3.1303 \(\int \frac{(A+B x) (a+c x^2)^2}{d+e x} \, dx\)

Optimal. Leaf size=169 \[ \frac{c x^3 \left (2 a B e^2-A c d e+B c d^2\right )}{3 e^3}-\frac{c x^2 \left (2 a e^2+c d^2\right ) (B d-A e)}{2 e^4}+\frac{x \left (B \left (a e^2+c d^2\right )^2-A c d e \left (2 a e^2+c d^2\right )\right )}{e^5}-\frac{\left (a e^2+c d^2\right )^2 (B d-A e) \log (d+e x)}{e^6}-\frac{c^2 x^4 (B d-A e)}{4 e^2}+\frac{B c^2 x^5}{5 e} \]

[Out]

((B*(c*d^2 + a*e^2)^2 - A*c*d*e*(c*d^2 + 2*a*e^2))*x)/e^5 - (c*(B*d - A*e)*(c*d^2 + 2*a*e^2)*x^2)/(2*e^4) + (c
*(B*c*d^2 - A*c*d*e + 2*a*B*e^2)*x^3)/(3*e^3) - (c^2*(B*d - A*e)*x^4)/(4*e^2) + (B*c^2*x^5)/(5*e) - ((B*d - A*
e)*(c*d^2 + a*e^2)^2*Log[d + e*x])/e^6

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Rubi [A]  time = 0.191967, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.045, Rules used = {772} \[ \frac{c x^3 \left (2 a B e^2-A c d e+B c d^2\right )}{3 e^3}-\frac{c x^2 \left (2 a e^2+c d^2\right ) (B d-A e)}{2 e^4}+\frac{x \left (B \left (a e^2+c d^2\right )^2-A c d e \left (2 a e^2+c d^2\right )\right )}{e^5}-\frac{\left (a e^2+c d^2\right )^2 (B d-A e) \log (d+e x)}{e^6}-\frac{c^2 x^4 (B d-A e)}{4 e^2}+\frac{B c^2 x^5}{5 e} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^2)/(d + e*x),x]

[Out]

((B*(c*d^2 + a*e^2)^2 - A*c*d*e*(c*d^2 + 2*a*e^2))*x)/e^5 - (c*(B*d - A*e)*(c*d^2 + 2*a*e^2)*x^2)/(2*e^4) + (c
*(B*c*d^2 - A*c*d*e + 2*a*B*e^2)*x^3)/(3*e^3) - (c^2*(B*d - A*e)*x^4)/(4*e^2) + (B*c^2*x^5)/(5*e) - ((B*d - A*
e)*(c*d^2 + a*e^2)^2*Log[d + e*x])/e^6

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{(A+B x) \left (a+c x^2\right )^2}{d+e x} \, dx &=\int \left (\frac{B \left (c d^2+a e^2\right )^2-A c d e \left (c d^2+2 a e^2\right )}{e^5}+\frac{c (-B d+A e) \left (c d^2+2 a e^2\right ) x}{e^4}-\frac{c \left (-B c d^2+A c d e-2 a B e^2\right ) x^2}{e^3}+\frac{c^2 (-B d+A e) x^3}{e^2}+\frac{B c^2 x^4}{e}+\frac{(-B d+A e) \left (c d^2+a e^2\right )^2}{e^5 (d+e x)}\right ) \, dx\\ &=\frac{\left (B \left (c d^2+a e^2\right )^2-A c d e \left (c d^2+2 a e^2\right )\right ) x}{e^5}-\frac{c (B d-A e) \left (c d^2+2 a e^2\right ) x^2}{2 e^4}+\frac{c \left (B c d^2-A c d e+2 a B e^2\right ) x^3}{3 e^3}-\frac{c^2 (B d-A e) x^4}{4 e^2}+\frac{B c^2 x^5}{5 e}-\frac{(B d-A e) \left (c d^2+a e^2\right )^2 \log (d+e x)}{e^6}\\ \end{align*}

Mathematica [A]  time = 0.105667, size = 174, normalized size = 1.03 \[ \frac{e x \left (B \left (60 a^2 e^4+20 a c e^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )+c^2 \left (20 d^2 e^2 x^2-30 d^3 e x+60 d^4-15 d e^3 x^3+12 e^4 x^4\right )\right )+5 A c e \left (12 a e^2 (e x-2 d)+c \left (6 d^2 e x-12 d^3-4 d e^2 x^2+3 e^3 x^3\right )\right )\right )-60 \left (a e^2+c d^2\right )^2 (B d-A e) \log (d+e x)}{60 e^6} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^2)/(d + e*x),x]

[Out]

