3.11.62 \(\int x^{10} (a+b x^4)^{5/4} \, dx\) [1062]

Optimal. Leaf size=148 \[ -\frac {35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}+\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}-\frac {35 a^4 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}+\frac {35 a^4 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}} \]

[Out]

-35/6144*a^3*x^3*(b*x^4+a)^(1/4)/b^2+5/1536*a^2*x^7*(b*x^4+a)^(1/4)/b+5/192*a*x^11*(b*x^4+a)^(1/4)+1/16*x^11*(
b*x^4+a)^(5/4)-35/4096*a^4*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(11/4)+35/4096*a^4*arctanh(b^(1/4)*x/(b*x^4+a)^
(1/4))/b^(11/4)

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Rubi [A]
time = 0.04, antiderivative size = 148, 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 = {285, 327, 338, 304, 209, 212} \begin {gather*} -\frac {35 a^4 \text {ArcTan}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}+\frac {35 a^4 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}-\frac {35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}+\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^10*(a + b*x^4)^(5/4),x]

[Out]

(-35*a^3*x^3*(a + b*x^4)^(1/4))/(6144*b^2) + (5*a^2*x^7*(a + b*x^4)^(1/4))/(1536*b) + (5*a*x^11*(a + b*x^4)^(1
/4))/192 + (x^11*(a + b*x^4)^(5/4))/16 - (35*a^4*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(4096*b^(11/4)) + (35*
a^4*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(4096*b^(11/4))

Rule 209

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

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 285

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 304

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

Rule 327

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]

Rule 338

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

Rubi steps

\begin {align*} \int x^{10} \left (a+b x^4\right )^{5/4} \, dx &=\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}+\frac {1}{16} (5 a) \int x^{10} \sqrt [4]{a+b x^4} \, dx\\ &=\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}+\frac {1}{192} \left (5 a^2\right ) \int \frac {x^{10}}{\left (a+b x^4\right )^{3/4}} \, dx\\ &=\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}-\frac {\left (35 a^3\right ) \int \frac {x^6}{\left (a+b x^4\right )^{3/4}} \, dx}{1536 b}\\ &=-\frac {35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}+\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}+\frac {\left (35 a^4\right ) \int \frac {x^2}{\left (a+b x^4\right )^{3/4}} \, dx}{2048 b^2}\\ &=-\frac {35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}+\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}+\frac {\left (35 a^4\right ) \text {Subst}\left (\int \frac {x^2}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{2048 b^2}\\ &=-\frac {35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}+\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}+\frac {\left (35 a^4\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{5/2}}-\frac {\left (35 a^4\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{5/2}}\\ &=-\frac {35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}+\frac {5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac {5}{192} a x^{11} \sqrt [4]{a+b x^4}+\frac {1}{16} x^{11} \left (a+b x^4\right )^{5/4}-\frac {35 a^4 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}+\frac {35 a^4 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}\\ \end {align*}

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Mathematica [A]
time = 0.40, size = 111, normalized size = 0.75 \begin {gather*} \frac {2 b^{3/4} x^3 \sqrt [4]{a+b x^4} \left (-35 a^3+20 a^2 b x^4+544 a b^2 x^8+384 b^3 x^{12}\right )-105 a^4 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )+105 a^4 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{12288 b^{11/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^10*(a + b*x^4)^(5/4),x]

[Out]

(2*b^(3/4)*x^3*(a + b*x^4)^(1/4)*(-35*a^3 + 20*a^2*b*x^4 + 544*a*b^2*x^8 + 384*b^3*x^12) - 105*a^4*ArcTan[(b^(
1/4)*x)/(a + b*x^4)^(1/4)] + 105*a^4*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(12288*b^(11/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x^{10} \left (b \,x^{4}+a \right )^{\frac {5}{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10*(b*x^4+a)^(5/4),x)

[Out]

int(x^10*(b*x^4+a)^(5/4),x)

