3.112 \(\int \frac{x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx\)

Optimal. Leaf size=319 \[ \frac{\left (a^2 d^2+a b c d+b^2 c^2\right ) (e+f x)^{n+1}}{b^3 d^3 f (n+1)}-\frac{a^4 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{b^3 (n+1) (b c-a d) (b e-a f)}+\frac{e (a d+b c) (e+f x)^{n+1}}{b^2 d^2 f^2 (n+1)}-\frac{(a d+b c) (e+f x)^{n+2}}{b^2 d^2 f^2 (n+2)}+\frac{c^4 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )}{d^3 (n+1) (b c-a d) (d e-c f)}+\frac{e^2 (e+f x)^{n+1}}{b d f^3 (n+1)}-\frac{2 e (e+f x)^{n+2}}{b d f^3 (n+2)}+\frac{(e+f x)^{n+3}}{b d f^3 (n+3)} \]

[Out]

(e^2*(e + f*x)^(1 + n))/(b*d*f^3*(1 + n)) + ((b*c + a*d)*e*(e + f*x)^(1 + n))/(b^2*d^2*f^2*(1 + n)) + ((b^2*c^
2 + a*b*c*d + a^2*d^2)*(e + f*x)^(1 + n))/(b^3*d^3*f*(1 + n)) - (2*e*(e + f*x)^(2 + n))/(b*d*f^3*(2 + n)) - ((
b*c + a*d)*(e + f*x)^(2 + n))/(b^2*d^2*f^2*(2 + n)) + (e + f*x)^(3 + n)/(b*d*f^3*(3 + n)) - (a^4*(e + f*x)^(1
+ n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/(b^3*(b*c - a*d)*(b*e - a*f)*(1 + n)) + (c
^4*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(e + f*x))/(d*e - c*f)])/(d^3*(b*c - a*d)*(d*e - c*
f)*(1 + n))

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Rubi [A]  time = 0.279362, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {180, 43, 68} \[ \frac{\left (a^2 d^2+a b c d+b^2 c^2\right ) (e+f x)^{n+1}}{b^3 d^3 f (n+1)}-\frac{a^4 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{b^3 (n+1) (b c-a d) (b e-a f)}+\frac{e (a d+b c) (e+f x)^{n+1}}{b^2 d^2 f^2 (n+1)}-\frac{(a d+b c) (e+f x)^{n+2}}{b^2 d^2 f^2 (n+2)}+\frac{c^4 (e+f x)^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )}{d^3 (n+1) (b c-a d) (d e-c f)}+\frac{e^2 (e+f x)^{n+1}}{b d f^3 (n+1)}-\frac{2 e (e+f x)^{n+2}}{b d f^3 (n+2)}+\frac{(e+f x)^{n+3}}{b d f^3 (n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(e + f*x)^n)/((a + b*x)*(c + d*x)),x]

[Out]

(e^2*(e + f*x)^(1 + n))/(b*d*f^3*(1 + n)) + ((b*c + a*d)*e*(e + f*x)^(1 + n))/(b^2*d^2*f^2*(1 + n)) + ((b^2*c^
2 + a*b*c*d + a^2*d^2)*(e + f*x)^(1 + n))/(b^3*d^3*f*(1 + n)) - (2*e*(e + f*x)^(2 + n))/(b*d*f^3*(2 + n)) - ((
b*c + a*d)*(e + f*x)^(2 + n))/(b^2*d^2*f^2*(2 + n)) + (e + f*x)^(3 + n)/(b*d*f^3*(3 + n)) - (a^4*(e + f*x)^(1
+ n)*Hypergeometric2F1[1, 1 + n, 2 + n, (b*(e + f*x))/(b*e - a*f)])/(b^3*(b*c - a*d)*(b*e - a*f)*(1 + n)) + (c
^4*(e + f*x)^(1 + n)*Hypergeometric2F1[1, 1 + n, 2 + n, (d*(e + f*x))/(d*e - c*f)])/(d^3*(b*c - a*d)*(d*e - c*
f)*(1 + n))

