3.540 \(\int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx\)

Optimal. Leaf size=143 \[ \frac{c x \left (a+b x^4\right )^{p+1} \, _2F_1\left (1,p+\frac{5}{4};\frac{5}{4};-\frac{b x^4}{a}\right )}{a}+\frac{d x^2 \left (a+b x^4\right )^{p+1} \, _2F_1\left (1,p+\frac{3}{2};\frac{3}{2};-\frac{b x^4}{a}\right )}{2 a}+\frac{e x^3 \left (a+b x^4\right )^{p+1} \, _2F_1\left (1,p+\frac{7}{4};\frac{7}{4};-\frac{b x^4}{a}\right )}{3 a}+\frac{f \left (a+b x^4\right )^{p+1}}{4 b (p+1)} \]

[Out]

(f*(a + b*x^4)^(1 + p))/(4*b*(1 + p)) + (c*x*(a + b*x^4)^(1 + p)*Hypergeometric2
F1[1, 5/4 + p, 5/4, -((b*x^4)/a)])/a + (d*x^2*(a + b*x^4)^(1 + p)*Hypergeometric
2F1[1, 3/2 + p, 3/2, -((b*x^4)/a)])/(2*a) + (e*x^3*(a + b*x^4)^(1 + p)*Hypergeom
etric2F1[1, 7/4 + p, 7/4, -((b*x^4)/a)])/(3*a)

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Rubi [A]  time = 0.275286, antiderivative size = 170, normalized size of antiderivative = 1.19, number of steps used = 12, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32 \[ c x \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+\frac{1}{2} d x^2 \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^4}{a}\right )+\frac{1}{3} e x^3 \left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )+\frac{f \left (a+b x^4\right )^{p+1}}{4 b (p+1)} \]

Antiderivative was successfully verified.

[In]  Int[(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^p,x]

[Out]

(f*(a + b*x^4)^(1 + p))/(4*b*(1 + p)) + (c*x*(a + b*x^4)^p*Hypergeometric2F1[1/4
, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (d*x^2*(a + b*x^4)^p*Hypergeometri
c2F1[1/2, -p, 3/2, -((b*x^4)/a)])/(2*(1 + (b*x^4)/a)^p) + (e*x^3*(a + b*x^4)^p*H
ypergeometric2F1[3/4, -p, 7/4, -((b*x^4)/a)])/(3*(1 + (b*x^4)/a)^p)

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Rubi in Sympy [A]  time = 36.2594, size = 136, normalized size = 0.95 \[ c x \left (1 + \frac{b x^{4}}{a}\right )^{- p} \left (a + b x^{4}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{- \frac{b x^{4}}{a}} \right )} + \frac{d x^{2} \left (1 + \frac{b x^{4}}{a}\right )^{- p} \left (a + b x^{4}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{1}{2} \\ \frac{3}{2} \end{matrix}\middle |{- \frac{b x^{4}}{a}} \right )}}{2} + \frac{e x^{3} \left (1 + \frac{b x^{4}}{a}\right )^{- p} \left (a + b x^{4}\right )^{p}{{}_{2}F_{1}\left (\begin{matrix} - p, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{- \frac{b x^{4}}{a}} \right )}}{3} + \frac{f \left (a + b x^{4}\right )^{p + 1}}{4 b \left (p + 1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**p,x)

[Out]

c*x*(1 + b*x**4/a)**(-p)*(a + b*x**4)**p*hyper((-p, 1/4), (5/4,), -b*x**4/a) + d
*x**2*(1 + b*x**4/a)**(-p)*(a + b*x**4)**p*hyper((-p, 1/2), (3/2,), -b*x**4/a)/2
+ e*x**3*(1 + b*x**4/a)**(-p)*(a + b*x**4)**p*hyper((-p, 3/4), (7/4,), -b*x**4/
a)/3 + f*(a + b*x**4)**(p + 1)/(4*b*(p + 1))

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Mathematica [A]  time = 0.162862, size = 184, normalized size = 1.29 \[ \frac{\left (a+b x^4\right )^p \left (\frac{b x^4}{a}+1\right )^{-p} \left (12 b c (p+1) x \, _2F_1\left (\frac{1}{4},-p;\frac{5}{4};-\frac{b x^4}{a}\right )+6 b d (p+1) x^2 \, _2F_1\left (\frac{1}{2},-p;\frac{3}{2};-\frac{b x^4}{a}\right )+4 b e x^3 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )+4 b e p x^3 \, _2F_1\left (\frac{3}{4},-p;\frac{7}{4};-\frac{b x^4}{a}\right )+3 b f x^4 \left (\frac{b x^4}{a}+1\right )^p+3 a f \left (\frac{b x^4}{a}+1\right )^p-3 a f\right )}{12 b (p+1)} \]

Antiderivative was successfully verified.

