3.8.91 \(\int \frac {(a+c x^4)^{3/2}}{x^3} \, dx\) [791]

Optimal. Leaf size=69 \[ \frac {3}{4} c x^2 \sqrt {a+c x^4}-\frac {\left (a+c x^4\right )^{3/2}}{2 x^2}+\frac {3}{4} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 69, 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 = {281, 283, 201, 223, 212} \begin {gather*} \frac {3}{4} c x^2 \sqrt {a+c x^4}-\frac {\left (a+c x^4\right )^{3/2}}{2 x^2}+\frac {3}{4} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + c*x^4)^(3/2)/x^3,x]

[Out]

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

Rule 201

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 283

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

Rubi steps

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

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Mathematica [A]
time = 0.17, size = 59, normalized size = 0.86 \begin {gather*} \frac {\left (-2 a+c x^4\right ) \sqrt {a+c x^4}}{4 x^2}+\frac {3}{4} a \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {a+c x^4}}{\sqrt {c} x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^4)^(3/2)/x^3,x]

[Out]

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

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Maple [A]
time = 0.14, size = 56, normalized size = 0.81

method	result	size
risch	\(-\frac {\sqrt {x^{4} c +a}\, \left (-x^{4} c +2 a \right )}{4 x^{2}}+\frac {3 a \sqrt {c}\, \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{4}\)	\(50\)
default	\(\frac {c \,x^{2} \sqrt {x^{4} c +a}}{4}+\frac {3 a \sqrt {c}\, \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{4}-\frac {a \sqrt {x^{4} c +a}}{2 x^{2}}\)	\(56\)
elliptic	\(\frac {c \,x^{2} \sqrt {x^{4} c +a}}{4}+\frac {3 a \sqrt {c}\, \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{4}-\frac {a \sqrt {x^{4} c +a}}{2 x^{2}}\)	\(56\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 94, normalized size = 1.36 \begin {gather*} -\frac {3}{8} \, a \sqrt {c} \log \left (-\frac {\sqrt {c} - \frac {\sqrt {c x^{4} + a}}{x^{2}}}{\sqrt {c} + \frac {\sqrt {c x^{4} + a}}{x^{2}}}\right ) - \frac {\sqrt {c x^{4} + a} a}{2 \, x^{2}} - \frac {\sqrt {c x^{4} + a} a c}{4 \, {\left (c - \frac {c x^{4} + a}{x^{4}}\right )} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.39, size = 120, normalized size = 1.74 \begin {gather*} \left [\frac {3 \, a \sqrt {c} x^{2} \log \left (-2 \, c x^{4} - 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) + 2 \, \sqrt {c x^{4} + a} {\left (c x^{4} - 2 \, a\right )}}{8 \, x^{2}}, -\frac {3 \, a \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {-c} x^{2}}{\sqrt {c x^{4} + a}}\right ) - \sqrt {c x^{4} + a} {\left (c x^{4} - 2 \, a\right )}}{4 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.26, size = 95, normalized size = 1.38 \begin {gather*} - \frac {a^{\frac {3}{2}}}{2 x^{2} \sqrt {1 + \frac {c x^{4}}{a}}} - \frac {\sqrt {a} c x^{2}}{4 \sqrt {1 + \frac {c x^{4}}{a}}} + \frac {3 a \sqrt {c} \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{4} + \frac {c^{2} x^{6}}{4 \sqrt {a} \sqrt {1 + \frac {c x^{4}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.16, size = 78, normalized size = 1.13 \begin {gather*} \frac {1}{4} \, \sqrt {c x^{4} + a} c x^{2} - \frac {3}{8} \, a \sqrt {c} \log \left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + a}\right )}^{2}\right ) + \frac {a^{2} \sqrt {c}}{{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + a}\right )}^{2} - a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,x^4+a\right )}^{3/2}}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + c*x^4)^(3/2)/x^3,x)

[Out]

int((a + c*x^4)^(3/2)/x^3, x)

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