\(\int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx\) [81]

Optimal result

Integrand size = 26, antiderivative size = 126 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {3 b x^2}{16 c^3 \sqrt {\pi }}-\frac {b x^4}{16 c \sqrt {\pi }}-\frac {3 x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{8 c^4 \pi }+\frac {x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{4 c^2 \pi }+\frac {3 (a+b \text {arcsinh}(c x))^2}{16 b c^5 \sqrt {\pi }} \]

[Out]

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

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {5812, 5783, 30} \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {3 (a+b \text {arcsinh}(c x))^2}{16 \sqrt {\pi } b c^5}+\frac {x^3 \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{4 \pi c^2}-\frac {3 x \sqrt {\pi c^2 x^2+\pi } (a+b \text {arcsinh}(c x))}{8 \pi c^4}+\frac {3 b x^2}{16 \sqrt {\pi } c^3}-\frac {b x^4}{16 \sqrt {\pi } c} \]

[In]

Int[(x^4*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

(3*b*x^2)/(16*c^3*Sqrt[Pi]) - (b*x^4)/(16*c*Sqrt[Pi]) - (3*x*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(8*c^
4*Pi) + (x^3*Sqrt[Pi + c^2*Pi*x^2]*(a + b*ArcSinh[c*x]))/(4*c^2*Pi) + (3*(a + b*ArcSinh[c*x])^2)/(16*b*c^5*Sqr
t[Pi])

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 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
]

Rubi steps \begin{align*} \text {integral}& = \frac {x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{4 c^2 \pi }-\frac {3 \int \frac {x^2 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx}{4 c^2}-\frac {b \int x^3 \, dx}{4 c \sqrt {\pi }} \\ & = -\frac {b x^4}{16 c \sqrt {\pi }}-\frac {3 x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{8 c^4 \pi }+\frac {x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{4 c^2 \pi }+\frac {3 \int \frac {a+b \text {arcsinh}(c x)}{\sqrt {\pi +c^2 \pi x^2}} \, dx}{8 c^4}+\frac {(3 b) \int x \, dx}{8 c^3 \sqrt {\pi }} \\ & = \frac {3 b x^2}{16 c^3 \sqrt {\pi }}-\frac {b x^4}{16 c \sqrt {\pi }}-\frac {3 x \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{8 c^4 \pi }+\frac {x^3 \sqrt {\pi +c^2 \pi x^2} (a+b \text {arcsinh}(c x))}{4 c^2 \pi }+\frac {3 (a+b \text {arcsinh}(c x))^2}{16 b c^5 \sqrt {\pi }} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {-48 a c x \sqrt {1+c^2 x^2}+32 a c^3 x^3 \sqrt {1+c^2 x^2}+24 b \text {arcsinh}(c x)^2+16 b \cosh (2 \text {arcsinh}(c x))-b \cosh (4 \text {arcsinh}(c x))+4 \text {arcsinh}(c x) (12 a-8 b \sinh (2 \text {arcsinh}(c x))+b \sinh (4 \text {arcsinh}(c x)))}{128 c^5 \sqrt {\pi }} \]

[In]

Integrate[(x^4*(a + b*ArcSinh[c*x]))/Sqrt[Pi + c^2*Pi*x^2],x]

[Out]

(-48*a*c*x*Sqrt[1 + c^2*x^2] + 32*a*c^3*x^3*Sqrt[1 + c^2*x^2] + 24*b*ArcSinh[c*x]^2 + 16*b*Cosh[2*ArcSinh[c*x]
] - b*Cosh[4*ArcSinh[c*x]] + 4*ArcSinh[c*x]*(12*a - 8*b*Sinh[2*ArcSinh[c*x]] + b*Sinh[4*ArcSinh[c*x]]))/(128*c
^5*Sqrt[Pi])

Maple [A] (verified)

