\(\int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx\) [542]

Optimal result

Integrand size = 35, antiderivative size = 304 \[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {i b f x \sqrt {d+i c d x} \sqrt {f-i c f x}}{3 \sqrt {1+c^2 x^2}}-\frac {b c f x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \sqrt {1+c^2 x^2}}+\frac {i b c^2 f x^3 \sqrt {d+i c d x} \sqrt {f-i c f x}}{9 \sqrt {1+c^2 x^2}}+\frac {1}{2} f x \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))-\frac {i f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 c}+\frac {f \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{4 b c \sqrt {1+c^2 x^2}} \]

[Out]

1/2*f*x*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)-1/3*I*f*(c^2*x^2+1)*(a+b*arcsinh(c*x))*(d+I*c*d
*x)^(1/2)*(f-I*c*f*x)^(1/2)/c+1/3*I*b*f*x*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)-1/4*b*c*f*x^2*
(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)+1/9*I*b*c^2*f*x^3*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c
^2*x^2+1)^(1/2)+1/4*f*(a+b*arcsinh(c*x))^2*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/b/c/(c^2*x^2+1)^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {5796, 5838, 5785, 5783, 30, 5798} \[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {f \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{4 b c \sqrt {c^2 x^2+1}}-\frac {i f \left (c^2 x^2+1\right ) \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{3 c}+\frac {1}{2} f x \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))-\frac {b c f x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{4 \sqrt {c^2 x^2+1}}+\frac {i b f x \sqrt {d+i c d x} \sqrt {f-i c f x}}{3 \sqrt {c^2 x^2+1}}+\frac {i b c^2 f x^3 \sqrt {d+i c d x} \sqrt {f-i c f x}}{9 \sqrt {c^2 x^2+1}} \]

[In]

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

[Out]

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

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 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]

Rule 5838

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

Mathematica [A] (verified)

Time = 2.94 (sec) , antiderivative size = 273, normalized size of antiderivative = 0.90 \[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\frac {12 a f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (-2 i+3 c x-2 i c^2 x^2\right )+36 a \sqrt {d} f^{3/2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )+\frac {9 b f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (2 \text {arcsinh}(c x)^2-\cosh (2 \text {arcsinh}(c x))+2 \text {arcsinh}(c x) \sinh (2 \text {arcsinh}(c x))\right )}{\sqrt {1+c^2 x^2}}+\frac {2 i b f \sqrt {d+i c d x} \sqrt {f-i c f x} \left (9 c x-3 \text {arcsinh}(c x) \left (3 \sqrt {1+c^2 x^2}+\cosh (3 \text {arcsinh}(c x))\right )+\sinh (3 \text {arcsinh}(c x))\right )}{\sqrt {1+c^2 x^2}}}{72 c} \]

[In]

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

[Out]

(12*a*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-2*I + 3*c*x - (2*I)*c^2*x^2) + 36*a*Sqrt[d]*f^(3/2)*Log[c*d*f*x
+ Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]] + (9*b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(2*ArcSinh
[c*x]^2 - Cosh[2*ArcSinh[c*x]] + 2*ArcSinh[c*x]*Sinh[2*ArcSinh[c*x]]))/Sqrt[1 + c^2*x^2] + ((2*I)*b*f*Sqrt[d +
I*c*d*x]*Sqrt[f - I*c*f*x]*(9*c*x - 3*ArcSinh[c*x]*(3*Sqrt[1 + c^2*x^2] + Cosh[3*ArcSinh[c*x]]) + Sinh[3*ArcS
inh[c*x]]))/Sqrt[1 + c^2*x^2])/(72*c)

Maple [F]

\[\int \left (-i c f x +f \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right ) \sqrt {i c d x +d}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int { \sqrt {i \, c d x + d} {\left (-i \, c f x + f\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} \,d x } \]

[In]

integrate((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int \sqrt {i d \left (c x - i\right )} \left (- i f \left (c x + i\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )\, dx \]

[In]

integrate((f-I*c*f*x)**(3/2)*(a+b*asinh(c*x))*(d+I*c*d*x)**(1/2),x)

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [F(-2)]

Exception generated. \[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \]

[In]

integrate((f-I*c*f*x)^(3/2)*(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,sageVARx):;OUTP
UT:Error: Bad Argument TypeError: Bad Argument TypeError: Bad Argument TypeDone

Mupad [F(-1)]

Timed out. \[ \int \sqrt {d+i c d x} (f-i c f x)^{3/2} (a+b \text {arcsinh}(c x)) \, dx=\int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

[In]

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

[Out]

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