3.1031 \(\int x^{12} (a+b x^4)^{3/4} \, dx\)

Optimal. Leaf size=149 \[ -\frac {45 a^4 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}}-\frac {45 a^4 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}}+\frac {45 a^3 x \left (a+b x^4\right )^{3/4}}{2048 b^3}-\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b} \]

[Out]

45/2048*a^3*x*(b*x^4+a)^(3/4)/b^3-9/512*a^2*x^5*(b*x^4+a)^(3/4)/b^2+1/64*a*x^9*(b*x^4+a)^(3/4)/b+1/16*x^13*(b*
x^4+a)^(3/4)-45/4096*a^4*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(13/4)-45/4096*a^4*arctanh(b^(1/4)*x/(b*x^4+a)^(1
/4))/b^(13/4)

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Rubi [A]  time = 0.06, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {279, 321, 240, 212, 206, 203} \[ -\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {45 a^3 x \left (a+b x^4\right )^{3/4}}{2048 b^3}-\frac {45 a^4 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}}-\frac {45 a^4 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b} \]

Antiderivative was successfully verified.

[In]

Int[x^12*(a + b*x^4)^(3/4),x]

[Out]

(45*a^3*x*(a + b*x^4)^(3/4))/(2048*b^3) - (9*a^2*x^5*(a + b*x^4)^(3/4))/(512*b^2) + (a*x^9*(a + b*x^4)^(3/4))/
(64*b) + (x^13*(a + b*x^4)^(3/4))/16 - (45*a^4*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(4096*b^(13/4)) - (45*a^
4*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(4096*b^(13/4))

Rule 203

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

Rule 206

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

Rule 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

Rule 240

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

Rule 279

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

Rule 321

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

Rubi steps

\begin {align*} \int x^{12} \left (a+b x^4\right )^{3/4} \, dx &=\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}+\frac {1}{16} (3 a) \int \frac {x^{12}}{\sqrt [4]{a+b x^4}} \, dx\\ &=\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}-\frac {\left (9 a^2\right ) \int \frac {x^8}{\sqrt [4]{a+b x^4}} \, dx}{64 b}\\ &=-\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}+\frac {\left (45 a^3\right ) \int \frac {x^4}{\sqrt [4]{a+b x^4}} \, dx}{512 b^2}\\ &=\frac {45 a^3 x \left (a+b x^4\right )^{3/4}}{2048 b^3}-\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}-\frac {\left (45 a^4\right ) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx}{2048 b^3}\\ &=\frac {45 a^3 x \left (a+b x^4\right )^{3/4}}{2048 b^3}-\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}-\frac {\left (45 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{2048 b^3}\\ &=\frac {45 a^3 x \left (a+b x^4\right )^{3/4}}{2048 b^3}-\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}-\frac {\left (45 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^3}-\frac {\left (45 a^4\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^3}\\ &=\frac {45 a^3 x \left (a+b x^4\right )^{3/4}}{2048 b^3}-\frac {9 a^2 x^5 \left (a+b x^4\right )^{3/4}}{512 b^2}+\frac {a x^9 \left (a+b x^4\right )^{3/4}}{64 b}+\frac {1}{16} x^{13} \left (a+b x^4\right )^{3/4}-\frac {45 a^4 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}}-\frac {45 a^4 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{13/4}}\\ \end {align*}

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Mathematica [C]  time = 0.10, size = 105, normalized size = 0.70 \[ \frac {x \left (a+b x^4\right )^{3/4} \left (\left (\frac {b x^4}{a}+1\right )^{3/4} \left (15 a^3-9 a^2 b x^4+8 a b^2 x^8+32 b^3 x^{12}\right )-15 a^3 \, _2F_1\left (-\frac {3}{4},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )\right )}{512 b^3 \left (\frac {b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^12*(a + b*x^4)^(3/4),x]

[Out]

(x*(a + b*x^4)^(3/4)*((1 + (b*x^4)/a)^(3/4)*(15*a^3 - 9*a^2*b*x^4 + 8*a*b^2*x^8 + 32*b^3*x^12) - 15*a^3*Hyperg
eometric2F1[-3/4, 1/4, 5/4, -((b*x^4)/a)]))/(512*b^3*(1 + (b*x^4)/a)^(3/4))