Time = 0.16 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.31

method	result	size
default	\(\frac {a \,x^{3} \sqrt {\pi \,c^{2} x^{2}+\pi }}{4 \pi \,c^{2}}-\frac {3 a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{4} \pi }+\frac {3 a \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{4} \sqrt {\pi \,c^{2}}}+\frac {b \left (4 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}-c^{4} x^{4}-6 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+3 c^{2} x^{2}+3 \operatorname {arcsinh}\left (c x \right )^{2}+3\right )}{16 c^{5} \sqrt {\pi }}\)	\(165\)
parts	\(\frac {a \,x^{3} \sqrt {\pi \,c^{2} x^{2}+\pi }}{4 \pi \,c^{2}}-\frac {3 a x \sqrt {\pi \,c^{2} x^{2}+\pi }}{8 c^{4} \pi }+\frac {3 a \ln \left (\frac {\pi \,c^{2} x}{\sqrt {\pi \,c^{2}}}+\sqrt {\pi \,c^{2} x^{2}+\pi }\right )}{8 c^{4} \sqrt {\pi \,c^{2}}}+\frac {b \left (4 \,\operatorname {arcsinh}\left (c x \right ) \sqrt {c^{2} x^{2}+1}\, x^{3} c^{3}-c^{4} x^{4}-6 \,\operatorname {arcsinh}\left (c x \right ) c x \sqrt {c^{2} x^{2}+1}+3 c^{2} x^{2}+3 \operatorname {arcsinh}\left (c x \right )^{2}+3\right )}{16 c^{5} \sqrt {\pi }}\)	\(165\)

[In]

int(x^4*(a+b*arcsinh(c*x))/(Pi*c^2*x^2+Pi)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/4*a*x^3/Pi/c^2*(Pi*c^2*x^2+Pi)^(1/2)-3/8*a/c^4*x/Pi*(Pi*c^2*x^2+Pi)^(1/2)+3/8*a/c^4*ln(Pi*c^2*x/(Pi*c^2)^(1/
2)+(Pi*c^2*x^2+Pi)^(1/2))/(Pi*c^2)^(1/2)+1/16*b*(4*arcsinh(c*x)*(c^2*x^2+1)^(1/2)*x^3*c^3-c^4*x^4-6*arcsinh(c*
x)*c*x*(c^2*x^2+1)^(1/2)+3*c^2*x^2+3*arcsinh(c*x)^2+3)/c^5/Pi^(1/2)

Fricas [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{\sqrt {\pi + \pi c^{2} x^{2}}} \,d x } \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="fricas")

[Out]

integral((b*x^4*arcsinh(c*x) + a*x^4)/sqrt(pi + pi*c^2*x^2), x)

Sympy [A] (verification not implemented)

Time = 2.08 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.47 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\frac {a \left (\begin {cases} \frac {x^{3} \sqrt {c^{2} x^{2} + 1}}{4 c^{2}} - \frac {3 x \sqrt {c^{2} x^{2} + 1}}{8 c^{4}} + \frac {3 \log {\left (2 c^{2} x + 2 \sqrt {c^{2} x^{2} + 1} \sqrt {c^{2}} \right )}}{8 c^{4} \sqrt {c^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x^{5}}{5} & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} + \frac {b \left (\begin {cases} - \frac {x^{4}}{16 c} + \frac {x^{3} \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{4 c^{2}} + \frac {3 x^{2}}{16 c^{3}} - \frac {3 x \sqrt {c^{2} x^{2} + 1} \operatorname {asinh}{\left (c x \right )}}{8 c^{4}} + \frac {3 \operatorname {asinh}^{2}{\left (c x \right )}}{16 c^{5}} & \text {for}\: c \neq 0 \\0 & \text {otherwise} \end {cases}\right )}{\sqrt {\pi }} \]

[In]

integrate(x**4*(a+b*asinh(c*x))/(pi*c**2*x**2+pi)**(1/2),x)

[Out]

a*Piecewise((x**3*sqrt(c**2*x**2 + 1)/(4*c**2) - 3*x*sqrt(c**2*x**2 + 1)/(8*c**4) + 3*log(2*c**2*x + 2*sqrt(c*
*2*x**2 + 1)*sqrt(c**2))/(8*c**4*sqrt(c**2)), Ne(c**2, 0)), (x**5/5, True))/sqrt(pi) + b*Piecewise((-x**4/(16*
c) + x**3*sqrt(c**2*x**2 + 1)*asinh(c*x)/(4*c**2) + 3*x**2/(16*c**3) - 3*x*sqrt(c**2*x**2 + 1)*asinh(c*x)/(8*c
**4) + 3*asinh(c*x)**2/(16*c**5), Ne(c, 0)), (0, True))/sqrt(pi)

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{\sqrt {\pi + \pi c^{2} x^{2}}} \,d x } \]

[In]

integrate(x^4*(a+b*arcsinh(c*x))/(pi*c^2*x^2+pi)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsinh(c*x) + a)*x^4/sqrt(pi + pi*c^2*x^2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{\sqrt {\pi +c^2 \pi x^2}} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {\Pi \,c^2\,x^2+\Pi }} \,d x \]

[In]

int((x^4*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2),x)

[Out]

int((x^4*(a + b*asinh(c*x)))/(Pi + Pi*c^2*x^2)^(1/2), x)