3.1079 \(\int (a+b x^4)^{7/4} \, dx\)

Optimal. Leaf size=96 \[ \frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}+\frac {7}{32} a x \left (a+b x^4\right )^{3/4}+\frac {1}{8} x \left (a+b x^4\right )^{7/4} \]

[Out]

7/32*a*x*(b*x^4+a)^(3/4)+1/8*x*(b*x^4+a)^(7/4)+21/64*a^2*arctan(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)+21/64*a^2*a
rctanh(b^(1/4)*x/(b*x^4+a)^(1/4))/b^(1/4)

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Rubi [A]  time = 0.03, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {195, 240, 212, 206, 203} \[ \frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}+\frac {7}{32} a x \left (a+b x^4\right )^{3/4}+\frac {1}{8} x \left (a+b x^4\right )^{7/4} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^4)^(7/4),x]

[Out]

(7*a*x*(a + b*x^4)^(3/4))/32 + (x*(a + b*x^4)^(7/4))/8 + (21*a^2*ArcTan[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^
(1/4)) + (21*a^2*ArcTanh[(b^(1/4)*x)/(a + b*x^4)^(1/4)])/(64*b^(1/4))

Rule 195

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 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]

Rubi steps

\begin {align*} \int \left (a+b x^4\right )^{7/4} \, dx &=\frac {1}{8} x \left (a+b x^4\right )^{7/4}+\frac {1}{8} (7 a) \int \left (a+b x^4\right )^{3/4} \, dx\\ &=\frac {7}{32} a x \left (a+b x^4\right )^{3/4}+\frac {1}{8} x \left (a+b x^4\right )^{7/4}+\frac {1}{32} \left (21 a^2\right ) \int \frac {1}{\sqrt [4]{a+b x^4}} \, dx\\ &=\frac {7}{32} a x \left (a+b x^4\right )^{3/4}+\frac {1}{8} x \left (a+b x^4\right )^{7/4}+\frac {1}{32} \left (21 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^4} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac {7}{32} a x \left (a+b x^4\right )^{3/4}+\frac {1}{8} x \left (a+b x^4\right )^{7/4}+\frac {1}{64} \left (21 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )+\frac {1}{64} \left (21 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {b} x^2} \, dx,x,\frac {x}{\sqrt [4]{a+b x^4}}\right )\\ &=\frac {7}{32} a x \left (a+b x^4\right )^{3/4}+\frac {1}{8} x \left (a+b x^4\right )^{7/4}+\frac {21 a^2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}+\frac {21 a^2 \tanh ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a+b x^4}}\right )}{64 \sqrt [4]{b}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 47, normalized size = 0.49 \[ \frac {a x \left (a+b x^4\right )^{3/4} \, _2F_1\left (-\frac {7}{4},\frac {1}{4};\frac {5}{4};-\frac {b x^4}{a}\right )}{\left (\frac {b x^4}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^4)^(7/4),x]

[Out]

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

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fricas [B]  time = 0.91, size = 202, normalized size = 2.10 \[ \frac {1}{32} \, {\left (4 \, b x^{5} + 11 \, a x\right )} {\left (b x^{4} + a\right )}^{\frac {3}{4}} + \frac {21}{32} \, \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} \arctan \left (-\frac {\left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} {\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} - \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} x \sqrt {\frac {\sqrt {b x^{4} + a} a^{12} + \sqrt {\frac {a^{8}}{b}} a^{8} b x^{2}}{x^{2}}}}{a^{8} x}\right ) + \frac {21}{128} \, \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} \log \left (\frac {9261 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} + \left (\frac {a^{8}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) - \frac {21}{128} \, \left (\frac {a^{8}}{b}\right )^{\frac {1}{4}} \log \left (\frac {9261 \, {\left ({\left (b x^{4} + a\right )}^{\frac {1}{4}} a^{6} - \left (\frac {a^{8}}{b}\right )^{\frac {3}{4}} b x\right )}}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(7/4),x, algorithm="fricas")

[Out]

1/32*(4*b*x^5 + 11*a*x)*(b*x^4 + a)^(3/4) + 21/32*(a^8/b)^(1/4)*arctan(-((a^8/b)^(1/4)*(b*x^4 + a)^(1/4)*a^6 -
(a^8/b)^(1/4)*x*sqrt((sqrt(b*x^4 + a)*a^12 + sqrt(a^8/b)*a^8*b*x^2)/x^2))/(a^8*x)) + 21/128*(a^8/b)^(1/4)*log
(9261*((b*x^4 + a)^(1/4)*a^6 + (a^8/b)^(3/4)*b*x)/x) - 21/128*(a^8/b)^(1/4)*log(9261*((b*x^4 + a)^(1/4)*a^6 -
(a^8/b)^(3/4)*b*x)/x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{4} + a\right )}^{\frac {7}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(7/4),x, algorithm="giac")

[Out]

integrate((b*x^4 + a)^(7/4), x)

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maple [F]  time = 0.16, size = 0, normalized size = 0.00 \[ \int \left (b \,x^{4}+a \right )^{\frac {7}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^4+a)^(7/4),x)

[Out]

int((b*x^4+a)^(7/4),x)

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maxima [A]  time = 3.03, size = 143, normalized size = 1.49 \[ -\frac {21}{128} \, a^{2} {\left (\frac {2 \, \arctan \left (\frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{b^{\frac {1}{4}} x}\right )}{b^{\frac {1}{4}}} + \frac {\log \left (-\frac {b^{\frac {1}{4}} - \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}{b^{\frac {1}{4}} + \frac {{\left (b x^{4} + a\right )}^{\frac {1}{4}}}{x}}\right )}{b^{\frac {1}{4}}}\right )} - \frac {\frac {7 \, {\left (b x^{4} + a\right )}^{\frac {3}{4}} a^{2} b}{x^{3}} - \frac {11 \, {\left (b x^{4} + a\right )}^{\frac {7}{4}} a^{2}}{x^{7}}}{32 \, {\left (b^{2} - \frac {2 \, {\left (b x^{4} + a\right )} b}{x^{4}} + \frac {{\left (b x^{4} + a\right )}^{2}}{x^{8}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^4+a)^(7/4),x, algorithm="maxima")

[Out]

-21/128*a^2*(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)) - 1/32*(7*(b*x^4 + a)^(3/4)*a^2*b/x^3 - 11*(b*x^4 + a)^(7/4)*a^2/x^7)/(b^2 - 2
*(b*x^4 + a)*b/x^4 + (b*x^4 + a)^2/x^8)

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mupad [B]  time = 1.06, size = 37, normalized size = 0.39 \[ \frac {x\,{\left (b\,x^4+a\right )}^{7/4}\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{4};\ \frac {5}{4};\ -\frac {b\,x^4}{a}\right )}{{\left (\frac {b\,x^4}{a}+1\right )}^{7/4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^4)^(7/4),x)

[Out]

(x*(a + b*x^4)^(7/4)*hypergeom([-7/4, 1/4], 5/4, -(b*x^4)/a))/((b*x^4)/a + 1)^(7/4)

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sympy [C]  time = 2.00, size = 37, normalized size = 0.39 \[ \frac {a^{\frac {7}{4}} x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**4+a)**(7/4),x)

[Out]

a**(7/4)*x*gamma(1/4)*hyper((-7/4, 1/4), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(5/4))

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