[In]  Integrate[(c + d*x + e*x^2 + f*x^3)*(a + b*x^4)^p,x]

[Out]

((a + b*x^4)^p*(-3*a*f + 3*a*f*(1 + (b*x^4)/a)^p + 3*b*f*x^4*(1 + (b*x^4)/a)^p +
12*b*c*(1 + p)*x*Hypergeometric2F1[1/4, -p, 5/4, -((b*x^4)/a)] + 6*b*d*(1 + p)*
x^2*Hypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a)] + 4*b*e*x^3*Hypergeometric2F1[
3/4, -p, 7/4, -((b*x^4)/a)] + 4*b*e*p*x^3*Hypergeometric2F1[3/4, -p, 7/4, -((b*x
^4)/a)]))/(12*b*(1 + p)*(1 + (b*x^4)/a)^p)

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \[ \int \left ( f{x}^{3}+e{x}^{2}+dx+c \right ) \left ( b{x}^{4}+a \right ) ^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^p,x)

[Out]

int((f*x^3+e*x^2+d*x+c)*(b*x^4+a)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (f x^{3} + e x^{2} + d x + c\right )}{\left (b x^{4} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x^3 + e*x^2 + d*x + c)*(b*x^4 + a)^p,x, algorithm="maxima")

[Out]

integrate((f*x^3 + e*x^2 + d*x + c)*(b*x^4 + a)^p, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \[{\rm integral}\left ({\left (f x^{3} + e x^{2} + d x + c\right )}{\left (b x^{4} + a\right )}^{p}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x^3 + e*x^2 + d*x + c)*(b*x^4 + a)^p,x, algorithm="fricas")

[Out]

integral((f*x^3 + e*x^2 + d*x + c)*(b*x^4 + a)^p, x)

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Sympy [A]  time = 142.287, size = 141, normalized size = 0.99 \[ \frac{a^{p} c x \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, - p \\ \frac{5}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{5}{4}\right )} + \frac{a^{p} d x^{2}{{}_{2}F_{1}\left (\begin{matrix} \frac{1}{2}, - p \\ \frac{3}{2} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{2} + \frac{a^{p} e x^{3} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{4}, - p \\ \frac{7}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac{7}{4}\right )} + f \left (\begin{cases} \frac{a^{p} x^{4}}{4} & \text{for}\: b = 0 \\\frac{\begin{cases} \frac{\left (a + b x^{4}\right )^{p + 1}}{p + 1} & \text{for}\: p \neq -1 \\\log{\left (a + b x^{4} \right )} & \text{otherwise} \end{cases}}{4 b} & \text{otherwise} \end{cases}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x**3+e*x**2+d*x+c)*(b*x**4+a)**p,x)

[Out]

a**p*c*x*gamma(1/4)*hyper((1/4, -p), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(
5/4)) + a**p*d*x**2*hyper((1/2, -p), (3/2,), b*x**4*exp_polar(I*pi)/a)/2 + a**p*
e*x**3*gamma(3/4)*hyper((3/4, -p), (7/4,), b*x**4*exp_polar(I*pi)/a)/(4*gamma(7/
4)) + f*Piecewise((a**p*x**4/4, Eq(b, 0)), (Piecewise(((a + b*x**4)**(p + 1)/(p
+ 1), Ne(p, -1)), (log(a + b*x**4), True))/(4*b), True))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int{\left (f x^{3} + e x^{2} + d x + c\right )}{\left (b x^{4} + a\right )}^{p}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate((f*x^3 + e*x^2 + d*x + c)*(b*x^4 + a)^p,x, algorithm="giac")

[Out]

integrate((f*x^3 + e*x^2 + d*x + c)*(b*x^4 + a)^p, x)