3.6.82 \(\int (d+i c d x)^{5/2} (f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x))^2 \, dx\) [582]

Optimal. Leaf size=548 \[ \frac {1}{108} b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}+\frac {245 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1152 \left (1+c^2 x^2\right )^2}+\frac {65 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1728 \left (1+c^2 x^2\right )}-\frac {115 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sinh ^{-1}(c x)}{1152 c \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c \sqrt {1+c^2 x^2}}-\frac {b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \left (1+c^2 x^2\right )^2}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{24 \left (1+c^2 x^2\right )}+\frac {5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \left (1+c^2 x^2\right )^{5/2}} \]

[Out]

1/108*b^2*x*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)+245/1152*b^2*x*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)/(c^2*x^2+1)
^2+65/1728*b^2*x*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)/(c^2*x^2+1)-115/1152*b^2*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5
/2)*arcsinh(c*x)/c/(c^2*x^2+1)^(5/2)-5/16*b*c*x^2*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))/(c^2*
x^2+1)^(5/2)+1/6*x*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))^2+5/16*x*(d+I*c*d*x)^(5/2)*(f-I*c*f*
x)^(5/2)*(a+b*arcsinh(c*x))^2/(c^2*x^2+1)^2+5/24*x*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))^2/(c
^2*x^2+1)+5/48*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))^3/b/c/(c^2*x^2+1)^(5/2)-5/48*b*(d+I*c*d*
x)^(5/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))/c/(c^2*x^2+1)^(1/2)-1/18*b*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(
a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/c

________________________________________________________________________________________

Rubi [A]
time = 0.39, antiderivative size = 548, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {5796, 5786, 5785, 5783, 5776, 327, 221, 5798, 201} \begin {gather*} \frac {5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \left (c^2 x^2+1\right )^{5/2}}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{24 \left (c^2 x^2+1\right )}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \left (c^2 x^2+1\right )^2}-\frac {b \sqrt {c^2 x^2+1} (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}-\frac {5 b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c \sqrt {c^2 x^2+1}}-\frac {5 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (c^2 x^2+1\right )^{5/2}}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {65 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1728 \left (c^2 x^2+1\right )}+\frac {245 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1152 \left (c^2 x^2+1\right )^2}-\frac {115 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sinh ^{-1}(c x)}{1152 c \left (c^2 x^2+1\right )^{5/2}}+\frac {1}{108} b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b^2*x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))/108 + (245*b^2*x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))/(115
2*(1 + c^2*x^2)^2) + (65*b^2*x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))/(1728*(1 + c^2*x^2)) - (115*b^2*(d + I
*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*ArcSinh[c*x])/(1152*c*(1 + c^2*x^2)^(5/2)) - (5*b*c*x^2*(d + I*c*d*x)^(5/2)*
(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x]))/(16*(1 + c^2*x^2)^(5/2)) - (5*b*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5
/2)*(a + b*ArcSinh[c*x]))/(48*c*Sqrt[1 + c^2*x^2]) - (b*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*Sqrt[1 + c^2*x
^2]*(a + b*ArcSinh[c*x]))/(18*c) + (x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x])^2)/6 + (5*x
*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x])^2)/(16*(1 + c^2*x^2)^2) + (5*x*(d + I*c*d*x)^(5/
2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x])^2)/(24*(1 + c^2*x^2)) + (5*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)
*(a + b*ArcSinh[c*x])^3)/(48*b*c*(1 + c^2*x^2)^(5/2))

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 5776

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*ArcS
inh[c*x])^n/(d*(m + 1))), x] - Dist[b*c*(n/(d*(m + 1))), Int[(d*x)^(m + 1)*((a + b*ArcSinh[c*x])^(n - 1)/Sqrt[
1 + c^2*x^2]), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && NeQ[m, -1]

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 5785

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

Rule 5786

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

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 5798

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/(2*c*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)
^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]

Rubi steps

\begin {align*} \int (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx &=\frac {\left ((d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx}{\left (1+c^2 x^2\right )^{5/2}}\\ &=\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {\left (5 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx}{6 \left (1+c^2 x^2\right )^{5/2}}-\frac {\left (b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int x \left (1+c^2 x^2\right )^2 \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{3 \left (1+c^2 x^2\right )^{5/2}}\\ &=-\frac {b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{24 \left (1+c^2 x^2\right )}+\frac {\left (5 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2 \, dx}{8 \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (1+c^2 x^2\right )^{5/2} \, dx}{18 \left (1+c^2 x^2\right )^{5/2}}-\frac {\left (5 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{12 \left (1+c^2 x^2\right )^{5/2}}\\ &=\frac {1}{108} b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}-\frac {5 b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c \sqrt {1+c^2 x^2}}-\frac {b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \left (1+c^2 x^2\right )^2}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{24 \left (1+c^2 x^2\right )}+\frac {\left (5 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{16 \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (5 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{108 \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (5 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \left (1+c^2 x^2\right )^{3/2} \, dx}{48 \left (1+c^2 x^2\right )^{5/2}}-\frac {\left (5 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int x \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{8 \left (1+c^2 x^2\right )^{5/2}}\\ &=\frac {1}{108} b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}+\frac {65 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1728 \left (1+c^2 x^2\right )}-\frac {5 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c \sqrt {1+c^2 x^2}}-\frac {b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \left (1+c^2 x^2\right )^2}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{24 \left (1+c^2 x^2\right )}+\frac {5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (5 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \sqrt {1+c^2 x^2} \, dx}{144 \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (5 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \sqrt {1+c^2 x^2} \, dx}{64 \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (5 b^2 c^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \frac {x^2}{\sqrt {1+c^2 x^2}} \, dx}{16 \left (1+c^2 x^2\right )^{5/2}}\\ &=\frac {1}{108} b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}+\frac {245 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1152 \left (1+c^2 x^2\right )^2}+\frac {65 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1728 \left (1+c^2 x^2\right )}-\frac {5 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c \sqrt {1+c^2 x^2}}-\frac {b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \left (1+c^2 x^2\right )^2}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{24 \left (1+c^2 x^2\right )}+\frac {5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (5 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{288 \left (1+c^2 x^2\right )^{5/2}}+\frac {\left (5 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{128 \left (1+c^2 x^2\right )^{5/2}}-\frac {\left (5 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}\right ) \int \frac {1}{\sqrt {1+c^2 x^2}} \, dx}{32 \left (1+c^2 x^2\right )^{5/2}}\\ &=\frac {1}{108} b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}+\frac {245 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1152 \left (1+c^2 x^2\right )^2}+\frac {65 b^2 x (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{1728 \left (1+c^2 x^2\right )}-\frac {115 b^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sinh ^{-1}(c x)}{1152 c \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{48 c \sqrt {1+c^2 x^2}}-\frac {b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )}{18 c}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 \left (1+c^2 x^2\right )^2}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{24 \left (1+c^2 x^2\right )}+\frac {5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{48 b c \left (1+c^2 x^2\right )^{5/2}}\\ \end {align*}