(e*x*(5*A*c*e*(12*a*e^2*(-2*d + e*x) + c*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3)) + B*(60*a^2*e^4 + 20
*a*c*e^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2) + c^2*(60*d^4 - 30*d^3*e*x + 20*d^2*e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4
))) - 60*(B*d - A*e)*(c*d^2 + a*e^2)^2*Log[d + e*x])/(60*e^6)

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Maple [A]  time = 0.004, size = 285, normalized size = 1.7 \begin{align*}{\frac{B{c}^{2}{x}^{5}}{5\,e}}+{\frac{A{c}^{2}{x}^{4}}{4\,e}}-{\frac{B{c}^{2}{x}^{4}d}{4\,{e}^{2}}}-{\frac{A{c}^{2}{x}^{3}d}{3\,{e}^{2}}}+{\frac{2\,aBc{x}^{3}}{3\,e}}+{\frac{B{c}^{2}{x}^{3}{d}^{2}}{3\,{e}^{3}}}+{\frac{aAc{x}^{2}}{e}}+{\frac{A{c}^{2}{x}^{2}{d}^{2}}{2\,{e}^{3}}}-{\frac{aBc{x}^{2}d}{{e}^{2}}}-{\frac{B{c}^{2}{x}^{2}{d}^{3}}{2\,{e}^{4}}}-2\,{\frac{Adacx}{{e}^{2}}}-{\frac{A{d}^{3}{c}^{2}x}{{e}^{4}}}+{\frac{{a}^{2}Bx}{e}}+2\,{\frac{aBc{d}^{2}x}{{e}^{3}}}+{\frac{B{c}^{2}{d}^{4}x}{{e}^{5}}}+{\frac{\ln \left ( ex+d \right ) A{a}^{2}}{e}}+2\,{\frac{\ln \left ( ex+d \right ) Aac{d}^{2}}{{e}^{3}}}+{\frac{{d}^{4}\ln \left ( ex+d \right ) A{c}^{2}}{{e}^{5}}}-{\frac{\ln \left ( ex+d \right ) B{a}^{2}d}{{e}^{2}}}-2\,{\frac{\ln \left ( ex+d \right ) Bac{d}^{3}}{{e}^{4}}}-{\frac{{d}^{5}\ln \left ( ex+d \right ) B{c}^{2}}{{e}^{6}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^2/(e*x+d),x)

[Out]

1/5*B*c^2*x^5/e+1/4/e*A*x^4*c^2-1/4/e^2*B*x^4*c^2*d-1/3/e^2*A*x^3*c^2*d+2/3/e*B*x^3*a*c+1/3/e^3*B*x^3*c^2*d^2+
1/e*A*x^2*a*c+1/2/e^3*A*x^2*c^2*d^2-1/e^2*B*x^2*a*c*d-1/2/e^4*B*x^2*c^2*d^3-2/e^2*A*a*c*d*x-1/e^4*A*c^2*d^3*x+
1/e*B*a^2*x+2/e^3*B*a*c*d^2*x+1/e^5*B*c^2*d^4*x+1/e*ln(e*x+d)*A*a^2+2/e^3*ln(e*x+d)*A*a*c*d^2+d^4/e^5*ln(e*x+d
)*A*c^2-1/e^2*ln(e*x+d)*B*a^2*d-2/e^4*ln(e*x+d)*B*a*c*d^3-d^5/e^6*ln(e*x+d)*B*c^2

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Maxima [A]  time = 1.01522, size = 327, normalized size = 1.93 \begin{align*} \frac{12 \, B c^{2} e^{4} x^{5} - 15 \,{\left (B c^{2} d e^{3} - A c^{2} e^{4}\right )} x^{4} + 20 \,{\left (B c^{2} d^{2} e^{2} - A c^{2} d e^{3} + 2 \, B a c e^{4}\right )} x^{3} - 30 \,{\left (B c^{2} d^{3} e - A c^{2} d^{2} e^{2} + 2 \, B a c d e^{3} - 2 \, A a c e^{4}\right )} x^{2} + 60 \,{\left (B c^{2} d^{4} - A c^{2} d^{3} e + 2 \, B a c d^{2} e^{2} - 2 \, A a c d e^{3} + B a^{2} e^{4}\right )} x}{60 \, e^{5}} - \frac{{\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )} \log \left (e x + d\right )}{e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d),x, algorithm="maxima")

[Out]

1/60*(12*B*c^2*e^4*x^5 - 15*(B*c^2*d*e^3 - A*c^2*e^4)*x^4 + 20*(B*c^2*d^2*e^2 - A*c^2*d*e^3 + 2*B*a*c*e^4)*x^3
 - 30*(B*c^2*d^3*e - A*c^2*d^2*e^2 + 2*B*a*c*d*e^3 - 2*A*a*c*e^4)*x^2 + 60*(B*c^2*d^4 - A*c^2*d^3*e + 2*B*a*c*
d^2*e^2 - 2*A*a*c*d*e^3 + B*a^2*e^4)*x)/e^5 - (B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3 + B
*a^2*d*e^4 - A*a^2*e^5)*log(e*x + d)/e^6

