3.11.75 \(\int \frac {(a+b x^4)^{5/4}}{x^4} \, dx\) [1075]

Optimal. Leaf size=99 \[ \frac {5}{6} b x \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3}-\frac {5 \sqrt {a} b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{6 \left (a+b x^4\right )^{3/4}} \]

[Out]

5/6*b*x*(b*x^4+a)^(1/4)-1/3*(b*x^4+a)^(5/4)/x^3-5/6*b^(3/2)*(1+a/b/x^4)^(3/4)*x^3*(cos(1/2*arccot(x^2*b^(1/2)/
a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x^2*b^(1/2)/a^(1/2))),2^(1/2)
)*a^(1/2)/(b*x^4+a)^(3/4)

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Rubi [A]
time = 0.03, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {283, 201, 243, 342, 281, 237} \begin {gather*} -\frac {5 \sqrt {a} b^{3/2} x^3 \left (\frac {a}{b x^4}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{6 \left (a+b x^4\right )^{3/4}}+\frac {5}{6} b x \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(5/4)/x^4,x]

[Out]

(5*b*x*(a + b*x^4)^(1/4))/6 - (a + b*x^4)^(5/4)/(3*x^3) - (5*Sqrt[a]*b^(3/2)*(1 + a/(b*x^4))^(3/4)*x^3*Ellipti
cF[ArcCot[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(6*(a + b*x^4)^(3/4))

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 237

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]))*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

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]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^4\right )^{5/4}}{x^4} \, dx &=-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3}+\frac {1}{3} (5 b) \int \sqrt [4]{a+b x^4} \, dx\\ &=\frac {5}{6} b x \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3}+\frac {1}{6} (5 a b) \int \frac {1}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac {5}{6} b x \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3}+\frac {\left (5 a b \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \int \frac {1}{\left (1+\frac {a}{b x^4}\right )^{3/4} x^3} \, dx}{6 \left (a+b x^4\right )^{3/4}}\\ &=\frac {5}{6} b x \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3}-\frac {\left (5 a b \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {x}{\left (1+\frac {a x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{x}\right )}{6 \left (a+b x^4\right )^{3/4}}\\ &=\frac {5}{6} b x \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3}-\frac {\left (5 a b \left (1+\frac {a}{b x^4}\right )^{3/4} x^3\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{x^2}\right )}{12 \left (a+b x^4\right )^{3/4}}\\ &=\frac {5}{6} b x \sqrt [4]{a+b x^4}-\frac {\left (a+b x^4\right )^{5/4}}{3 x^3}-\frac {5 \sqrt {a} b^{3/2} \left (1+\frac {a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{6 \left (a+b x^4\right )^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 52, normalized size = 0.53 \begin {gather*} -\frac {a \sqrt [4]{a+b x^4} \, _2F_1\left (-\frac {5}{4},-\frac {3}{4};\frac {1}{4};-\frac {b x^4}{a}\right )}{3 x^3 \sqrt [4]{1+\frac {b x^4}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(5/4)/x^4,x]

[Out]

-1/3*(a*(a + b*x^4)^(1/4)*Hypergeometric2F1[-5/4, -3/4, 1/4, -((b*x^4)/a)])/(x^3*(1 + (b*x^4)/a)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (b \,x^{4}+a \right )^{\frac {5}{4}}}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(5/4)/x^4,x)

[Out]

int((b*x^4+a)^(5/4)/x^4,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^4 + a)^(5/4)/x^4, x)

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Fricas [F]
time = 0.09, size = 15, normalized size = 0.15 \begin {gather*} {\rm integral}\left (\frac {{\left (b x^{4} + a\right )}^{\frac {5}{4}}}{x^{4}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x^4 + a)^(5/4)/x^4, x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.62, size = 42, normalized size = 0.42 \begin {gather*} \frac {a^{\frac {5}{4}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 x^{3} \Gamma \left (\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(5/4)/x**4,x)

[Out]

a**(5/4)*gamma(-3/4)*hyper((-5/4, -3/4), (1/4,), b*x**4*exp_polar(I*pi)/a)/(4*x**3*gamma(1/4))

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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((b*x^4+a)^(5/4)/x^4,x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)/x^4, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,x^4+a\right )}^{5/4}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(5/4)/x^4,x)

[Out]

int((a + b*x^4)^(5/4)/x^4, x)

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