\(\int (1+b x^4)^p \, dx\) [185]

Optimal result

Integrand size = 9, antiderivative size = 18 \[ \int \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right ) \]

[Out]

x*hypergeom([1/4, -p],[5/4],-b*x^4)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {251} \[ \int \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right ) \]

[In]

Int[(1 + b*x^4)^p,x]

[Out]

x*Hypergeometric2F1[1/4, -p, 5/4, -(b*x^4)]

Rule 251

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)],
x] /; FreeQ[{a, b, n, p}, x] &&  !IGtQ[p, 0] &&  !IntegerQ[1/n] &&  !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p
] || GtQ[a, 0])

Rubi steps \begin{align*} \text {integral}& = x \, _2F_1\left (\frac {1}{4},-p;\frac {5}{4};-b x^4\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \left (1+b x^4\right )^p \, dx=x \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-b x^4\right ) \]

[In]

Integrate[(1 + b*x^4)^p,x]

[Out]

x*Hypergeometric2F1[1/4, -p, 5/4, -(b*x^4)]

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

method	result	size
meijerg	\(x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},-p ;\frac {5}{4};-b \,x^{4}\right )\)	\(17\)

[In]

int((b*x^4+1)^p,x,method=_RETURNVERBOSE)

[Out]

x*hypergeom([1/4,-p],[5/4],-b*x^4)

Fricas [F]

\[ \int \left (1+b x^4\right )^p \, dx=\int { {\left (b x^{4} + 1\right )}^{p} \,d x } \]

[In]

integrate((b*x^4+1)^p,x, algorithm="fricas")

[Out]

integral((b*x^4 + 1)^p, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.08 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.61 \[ \int \left (1+b x^4\right )^p \, dx=\frac {x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {b x^{4} e^{i \pi }} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} \]

[In]

integrate((b*x**4+1)**p,x)

[Out]

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

Maxima [F]

\[ \int \left (1+b x^4\right )^p \, dx=\int { {\left (b x^{4} + 1\right )}^{p} \,d x } \]

[In]

integrate((b*x^4+1)^p,x, algorithm="maxima")

[Out]

integrate((b*x^4 + 1)^p, x)

Giac [F]

\[ \int \left (1+b x^4\right )^p \, dx=\int { {\left (b x^{4} + 1\right )}^{p} \,d x } \]

[In]

integrate((b*x^4+1)^p,x, algorithm="giac")

[Out]

integrate((b*x^4 + 1)^p, x)

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \left (1+b x^4\right )^p \, dx=x\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{4},-p;\ \frac {5}{4};\ -b\,x^4\right ) \]

[In]

int((b*x^4 + 1)^p,x)

[Out]

x*hypergeom([1/4, -p], 5/4, -b*x^4)