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fricas [B]  time = 0.90, size = 250, normalized size = 1.68 \[ -\frac {180 \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} \arctan \left (-\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}} \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} a^{12} b^{3} - \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} x \sqrt {\frac {\sqrt {\frac {a^{16}}{b^{13}}} a^{16} b^{7} x^{2} + \sqrt {b x^{4} + a} a^{24}}{x^{2}}}}{a^{16} x}\right ) + 45 \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} \log \left (\frac {91125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{12} + \left (\frac {a^{16}}{b^{13}}\right )^{\frac {3}{4}} b^{10} x\right )}}{x}\right ) - 45 \, \left (\frac {a^{16}}{b^{13}}\right )^{\frac {1}{4}} b^{3} \log \left (\frac {91125 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{12} - \left (\frac {a^{16}}{b^{13}}\right )^{\frac {3}{4}} b^{10} x\right )}}{x}\right ) - 4 \, {\left (128 \, b^{3} x^{13} + 32 \, a b^{2} x^{9} - 36 \, a^{2} b x^{5} + 45 \, a^{3} x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}}}{8192 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b*x^4+a)^(3/4),x, algorithm="fricas")

[Out]

-1/8192*(180*(a^16/b^13)^(1/4)*b^3*arctan(-((b*x^4 + a)^(1/4)*(a^16/b^13)^(1/4)*a^12*b^3 - (a^16/b^13)^(1/4)*b
^3*x*sqrt((sqrt(a^16/b^13)*a^16*b^7*x^2 + sqrt(b*x^4 + a)*a^24)/x^2))/(a^16*x)) + 45*(a^16/b^13)^(1/4)*b^3*log
(91125*((b*x^4 + a)^(1/4)*a^12 + (a^16/b^13)^(3/4)*b^10*x)/x) - 45*(a^16/b^13)^(1/4)*b^3*log(91125*((b*x^4 + a
)^(1/4)*a^12 - (a^16/b^13)^(3/4)*b^10*x)/x) - 4*(128*b^3*x^13 + 32*a*b^2*x^9 - 36*a^2*b*x^5 + 45*a^3*x)*(b*x^4
+ a)^(3/4))/b^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{4} + a\right )}^{\frac {3}{4}} x^{12}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b*x^4+a)^(3/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(3/4)*x^12, x)

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{4}+a \right )^{\frac {3}{4}} x^{12}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12*(b*x^4+a)^(3/4),x)

[Out]

int(x^12*(b*x^4+a)^(3/4),x)

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maxima [A]  time = 3.08, size = 225, normalized size = 1.51 \[ \frac {45 \, a^{4} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )}}{8192 \, b^{3}} + \frac {\frac {15 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{4} b^{3}}{x^{3}} + \frac {239 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{4} b^{2}}{x^{7}} - \frac {171 \, {\left (b x^{4} + a\right )}^{\frac {11}{4}} a^{4} b}{x^{11}} + \frac {45 \, {\left (b x^{4} + a\right )}^{\frac {15}{4}} a^{4}}{x^{15}}}{2048 \, {\left (b^{7} - \frac {4 \, {\left (b x^{4} + a\right )} b^{6}}{x^{4}} + \frac {6 \, {\left (b x^{4} + a\right )}^{2} b^{5}}{x^{8}} - \frac {4 \, {\left (b x^{4} + a\right )}^{3} b^{4}}{x^{12}} + \frac {{\left (b x^{4} + a\right )}^{4} b^{3}}{x^{16}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(b*x^4+a)^(3/4),x, algorithm="maxima")

[Out]

45/8192*a^4*(2*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(1/4) + log(-(b^(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) +
(b*x^4 + a)^(1/4)/x))/b^(1/4))/b^3 + 1/2048*(15*(b*x^4 + a)^(3/4)*a^4*b^3/x^3 + 239*(b*x^4 + a)^(7/4)*a^4*b^2
/x^7 - 171*(b*x^4 + a)^(11/4)*a^4*b/x^11 + 45*(b*x^4 + a)^(15/4)*a^4/x^15)/(b^7 - 4*(b*x^4 + a)*b^6/x^4 + 6*(b
*x^4 + a)^2*b^5/x^8 - 4*(b*x^4 + a)^3*b^4/x^12 + (b*x^4 + a)^4*b^3/x^16)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^{12}\,{\left (b\,x^4+a\right )}^{3/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12*(a + b*x^4)^(3/4),x)

[Out]

int(x^12*(a + b*x^4)^(3/4), x)

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sympy [C]  time = 10.40, size = 39, normalized size = 0.26 \[ \frac {a^{\frac {3}{4}} x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {17}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**12*(b*x**4+a)**(3/4),x)

[Out]

a**(3/4)*x**13*gamma(13/4)*hyper((-3/4, 13/4), (17/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(17/4))

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