1612776473e+26]Warning

, choosing root of [1,0,%%%{-8,[2,4,2]%%%}+%%%{-6,[0,0,0]%%%},%%%{-32,[2,4,2]%%%}+%%%{-8,[0,0,0]%%%},%%%{16,[4

,8,4]%%%}+%%%{-24,[2,4,2]%%%}+%%%{-3,[0,0,0]%%%}] at parameters values [-80,-23,65]Warning, need to choose a b

ranch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming [c,d,t_no

step]=[-35,-31,-9]Warning, choosing root of [1,0,%%%{-8,[2,4,2]%%%}+%%%{-6,[0,0,0]%%%},%%%{-32,[2,4,2]%%%}+%%%

{-8,[0,0,0]%%%},%%%{16,[4,8,4]%%%}+%%%{-24,[2,4,2]%%%}+%%%{-3,[0,0,0]%%%}] at parameters values [-44,-22,93]Wa

rning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.The choice wa

s done assuming [c,d,t_nostep]=[-90,-5,-92]Warning, need to choose a branch for the root of a polynomial with

parameters. This might be wrong.The choice was done assuming [c,d,t_nostep]=[-96,-85,-8]schur row 1 1.20095e-0

7Francis algorithm not precise enough for[1.0,0.0,-2.4631345152e+14,-4.9262690304e+14,1.51675790999e+28]ext_re

duce Error: Bad Argument Typeext_reduce Error: Bad Argument TypeWarning, need to choose a branch for the root

of a polynomial with parameters. This might be wrong.The choice was done assuming [c,f,t_nostep]=[10,-24,-7]Wa

rning, choosing root of [1,0,%%%{-8,[2,4,2]%%%}+%%%{-6,[0,0,0]%%%},%%%{-32,[2,4,2]%%%}+%%%{-8,[0,0,0]%%%},%%%{

16,[4,8,4]%%%}+%%%{-24,[2,4,2]%%%}+%%%{-3,[0,0,0]%%%}] at parameters values [-24,-17,41]Warning, need to choos

e a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming [c,f

,t_nostep]=[82,83,53]schur row 3 2.05953e-07Warning, choosing root of [1,0,%%%{-8,[2,4,2]%%%}+%%%{-6,[0,0,0]%%

%},%%%{-32,[2,4,2]%%%}+%%%{-8,[0,0,0]%%%},%%%{16,[4,8,4]%%%}+%%%{-24,[2,4,2]%%%}+%%%{-3,[0,0,0]%%%}] at parame

ters values [-51,-42,-65]Warning, need to choose a branch for the root of a polynomial with parameters. This m

ight be wrong.The choice was done assuming [c,f,t_nostep]=[-93,50,64]schur row 3 -1.05315e-07Warning, need to

choose a branch for the root of a polynomial with parameters. This might be wrong.The choice was done assuming

[c,f,t_nostep]=[-26,-23,30]Warning, need to choose a branch for the root of a polynomial with parameters. Thi

s might be wrong.The choice was done assuming [c,f,t_nostep]=[90,62,-68]Warning, choosing root of [1,0,%%%{-8,

[2,4,2]%%%}+%%%{-6,[0,0,0]%%%},%%%{-32,[2,4,2]%%%}+%%%{-8,[0,0,0]%%%},%%%{16,[4,8,4]%%%}+%%%{-24,[2,4,2]%%%}+%

%%{-3,[0,0,0]%%%}] at parameters values [-7,81,82]Warning, need to choose a branch for the root of a polynomia

l with parameters. This might be wrong.The choice was done assuming [c,f,t_nostep]=[-11,3,-26]Warning, choosin

g root of [1,0,%%%{-8,[2,4,2]%%%}+%%%{-6,[0,0,0]%%%},%%%{-32,[2,4,2]%%%}+%%%{-8,[0,0,0]%%%},%%%{16,[4,8,4]%%%}

+%%%{-24,[2,4,2]%%%}+%%%{-3,[0,0,0]%%%}] at parameters values [-11,-50,-53]Warning, need to choose a branch fo

r the root of a polynomial with parameters. This might be wrong.The choice was done assuming [c,f,t_nostep]=[-

78,84,2]schur row 3 -6.60848e-07Warning, need to choose a branch for the root of a polynomial with parameters.

This might be wrong.The choice was done assuming [c,f,t_nostep]=[-59,-23,-54]schur row 1 5.11765e-07Francis a

lgorithm not precise enough for[1.0,0.0,-2.27244234819e+13,-4.54488469638e+13,1.29099855646e+26]ext_reduce Err

or: Bad Argument Typeext_reduce Error: Bad Argument TypeEvaluation time: 4.24Done

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

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)^(5/2),x)

[Out]

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

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maxima [A] time = 0.53, size = 159, normalized size = 0.78 \[ \frac {1}{6} \, b c {\left (\frac {1}{c^{4} d^{\frac {5}{2}} f^{\frac {5}{2}} x^{2} + c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}} - \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \]

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)^(5/2),x, algorithm="maxima")

[Out]

1/6*b*c*(1/(c^4*d^(5/2)*f^(5/2)*x^2 + c^2*d^(5/2)*f^(5/2)) - 2*log(c^2*x^2 + 1)/(c^2*d^(5/2)*f^(5/2))) + 1/3*b

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

*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d^2*f^2))

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

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)^(5/2)),x)

[Out]

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

[Out]

Timed out

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