3.6.43 \(\int \frac {(f-i c f x)^{3/2} (a+b \sinh ^{-1}(c x))}{\sqrt {d+i c d x}} \, dx\) [543]

Optimal. Leaf size=266 \[ \frac {2 i b f^2 x \sqrt {1+c^2 x^2}}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {b c f^2 x^2 \sqrt {1+c^2 x^2}}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i f^2 \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {f^2 x \left (1+c^2 x^2\right ) \left (a+b \sinh ^{-1}(c x)\right )}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 f^2 \sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.31, antiderivative size = 266, 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, 5838, 5783, 5798, 8, 5812, 30} \begin {gather*} \frac {3 f^2 \sqrt {c^2 x^2+1} \left (a+b \sinh ^{-1}(c x)\right )^2}{4 b 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 \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 )}{c \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}}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b f^2 x \sqrt {c^2 x^2+1}}{\sqrt {d+i c d x} \sqrt {f-i c f x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*I)*b*f^2*x*Sqrt[1 + c^2*x^2])/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (b*c*f^2*x^2*Sqrt[1 + c^2*x^2])/(4*S
qrt[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]))/(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*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]) + (3*f^2*S
qrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/(4*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 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 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 5812

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

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

________________________________________________________________________________________

Mathematica [A]
time = 0.60, size = 344, normalized size = 1.29 \begin {gather*} \frac {16 i b c f x \sqrt {d+i c d x} \sqrt {f-i c f x}-16 i a f \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}-4 a c f x \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}-4 b f (4 i+c x) \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2} \sinh ^{-1}(c x)+6 b f \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh ^{-1}(c x)^2+b f \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh \left (2 \sinh ^{-1}(c x)\right )+12 a \sqrt {d} 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 )}{8 c d \sqrt {1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((16*I)*b*c*f*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (16*I)*a*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 +
c^2*x^2] - 4*a*c*f*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - 4*b*f*(4*I + c*x)*Sqrt[d + I*c*d*
x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2]*ArcSinh[c*x] + 6*b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2
+ b*f*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] + 12*a*Sqrt[d]*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]])/(8*c*d*Sqrt[1 + c^2*x^2])

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________