3.8.87 \(\int \frac {(a+c x^4)^{3/2}}{x^5} \, dx\) [787]

Optimal. Leaf size=63 \[ \frac {3}{4} c \sqrt {a+c x^4}-\frac {\left (a+c x^4\right )^{3/2}}{4 x^4}-\frac {3}{4} \sqrt {a} c \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right ) \]

[Out]

-1/4*(c*x^4+a)^(3/2)/x^4-3/4*c*arctanh((c*x^4+a)^(1/2)/a^(1/2))*a^(1/2)+3/4*c*(c*x^4+a)^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {272, 43, 52, 65, 214} \begin {gather*} -\frac {\left (a+c x^4\right )^{3/2}}{4 x^4}+\frac {3}{4} c \sqrt {a+c x^4}-\frac {3}{4} \sqrt {a} c \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x^5,x]

[Out]

(3*c*Sqrt[a + c*x^4])/4 - (a + c*x^4)^(3/2)/(4*x^4) - (3*Sqrt[a]*c*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/4

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + 1))), x] - Dist[d*(n/(b*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \frac {\left (a+c x^4\right )^{3/2}}{x^5} \, dx &=\frac {1}{4} \text {Subst}\left (\int \frac {(a+c x)^{3/2}}{x^2} \, dx,x,x^4\right )\\ &=-\frac {\left (a+c x^4\right )^{3/2}}{4 x^4}+\frac {1}{8} (3 c) \text {Subst}\left (\int \frac {\sqrt {a+c x}}{x} \, dx,x,x^4\right )\\ &=\frac {3}{4} c \sqrt {a+c x^4}-\frac {\left (a+c x^4\right )^{3/2}}{4 x^4}+\frac {1}{8} (3 a c) \text {Subst}\left (\int \frac {1}{x \sqrt {a+c x}} \, dx,x,x^4\right )\\ &=\frac {3}{4} c \sqrt {a+c x^4}-\frac {\left (a+c x^4\right )^{3/2}}{4 x^4}+\frac {1}{4} (3 a) \text {Subst}\left (\int \frac {1}{-\frac {a}{c}+\frac {x^2}{c}} \, dx,x,\sqrt {a+c x^4}\right )\\ &=\frac {3}{4} c \sqrt {a+c x^4}-\frac {\left (a+c x^4\right )^{3/2}}{4 x^4}-\frac {3}{4} \sqrt {a} c \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.08, size = 57, normalized size = 0.90 \begin {gather*} \frac {\sqrt {a+c x^4} \left (-a+2 c x^4\right )}{4 x^4}-\frac {3}{4} \sqrt {a} c \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {a}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x^5,x]

[Out]

(Sqrt[a + c*x^4]*(-a + 2*c*x^4))/(4*x^4) - (3*Sqrt[a]*c*ArcTanh[Sqrt[a + c*x^4]/Sqrt[a]])/4

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Maple [A]
time = 0.16, size = 58, normalized size = 0.92

method	result	size
default	\(\frac {c \sqrt {x^{4} c +a}}{2}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{4}-\frac {a \sqrt {x^{4} c +a}}{4 x^{4}}\)	\(58\)
risch	\(\frac {c \sqrt {x^{4} c +a}}{2}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{4}-\frac {a \sqrt {x^{4} c +a}}{4 x^{4}}\)	\(58\)
elliptic	\(\frac {c \sqrt {x^{4} c +a}}{2}-\frac {3 \sqrt {a}\, c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {x^{4} c +a}}{x^{2}}\right )}{4}-\frac {a \sqrt {x^{4} c +a}}{4 x^{4}}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^4+a)^(3/2)/x^5,x,method=_RETURNVERBOSE)

[Out]

1/2*c*(c*x^4+a)^(1/2)-3/4*a^(1/2)*c*ln((2*a+2*a^(1/2)*(c*x^4+a)^(1/2))/x^2)-1/4*a/x^4*(c*x^4+a)^(1/2)