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{x^4 (e+f x)^n}{(a+b x) (c+d x)} \, dx &=\int \left (\frac{\left (b^2 c^2+a b c d+a^2 d^2\right ) (e+f x)^n}{b^3 d^3}-\frac{(b c+a d) x (e+f x)^n}{b^2 d^2}+\frac{x^2 (e+f x)^n}{b d}+\frac{a^4 (e+f x)^n}{b^3 (b c-a d) (a+b x)}+\frac{c^4 (e+f x)^n}{d^3 (-b c+a d) (c+d x)}\right ) \, dx\\ &=\frac{\left (b^2 c^2+a b c d+a^2 d^2\right ) (e+f x)^{1+n}}{b^3 d^3 f (1+n)}+\frac{\int x^2 (e+f x)^n \, dx}{b d}+\frac{a^4 \int \frac{(e+f x)^n}{a+b x} \, dx}{b^3 (b c-a d)}-\frac{c^4 \int \frac{(e+f x)^n}{c+d x} \, dx}{d^3 (b c-a d)}-\frac{(b c+a d) \int x (e+f x)^n \, dx}{b^2 d^2}\\ &=\frac{\left (b^2 c^2+a b c d+a^2 d^2\right ) (e+f x)^{1+n}}{b^3 d^3 f (1+n)}-\frac{a^4 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{b (e+f x)}{b e-a f}\right )}{b^3 (b c-a d) (b e-a f) (1+n)}+\frac{c^4 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{d (e+f x)}{d e-c f}\right )}{d^3 (b c-a d) (d e-c f) (1+n)}+\frac{\int \left (\frac{e^2 (e+f x)^n}{f^2}-\frac{2 e (e+f x)^{1+n}}{f^2}+\frac{(e+f x)^{2+n}}{f^2}\right ) \, dx}{b d}-\frac{(b c+a d) \int \left (-\frac{e (e+f x)^n}{f}+\frac{(e+f x)^{1+n}}{f}\right ) \, dx}{b^2 d^2}\\ &=\frac{e^2 (e+f x)^{1+n}}{b d f^3 (1+n)}+\frac{(b c+a d) e (e+f x)^{1+n}}{b^2 d^2 f^2 (1+n)}+\frac{\left (b^2 c^2+a b c d+a^2 d^2\right ) (e+f x)^{1+n}}{b^3 d^3 f (1+n)}-\frac{2 e (e+f x)^{2+n}}{b d f^3 (2+n)}-\frac{(b c+a d) (e+f x)^{2+n}}{b^2 d^2 f^2 (2+n)}+\frac{(e+f x)^{3+n}}{b d f^3 (3+n)}-\frac{a^4 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{b (e+f x)}{b e-a f}\right )}{b^3 (b c-a d) (b e-a f) (1+n)}+\frac{c^4 (e+f x)^{1+n} \, _2F_1\left (1,1+n;2+n;\frac{d (e+f x)}{d e-c f}\right )}{d^3 (b c-a d) (d e-c f) (1+n)}\\ \end{align*}

Mathematica [A]  time = 1.32229, size = 285, normalized size = 0.89 \[ \frac{(e+f x)^{n+1} \left (\frac{b^3 c^4 f^3 \left (n^2+5 n+6\right ) \, _2F_1\left (1,n+1;n+2;\frac{d (e+f x)}{d e-c f}\right )-(b c-a d) (c f-d e) \left (a^2 d^2 f^2 \left (n^2+5 n+6\right )+a b d f (n+3) (c f (n+2)+d (e-f (n+1) x))+b^2 \left (c^2 f^2 \left (n^2+5 n+6\right )+c d f (n+3) (e-f (n+1) x)+d^2 \left (2 e^2-2 e f (n+1) x+f^2 \left (n^2+3 n+2\right ) x^2\right )\right )\right )}{f^3 (n+2) (n+3) (a d-b c) (c f-d e)}-\frac{a^4 d^3 \, _2F_1\left (1,n+1;n+2;\frac{b (e+f x)}{b e-a f}\right )}{(b c-a d) (b e-a f)}\right )}{b^3 d^3 (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^4*(e + f*x)^n)/((a + b*x)*(c + d*x)),x]

[Out]

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

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Maple [F]  time = 0.089, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{n}{x}^{4}}{ \left ( bx+a \right ) \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(f*x+e)^n/(b*x+a)/(d*x+c),x)

[Out]

int(x^4*(f*x+e)^n/(b*x+a)/(d*x+c),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n} x^{4}}{{\left (b x + a\right )}{\left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(f*x+e)^n/(b*x+a)/(d*x+c),x, algorithm="maxima")

[Out]

integrate((f*x + e)^n*x^4/((b*x + a)*(d*x + c)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (f x + e\right )}^{n} x^{4}}{b d x^{2} + a c +{\left (b c + a d\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(f*x+e)^n/(b*x+a)/(d*x+c),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4*(f*x+e)**n/(b*x+a)/(d*x+c),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{n} x^{4}}{{\left (b x + a\right )}{\left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(f*x+e)^n/(b*x+a)/(d*x+c),x, algorithm="giac")

[Out]

integrate((f*x + e)^n*x^4/((b*x + a)*(d*x + c)), x)