3.6.41 \(\int (d+i c d x)^{3/2} (f-i c f x)^{3/2} (a+b \sinh ^{-1}(c x)) \, dx\) [541]

Optimal. Leaf size=247 \[ -\frac {5 b c x^2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (1+c^2 x^2\right )^{3/2}}-\frac {b c^3 x^4 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (1+c^2 x^2\right )^{3/2}}+\frac {1}{4} x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )+\frac {3 x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \left (1+c^2 x^2\right )}+\frac {3 (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \left (1+c^2 x^2\right )^{3/2}} \]

[Out]

-5/16*b*c*x^2*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)/(c^2*x^2+1)^(3/2)-1/16*b*c^3*x^4*(d+I*c*d*x)^(3/2)*(f-I*c*f*
x)^(3/2)/(c^2*x^2+1)^(3/2)+1/4*x*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))+3/8*x*(d+I*c*d*x)^(3/2
)*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))/(c^2*x^2+1)+3/16*(d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))
^2/b/c/(c^2*x^2+1)^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.16, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {5796, 5786, 5785, 5783, 30, 14} \begin {gather*} \frac {3 x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )}{8 \left (c^2 x^2+1\right )}+\frac {3 (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{16 b c \left (c^2 x^2+1\right )^{3/2}}+\frac {1}{4} x (d+i c d x)^{3/2} (f-i c f x)^{3/2} \left (a+b \sinh ^{-1}(c x)\right )-\frac {5 b c x^2 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (c^2 x^2+1\right )^{3/2}}-\frac {b c^3 x^4 (d+i c d x)^{3/2} (f-i c f x)^{3/2}}{16 \left (c^2 x^2+1\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(-5*b*c*x^2*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))/(16*(1 + c^2*x^2)^(3/2)) - (b*c^3*x^4*(d + I*c*d*x)^(3/2)
*(f - I*c*f*x)^(3/2))/(16*(1 + c^2*x^2)^(3/2)) + (x*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x
]))/4 + (3*x*(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x]))/(8*(1 + c^2*x^2)) + (3*(d + I*c*d*x
)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x])^2)/(16*b*c*(1 + c^2*x^2)^(3/2))

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

________________________________________________________________________________________

Mathematica [A]
time = 0.53, size = 352, normalized size = 1.43 \begin {gather*} \frac {80 a c d f x \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+32 a c^3 d f x^3 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+24 b d f \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2-16 b d f \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )-b d f \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (4 \sinh ^{-1}(c x)\right )+48 a d^{3/2} f^{3/2} \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 )+4 b d f \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (8 \sinh \left (2 \sinh ^{-1}(c x)\right )+\sinh \left (4 \sinh ^{-1}(c x)\right )\right )}{128 c \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2)*(a + b*ArcSinh[c*x]),x]

[Out]

(80*a*c*d*f*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 32*a*c^3*d*f*x^3*Sqrt[d + I*c*d*x]*Sqrt[
f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 24*b*d*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2 - 16*b*d*f*Sqrt[d
+ I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] - b*d*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[4*ArcSinh[
c*x]] + 48*a*d^(3/2)*f^(3/2)*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]] + 4*b*d*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]*(8*Sinh[2*ArcSinh[c*x]] + Sinh[4*ArcSinh[c*x]])
)/(128*c*Sqrt[1 + c^2*x^2])

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (i c d x +d \right )^{\frac {3}{2}} \left (-i c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x)),x)

[Out]

int((d+I*c*d*x)^(3/2)*(f-I*c*f*x)^(3/2)*(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)^(3/2)*(f-I*c*f*x)^(3/2)*(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)^(3/2)*(f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d+I*c*d*x)**(3/2)*(f-I*c*f*x)**(3/2)*(a+b*asinh(c*x)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3003 deep

________________________________________________________________________________________

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)^(3/2)*(f-I*c*f*x)^(3/2)*(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 )}^{3/2}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2} \,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)^(3/2)*(f - c*f*x*1i)^(3/2),x)

[Out]

int((a + b*asinh(c*x))*(d + c*d*x*1i)^(3/2)*(f - c*f*x*1i)^(3/2), x)

________________________________________________________________________________________