3.1062 \(\int x^{10} (a+b x^4)^{5/4} \, dx\)

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

[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))

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Rubi [A]  time = 0.0641264, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {279, 321, 331, 298, 203, 206} \[ -\frac{35 a^3 x^3 \sqrt [4]{a+b x^4}}{6144 b^2}-\frac{35 a^4 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}+\frac{35 a^4 \tanh ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{4096 b^{11/4}}+\frac{5 a^2 x^7 \sqrt [4]{a+b x^4}}{1536 b}+\frac{1}{16} x^{11} \left (a+b x^4\right )^{5/4}+\frac{5}{192} a x^{11} \sqrt [4]{a+b x^4} \]

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 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]

Rule 331

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]

Rule 298

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 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])

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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}

Mathematica [C]  time = 0.0717602, size = 81, normalized size = 0.55 \[ \frac{x^3 \sqrt [4]{a+b x^4} \left (\frac{7 a^3 \, _2F_1\left (-\frac{5}{4},\frac{3}{4};\frac{7}{4};-\frac{b x^4}{a}\right )}{\sqrt [4]{\frac{b x^4}{a}+1}}-\left (7 a-12 b x^4\right ) \left (a+b x^4\right )^2\right )}{192 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10*(a + b*x^4)^(5/4),x]

[Out]

(x^3*(a + b*x^4)^(1/4)*(-((7*a - 12*b*x^4)*(a + b*x^4)^2) + (7*a^3*Hypergeometric2F1[-5/4, 3/4, 7/4, -((b*x^4)
/a)])/(1 + (b*x^4)/a)^(1/4)))/(192*b^2)

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Maple [F]  time = 0.028, size = 0, normalized size = 0. \begin{align*} \int{x}^{10} \left ( b{x}^{4}+a \right ) ^{{\frac{5}{4}}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.64754, size = 605, normalized size = 4.09 \begin{align*} -\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{align*}

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]  time = 9.25602, size = 39, normalized size = 0.26 \begin{align*} \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{align*}

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 [B]  time = 1.16938, size = 459, normalized size = 3.1 \begin{align*} \frac{1}{49152} \,{\left (\frac{8 \,{\left (\frac{399 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}{\left (b + \frac{a}{x^{4}}\right )} b^{2}}{x} - \frac{105 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}} b^{3}}{x} + \frac{125 \,{\left (b^{2} x^{8} + 2 \, a b x^{4} + a^{2}\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}} b}{x^{9}} - \frac{35 \,{\left (b^{3} x^{12} + 3 \, a b^{2} x^{8} + 3 \, a^{2} b x^{4} + a^{3}\right )}{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x^{13}}\right )} x^{16}}{a^{4} b^{2}} + \frac{210 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} + \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b^{3}} + \frac{210 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{2} \left (-b\right )^{\frac{1}{4}} - \frac{2 \,{\left (b x^{4} + a\right )}^{\frac{1}{4}}}{x}\right )}}{2 \, \left (-b\right )^{\frac{1}{4}}}\right )}{b^{3}} + \frac{105 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} + \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b^{3}} - \frac{105 \, \sqrt{2} \left (-b\right )^{\frac{1}{4}} \log \left (\sqrt{-b} - \frac{\sqrt{2}{\left (b x^{4} + a\right )}^{\frac{1}{4}} \left (-b\right )^{\frac{1}{4}}}{x} + \frac{\sqrt{b x^{4} + a}}{x^{2}}\right )}{b^{3}}\right )} a^{4} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/49152*(8*(399*(b*x^4 + a)^(1/4)*(b + a/x^4)*b^2/x - 105*(b*x^4 + a)^(1/4)*b^3/x + 125*(b^2*x^8 + 2*a*b*x^4 +
a^2)*(b*x^4 + a)^(1/4)*b/x^9 - 35*(b^3*x^12 + 3*a*b^2*x^8 + 3*a^2*b*x^4 + a^3)*(b*x^4 + a)^(1/4)/x^13)*x^16/(
a^4*b^2) + 210*sqrt(2)*(-b)^(1/4)*arctan(1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) + 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/
b^3 + 210*sqrt(2)*(-b)^(1/4)*arctan(-1/2*sqrt(2)*(sqrt(2)*(-b)^(1/4) - 2*(b*x^4 + a)^(1/4)/x)/(-b)^(1/4))/b^3
+ 105*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) + sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b^3 - 10
5*sqrt(2)*(-b)^(1/4)*log(sqrt(-b) - sqrt(2)*(b*x^4 + a)^(1/4)*(-b)^(1/4)/x + sqrt(b*x^4 + a)/x^2)/b^3)*a^4