3.579 \(\int \frac {(f-i c f x)^{3/2} (a+b \sinh ^{-1}(c x))^2}{\sqrt {d+i c d x}} \, dx\)

Optimal. Leaf size=436 \[ \frac {f^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i f^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {f^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {b c f^2 x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {4 i b f^2 x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {4 i b^2 f^2 \left (c^2 x^2+1\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {b^2 f^2 x \left (c^2 x^2+1\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {b^2 f^2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[Out]

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

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Rubi [A]  time = 0.63, antiderivative size = 436, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.243, Rules used = {5712, 5831, 3317, 3296, 2638, 3311, 32, 2635, 8} \[ \frac {f^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^3}{2 b c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i f^2 \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {f^2 x \left (c^2 x^2+1\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {b c f^2 x^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {4 i b f^2 x \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )}{\sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {4 i b^2 f^2 \left (c^2 x^2+1\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {b^2 f^2 x \left (c^2 x^2+1\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {b^2 f^2 \sqrt {c^2 x^2+1} \sinh ^{-1}(c x)}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 3317

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] ||
IGtQ[m, 0] || NeQ[a^2 - b^2, 0])

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]

Rule 5831

Int[(((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol]
:> Dist[1/(c^(m + 1)*Sqrt[d]), Subst[Int[(a + b*x)^n*(c*f + g*Sinh[x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{
a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && GtQ[d, 0] && (GtQ[m, 0] || IGtQ[n, 0])

Rubi steps

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

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Mathematica [A]  time = 2.26, size = 532, normalized size = 1.22 \[ \frac {12 a^2 \sqrt {d} f^{3/2} \sqrt {c^2 x^2+1} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )-16 i a^2 f \sqrt {c^2 x^2+1} \sqrt {d+i c d x} \sqrt {f-i c f x}-4 a^2 c f x \sqrt {c^2 x^2+1} \sqrt {d+i c d x} \sqrt {f-i c f x}+2 b f \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2 \left (6 a-8 i b \sqrt {c^2 x^2+1}-b \sinh \left (2 \sinh ^{-1}(c x)\right )\right )+2 b f \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x) \left (b \cosh \left (2 \sinh ^{-1}(c x)\right )+4 i \left (4 b c x+a (-4+i c x) \sqrt {c^2 x^2+1}\right )\right )+32 i a b c f x \sqrt {d+i c d x} \sqrt {f-i c f x}+2 a b f \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )-32 i b^2 f \sqrt {c^2 x^2+1} \sqrt {d+i c d x} \sqrt {f-i c f x}+4 b^2 f \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^3-b^2 f \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh \left (2 \sinh ^{-1}(c x)\right )}{8 c d \sqrt {c^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((32*I)*a*b*c*f*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (16*I)*a^2*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[
1 + c^2*x^2] - (32*I)*b^2*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - 4*a^2*c*f*x*Sqrt[d + I*c*d
*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 4*b^2*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^3 + 2*a*b*f
*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] + 2*b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[
c*x]*((4*I)*(4*b*c*x + a*(-4 + I*c*x)*Sqrt[1 + c^2*x^2]) + b*Cosh[2*ArcSinh[c*x]]) + 12*a^2*Sqrt[d]*f^(3/2)*Sq
rt[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]] - b^2*f*Sqrt[d + I*c*d*x]*S
qrt[f - I*c*f*x]*Sinh[2*ArcSinh[c*x]] + 2*b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2*(6*a - (8*I)*
b*Sqrt[1 + c^2*x^2] - b*Sinh[2*ArcSinh[c*x]]))/(8*c*d*Sqrt[1 + c^2*x^2])

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (b^{2} c f x + i \, b^{2} f\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )^{2} + 2 \, {\left (a b c f x + i \, a b f\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (a^{2} c f x + i \, a^{2} f\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{c d x - i \, d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-i \, c f x + f\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt {i \, c d x + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.35, size = 0, normalized size = 0.00 \[ \int \frac {\left (-i c f x +f \right )^{\frac {3}{2}} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\sqrt {i c d x +d}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f-I*c*f*x)^(3/2)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(1/2),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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}}{\sqrt {d+c\,d\,x\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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