3.6.46 \(\int (d+i c d x)^{5/2} (f-i c f x)^{5/2} (a+b \sinh ^{-1}(c x)) \, dx\) [546]

Optimal. Leaf size=344 \[ -\frac {25 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b c^3 x^4 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (1+c^2 x^2\right )^{5/2}}-\frac {b (d+i c d x)^{5/2} (f-i c f x)^{5/2} \sqrt {1+c^2 x^2}}{36 c}+\frac {1}{6} x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{16 \left (1+c^2 x^2\right )^2}+\frac {5 x (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )}{24 \left (1+c^2 x^2\right )}+\frac {5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \left (1+c^2 x^2\right )^{5/2}} \]

[Out]

-25/96*b*c*x^2*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)/(c^2*x^2+1)^(5/2)-5/96*b*c^3*x^4*(d+I*c*d*x)^(5/2)*(f-I*c*f
*x)^(5/2)/(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))+5/16*x*(d+I*c*d*x)^(5
/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))/(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))/(c^2*x^2+1)+5/32*(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(a+b*arcsinh(c*x))^2/b/c/(c^2*x^2+1)^(5/2)-1/36*b*
(d+I*c*d*x)^(5/2)*(f-I*c*f*x)^(5/2)*(c^2*x^2+1)^(1/2)/c

________________________________________________________________________________________

Rubi [A]
time = 0.20, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {5796, 5786, 5785, 5783, 30, 14, 267} \begin {gather*} \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 )}{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 )}{16 \left (c^2 x^2+1\right )^2}+\frac {5 (d+i c d x)^{5/2} (f-i c f x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{32 b c \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 )-\frac {25 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (c^2 x^2+1\right )^{5/2}}-\frac {b \sqrt {c^2 x^2+1} (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{36 c}-\frac {5 b c^3 x^4 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (c^2 x^2+1\right )^{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]),x]

[Out]

(-25*b*c*x^2*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))/(96*(1 + c^2*x^2)^(5/2)) - (5*b*c^3*x^4*(d + I*c*d*x)^(5
/2)*(f - I*c*f*x)^(5/2))/(96*(1 + c^2*x^2)^(5/2)) - (b*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*Sqrt[1 + c^2*x^
2])/(36*c) + (x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a + b*ArcSinh[c*x]))/6 + (5*x*(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)^2) + (5*x*(d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2)*(a
+ b*ArcSinh[c*x]))/(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])^2)/(32
*b*c*(1 + c^2*x^2)^(5/2))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

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

Rule 267

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 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]

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 ) \, 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 ) \, 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 )+\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 ) \, 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 \, dx}{6 \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}}{36 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 )+\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 )}{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 ) \, dx}{8 \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 ) \, dx}{24 \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}}{36 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 )+\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 )}{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 )}{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 {a+b \sinh ^{-1}(c x)}{\sqrt {1+c^2 x^2}} \, dx}{16 \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 \left (x+c^2 x^3\right ) \, dx}{24 \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 \, dx}{16 \left (1+c^2 x^2\right )^{5/2}}\\ &=-\frac {25 b c x^2 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \left (1+c^2 x^2\right )^{5/2}}-\frac {5 b c^3 x^4 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}{96 \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}}{36 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 )+\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 )}{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 )}{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 )^2}{32 b c \left (1+c^2 x^2\right )^{5/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.75, size = 481, normalized size = 1.40 \begin {gather*} \frac {1584 a c d^2 f^2 x \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+1248 a 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}+384 a 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}+360 b d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2-270 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 )-27 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 )-2 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 )+720 a 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 )+12 b d^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (45 \sinh \left (2 \sinh ^{-1}(c x)\right )+9 \sinh \left (4 \sinh ^{-1}(c x)\right )+\sinh \left (6 \sinh ^{-1}(c x)\right )\right )}{2304 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]),x]

[Out]

(1584*a*c*d^2*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 1248*a*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] + 384*a*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] + 360*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2 - 270*b*d^2*f^2*Sqrt[d + I*c*d*x
]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] - 27*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[4*ArcSinh[c*x
]] - 2*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[6*ArcSinh[c*x]] + 720*a*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]] + 12*b*d^2*f^2*Sqrt[d + I*c*d*x]*Sqrt
[f - I*c*f*x]*ArcSinh[c*x]*(45*Sinh[2*ArcSinh[c*x]] + 9*Sinh[4*ArcSinh[c*x]] + Sinh[6*ArcSinh[c*x]]))/(2304*c*
Sqrt[1 + c^2*x^2])

________________________________________________________________________________________

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

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)),x)

[Out]

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

________________________________________________________________________________________

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)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)),x)

[Out]

Timed out

________________________________________________________________________________________

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)),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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________