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Mathematica [A]
time = 1.46, size = 735, normalized size = 1.34 \begin {gather*} \frac {9504 a^2 c d^2 f^2 x \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+7488 a^2 c^3 d^2 f^2 x^3 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+2304 a^2 c^5 d^2 f^2 x^5 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+1440 b^2 d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^3-3240 a b d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )-324 a b d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (4 \sinh ^{-1}(c x)\right )-24 a b d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (6 \sinh ^{-1}(c x)\right )+4320 a^2 d^{5/2} f^{5/2} \sqrt {1+c^2 x^2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )+1620 b^2 d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh \left (2 \sinh ^{-1}(c x)\right )+81 b^2 d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh \left (4 \sinh ^{-1}(c x)\right )+4 b^2 d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh \left (6 \sinh ^{-1}(c x)\right )-12 b d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (270 b \cosh \left (2 \sinh ^{-1}(c x)\right )+27 b \cosh \left (4 \sinh ^{-1}(c x)\right )+2 b \cosh \left (6 \sinh ^{-1}(c x)\right )-540 a \sinh \left (2 \sinh ^{-1}(c x)\right )-108 a \sinh \left (4 \sinh ^{-1}(c x)\right )-12 a \sinh \left (6 \sinh ^{-1}(c x)\right )\right )+72 b d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2 \left (60 a+45 b \sinh \left (2 \sinh ^{-1}(c x)\right )+9 b \sinh \left (4 \sinh ^{-1}(c x)\right )+b \sinh \left (6 \sinh ^{-1}(c x)\right )\right )}{13824 c \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(9504*a^2*c*d^2*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 7488*a^2*c^3*d^2*f^2*x^3*Sqrt[d
+ I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 2304*a^2*c^5*d^2*f^2*x^5*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*
Sqrt[1 + c^2*x^2] + 1440*b^2*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^3 - 3240*a*b*d^2*f^2*Sqr
t[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] - 324*a*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Co
sh[4*ArcSinh[c*x]] - 24*a*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[6*ArcSinh[c*x]] + 4320*a^2*d^(5/2
)*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]] + 1620*b^2*d^2*
f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[2*ArcSinh[c*x]] + 81*b^2*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f
*x]*Sinh[4*ArcSinh[c*x]] + 4*b^2*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[6*ArcSinh[c*x]] - 12*b*d^2*f
^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]*(270*b*Cosh[2*ArcSinh[c*x]] + 27*b*Cosh[4*ArcSinh[c*x]] +
2*b*Cosh[6*ArcSinh[c*x]] - 540*a*Sinh[2*ArcSinh[c*x]] - 108*a*Sinh[4*ArcSinh[c*x]] - 12*a*Sinh[6*ArcSinh[c*x]]
) + 72*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2*(60*a + 45*b*Sinh[2*ArcSinh[c*x]] + 9*b*Si
nh[4*ArcSinh[c*x]] + b*Sinh[6*ArcSinh[c*x]]))/(13824*c*Sqrt[1 + c^2*x^2])

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (i c d x +d \right )^{\frac {5}{2}} \left (-i c f x +f \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )^{2}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))^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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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*c^4*d^2*f^2*x^4 + 2*b^2*c^2*d^2*f^2*x^2 + b^2*d^2*f^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(
c*x + sqrt(c^2*x^2 + 1))^2 + 2*(a*b*c^4*d^2*f^2*x^4 + 2*a*b*c^2*d^2*f^2*x^2 + a*b*d^2*f^2)*sqrt(I*c*d*x + d)*s
qrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) + (a^2*c^4*d^2*f^2*x^4 + 2*a^2*c^2*d^2*f^2*x^2 + a^2*d^2*f^2)*s
qrt(I*c*d*x + d)*sqrt(-I*c*f*x + f), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**(5/2)*(f-I*c*f*x)**(5/2)*(a+b*asinh(c*x))**2,x)

[Out]

Timed out

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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