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Maxima [A]
time = 0.50, size = 66, normalized size = 1.05 \begin {gather*} \frac {3}{8} \, \sqrt {a} c \log \left (\frac {\sqrt {c x^{4} + a} - \sqrt {a}}{\sqrt {c x^{4} + a} + \sqrt {a}}\right ) + \frac {1}{2} \, \sqrt {c x^{4} + a} c - \frac {\sqrt {c x^{4} + a} a}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^5,x, algorithm="maxima")

[Out]

3/8*sqrt(a)*c*log((sqrt(c*x^4 + a) - sqrt(a))/(sqrt(c*x^4 + a) + sqrt(a))) + 1/2*sqrt(c*x^4 + a)*c - 1/4*sqrt(
c*x^4 + a)*a/x^4

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Fricas [A]
time = 0.36, size = 121, normalized size = 1.92 \begin {gather*} \left [\frac {3 \, \sqrt {a} c x^{4} \log \left (\frac {c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {a} + 2 \, a}{x^{4}}\right ) + 2 \, {\left (2 \, c x^{4} - a\right )} \sqrt {c x^{4} + a}}{8 \, x^{4}}, \frac {3 \, \sqrt {-a} c x^{4} \arctan \left (\frac {\sqrt {c x^{4} + a} \sqrt {-a}}{a}\right ) + {\left (2 \, c x^{4} - a\right )} \sqrt {c x^{4} + a}}{4 \, x^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^5,x, algorithm="fricas")

[Out]

[1/8*(3*sqrt(a)*c*x^4*log((c*x^4 - 2*sqrt(c*x^4 + a)*sqrt(a) + 2*a)/x^4) + 2*(2*c*x^4 - a)*sqrt(c*x^4 + a))/x^
4, 1/4*(3*sqrt(-a)*c*x^4*arctan(sqrt(c*x^4 + a)*sqrt(-a)/a) + (2*c*x^4 - a)*sqrt(c*x^4 + a))/x^4]

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Sympy [A]
time = 1.32, size = 95, normalized size = 1.51 \begin {gather*} - \frac {3 \sqrt {a} c \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {c} x^{2}} \right )}}{4} - \frac {a^{2}}{4 \sqrt {c} x^{6} \sqrt {\frac {a}{c x^{4}} + 1}} + \frac {a \sqrt {c}}{4 x^{2} \sqrt {\frac {a}{c x^{4}} + 1}} + \frac {c^{\frac {3}{2}} x^{2}}{2 \sqrt {\frac {a}{c x^{4}} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**4+a)**(3/2)/x**5,x)

[Out]

-3*sqrt(a)*c*asinh(sqrt(a)/(sqrt(c)*x**2))/4 - a**2/(4*sqrt(c)*x**6*sqrt(a/(c*x**4) + 1)) + a*sqrt(c)/(4*x**2*
sqrt(a/(c*x**4) + 1)) + c**(3/2)*x**2/(2*sqrt(a/(c*x**4) + 1))

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Giac [A]
time = 1.73, size = 63, normalized size = 1.00 \begin {gather*} \frac {\frac {3 \, a c^{2} \arctan \left (\frac {\sqrt {c x^{4} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + 2 \, \sqrt {c x^{4} + a} c^{2} - \frac {\sqrt {c x^{4} + a} a c}{x^{4}}}{4 \, c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^4+a)^(3/2)/x^5,x, algorithm="giac")

[Out]

1/4*(3*a*c^2*arctan(sqrt(c*x^4 + a)/sqrt(-a))/sqrt(-a) + 2*sqrt(c*x^4 + a)*c^2 - sqrt(c*x^4 + a)*a*c/x^4)/c

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Mupad [B]
time = 1.36, size = 48, normalized size = 0.76 \begin {gather*} \frac {c\,\sqrt {c\,x^4+a}}{2}-\frac {a\,\sqrt {c\,x^4+a}}{4\,x^4}-\frac {3\,\sqrt {a}\,c\,\mathrm {atanh}\left (\frac {\sqrt {c\,x^4+a}}{\sqrt {a}}\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(3/2)/x^5,x)

[Out]

(c*(a + c*x^4)^(1/2))/2 - (a*(a + c*x^4)^(1/2))/(4*x^4) - (3*a^(1/2)*c*atanh((a + c*x^4)^(1/2)/a^(1/2)))/4

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