Integrand size = 25, antiderivative size = 143 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\frac {f \left (a+b x^4\right )^{1+p}}{4 b (1+p)}+\frac {c x \left (a+b x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {5}{4}+p,\frac {5}{4},-\frac {b x^4}{a}\right )}{a}+\frac {d x^2 \left (a+b x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {3}{2}+p,\frac {3}{2},-\frac {b x^4}{a}\right )}{2 a}+\frac {e x^3 \left (a+b x^4\right )^{1+p} \operatorname {Hypergeometric2F1}\left (1,\frac {7}{4}+p,\frac {7}{4},-\frac {b x^4}{a}\right )}{3 a} \]
1/4*f*(b*x^4+a)^(p+1)/b/(p+1)+c*x*(b*x^4+a)^(p+1)*hypergeom([1, 5/4+p],[5/ 4],-b*x^4/a)/a+1/2*d*x^2*(b*x^4+a)^(p+1)*hypergeom([1, 3/2+p],[3/2],-b*x^4 /a)/a+1/3*e*x^3*(b*x^4+a)^(p+1)*hypergeom([1, 7/4+p],[7/4],-b*x^4/a)/a
Time = 0.73 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.03 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\frac {1}{12} \left (a+b x^4\right )^p \left (\frac {3 f \left (a+b x^4\right )}{b (1+p)}+12 c x \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+6 d x^2 \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )+4 e x^3 \left (1+\frac {b x^4}{a}\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )\right ) \]
((a + b*x^4)^p*((3*f*(a + b*x^4))/(b*(1 + p)) + (12*c*x*Hypergeometric2F1[ 1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (6*d*x^2*Hypergeometric2F 1[1/2, -p, 3/2, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (4*e*x^3*Hypergeometric 2F1[3/4, -p, 7/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p))/12
Time = 0.35 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2424, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \left (a+b x^4\right )^p \left (c+d x+e x^2+f x^3\right ) \, dx\) |
\(\Big \downarrow \) 2424 |
\(\displaystyle \int \left (\left (c+e x^2\right ) \left (a+b x^4\right )^p+x \left (d+f x^2\right ) \left (a+b x^4\right )^p\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle c x \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},-p,\frac {5}{4},-\frac {b x^4}{a}\right )+\frac {1}{2} d x^2 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-p,\frac {3}{2},-\frac {b x^4}{a}\right )+\frac {1}{3} e x^3 \left (a+b x^4\right )^p \left (\frac {b x^4}{a}+1\right )^{-p} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},-p,\frac {7}{4},-\frac {b x^4}{a}\right )+\frac {f \left (a+b x^4\right )^{p+1}}{4 b (p+1)}\) |
(f*(a + b*x^4)^(1 + p))/(4*b*(1 + p)) + (c*x*(a + b*x^4)^p*Hypergeometric2 F1[1/4, -p, 5/4, -((b*x^4)/a)])/(1 + (b*x^4)/a)^p + (d*x^2*(a + b*x^4)^p*H ypergeometric2F1[1/2, -p, 3/2, -((b*x^4)/a)])/(2*(1 + (b*x^4)/a)^p) + (e*x ^3*(a + b*x^4)^p*Hypergeometric2F1[3/4, -p, 7/4, -((b*x^4)/a)])/(3*(1 + (b *x^4)/a)^p)
3.6.52.3.1 Defintions of rubi rules used
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Module[{q = Expon[Pq, x], j, k}, Int[Sum[x^j*Sum[Coeff[Pq, x, j + k*(n/2)]*x^(k*(n/2)), {k, 0, 2 *((q - j)/n) + 1}]*(a + b*x^n)^p, {j, 0, n/2 - 1}], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && !PolyQ[Pq, x^(n/2)]
\[\int \left (f \,x^{3}+e \,x^{2}+d x +c \right ) \left (b \,x^{4}+a \right )^{p}d x\]
\[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int { {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p} \,d x } \]
Time = 20.54 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.99 \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\frac {a^{p} c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, - p \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {5}{4}\right )} + \frac {a^{p} d x^{2} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, - p \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{2} + \frac {a^{p} e x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, - p \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {7}{4}\right )} + f \left (\begin {cases} \frac {a^{p} x^{4}}{4} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b x^{4}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b x^{4} \right )} & \text {otherwise} \end {cases}}{4 b} & \text {otherwise} \end {cases}\right ) \]
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))
\[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int { {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p} \,d x } \]
\[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int { {\left (f x^{3} + e x^{2} + d x + c\right )} {\left (b x^{4} + a\right )}^{p} \,d x } \]
Timed out. \[ \int \left (c+d x+e x^2+f x^3\right ) \left (a+b x^4\right )^p \, dx=\int {\left (b\,x^4+a\right )}^p\,\left (f\,x^3+e\,x^2+d\,x+c\right ) \,d x \]