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Maxima [A]
time = 0.50, size = 229, normalized size = 1.55 \begin {gather*} -\frac {\frac {105 \, {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} b^{3}}{x} - \frac {399 \, {\left (b x^{4} + a\right )}^{\frac {5}{4}} a^{4} b^{2}}{x^{5}} - \frac {125 \, {\left (b x^{4} + a\right )}^{\frac {9}{4}} a^{4} b}{x^{9}} + \frac {35 \, {\left (b x^{4} + a\right )}^{\frac {13}{4}} a^{4}}{x^{13}}}{6144 \, {\left (b^{6} - \frac {4 \, {\left (b x^{4} + a\right )} b^{5}}{x^{4}} + \frac {6 \, {\left (b x^{4} + a\right )}^{2} b^{4}}{x^{8}} - \frac {4 \, {\left (b x^{4} + a\right )}^{3} b^{3}}{x^{12}} + \frac {{\left (b x^{4} + a\right )}^{4} b^{2}}{x^{16}}\right )}} + \frac {35 \, {\left (\frac {2 \, a^{4} \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {3}{4}}} - \frac {a^{4} \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 {3}{4}}}\right )}}{8192 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

-1/6144*(105*(b*x^4 + a)^(1/4)*a^4*b^3/x - 399*(b*x^4 + a)^(5/4)*a^4*b^2/x^5 - 125*(b*x^4 + a)^(9/4)*a^4*b/x^9
+ 35*(b*x^4 + a)^(13/4)*a^4/x^13)/(b^6 - 4*(b*x^4 + a)*b^5/x^4 + 6*(b*x^4 + a)^2*b^4/x^8 - 4*(b*x^4 + a)^3*b^
3/x^12 + (b*x^4 + a)^4*b^2/x^16) + 35/8192*(2*a^4*arctan((b*x^4 + a)^(1/4)/(b^(1/4)*x))/b^(3/4) - a^4*log(-(b^
(1/4) - (b*x^4 + a)^(1/4)/x)/(b^(1/4) + (b*x^4 + a)^(1/4)/x))/b^(3/4))/b^2

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 249 vs. \(2 (116) = 232\).
time = 0.40, size = 249, normalized size = 1.68 \begin {gather*} -\frac {420 \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{2} \arctan \left (-\frac {\left (\frac {a^{16}}{b^{11}}\right )^{\frac {3}{4}} {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} b^{8} - \left (\frac {a^{16}}{b^{11}}\right )^{\frac {3}{4}} b^{8} x \sqrt {\frac {\sqrt {b x^{4} + a} a^{8} + \sqrt {\frac {a^{16}}{b^{11}}} b^{6} x^{2}}{x^{2}}}}{a^{16} x}\right ) - 105 \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {35 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} + \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{3} x\right )}}{x}\right ) + 105 \, \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{2} \log \left (\frac {35 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{4} - \left (\frac {a^{16}}{b^{11}}\right )^{\frac {1}{4}} b^{3} x\right )}}{x}\right ) - 4 \, {\left (384 \, b^{3} x^{15} + 544 \, a b^{2} x^{11} + 20 \, a^{2} b x^{7} - 35 \, a^{3} x^{3}\right )} {\left (b x^{4} + a\right )}^{\frac {1}{4}}}{24576 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

-1/24576*(420*(a^16/b^11)^(1/4)*b^2*arctan(-((a^16/b^11)^(3/4)*(b*x^4 + a)^(1/4)*a^4*b^8 - (a^16/b^11)^(3/4)*b
^8*x*sqrt((sqrt(b*x^4 + a)*a^8 + sqrt(a^16/b^11)*b^6*x^2)/x^2))/(a^16*x)) - 105*(a^16/b^11)^(1/4)*b^2*log(35*(
(b*x^4 + a)^(1/4)*a^4 + (a^16/b^11)^(1/4)*b^3*x)/x) + 105*(a^16/b^11)^(1/4)*b^2*log(35*((b*x^4 + a)^(1/4)*a^4
- (a^16/b^11)^(1/4)*b^3*x)/x) - 4*(384*b^3*x^15 + 544*a*b^2*x^11 + 20*a^2*b*x^7 - 35*a^3*x^3)*(b*x^4 + a)^(1/4
))/b^2

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Sympy [C] Result contains complex when optimal does not.
time = 10.41, size = 39, normalized size = 0.26 \begin {gather*} \frac {a^{\frac {5}{4}} x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {15}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10*(b*x**4+a)**(5/4),x)

[Out]

a**(5/4)*x**11*gamma(11/4)*hyper((-5/4, 11/4), (15/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(15/4))

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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(x^10*(b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(5/4)*x^10, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{10}\,{\left (b\,x^4+a\right )}^{5/4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10*(a + b*x^4)^(5/4),x)

[Out]

int(x^10*(a + b*x^4)^(5/4), x)

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