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Fricas [A]  time = 1.76948, size = 509, normalized size = 3.01 \begin{align*} \frac{12 \, B c^{2} e^{5} x^{5} - 15 \,{\left (B c^{2} d e^{4} - A c^{2} e^{5}\right )} x^{4} + 20 \,{\left (B c^{2} d^{2} e^{3} - A c^{2} d e^{4} + 2 \, B a c e^{5}\right )} x^{3} - 30 \,{\left (B c^{2} d^{3} e^{2} - A c^{2} d^{2} e^{3} + 2 \, B a c d e^{4} - 2 \, A a c e^{5}\right )} x^{2} + 60 \,{\left (B c^{2} d^{4} e - A c^{2} d^{3} e^{2} + 2 \, B a c d^{2} e^{3} - 2 \, A a c d e^{4} + B a^{2} e^{5}\right )} x - 60 \,{\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d),x, algorithm="fricas")

[Out]

1/60*(12*B*c^2*e^5*x^5 - 15*(B*c^2*d*e^4 - A*c^2*e^5)*x^4 + 20*(B*c^2*d^2*e^3 - A*c^2*d*e^4 + 2*B*a*c*e^5)*x^3
 - 30*(B*c^2*d^3*e^2 - A*c^2*d^2*e^3 + 2*B*a*c*d*e^4 - 2*A*a*c*e^5)*x^2 + 60*(B*c^2*d^4*e - A*c^2*d^3*e^2 + 2*
B*a*c*d^2*e^3 - 2*A*a*c*d*e^4 + B*a^2*e^5)*x - 60*(B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3
 + B*a^2*d*e^4 - A*a^2*e^5)*log(e*x + d))/e^6

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Sympy [A]  time = 1.26524, size = 204, normalized size = 1.21 \begin{align*} \frac{B c^{2} x^{5}}{5 e} - \frac{x^{4} \left (- A c^{2} e + B c^{2} d\right )}{4 e^{2}} + \frac{x^{3} \left (- A c^{2} d e + 2 B a c e^{2} + B c^{2} d^{2}\right )}{3 e^{3}} - \frac{x^{2} \left (- 2 A a c e^{3} - A c^{2} d^{2} e + 2 B a c d e^{2} + B c^{2} d^{3}\right )}{2 e^{4}} + \frac{x \left (- 2 A a c d e^{3} - A c^{2} d^{3} e + B a^{2} e^{4} + 2 B a c d^{2} e^{2} + B c^{2} d^{4}\right )}{e^{5}} - \frac{\left (- A e + B d\right ) \left (a e^{2} + c d^{2}\right )^{2} \log{\left (d + e x \right )}}{e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**2/(e*x+d),x)

[Out]

B*c**2*x**5/(5*e) - x**4*(-A*c**2*e + B*c**2*d)/(4*e**2) + x**3*(-A*c**2*d*e + 2*B*a*c*e**2 + B*c**2*d**2)/(3*
e**3) - x**2*(-2*A*a*c*e**3 - A*c**2*d**2*e + 2*B*a*c*d*e**2 + B*c**2*d**3)/(2*e**4) + x*(-2*A*a*c*d*e**3 - A*
c**2*d**3*e + B*a**2*e**4 + 2*B*a*c*d**2*e**2 + B*c**2*d**4)/e**5 - (-A*e + B*d)*(a*e**2 + c*d**2)**2*log(d +
e*x)/e**6

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Giac [A]  time = 1.38469, size = 329, normalized size = 1.95 \begin{align*} -{\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{60} \,{\left (12 \, B c^{2} x^{5} e^{4} - 15 \, B c^{2} d x^{4} e^{3} + 20 \, B c^{2} d^{2} x^{3} e^{2} - 30 \, B c^{2} d^{3} x^{2} e + 60 \, B c^{2} d^{4} x + 15 \, A c^{2} x^{4} e^{4} - 20 \, A c^{2} d x^{3} e^{3} + 30 \, A c^{2} d^{2} x^{2} e^{2} - 60 \, A c^{2} d^{3} x e + 40 \, B a c x^{3} e^{4} - 60 \, B a c d x^{2} e^{3} + 120 \, B a c d^{2} x e^{2} + 60 \, A a c x^{2} e^{4} - 120 \, A a c d x e^{3} + 60 \, B a^{2} x e^{4}\right )} e^{\left (-5\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d),x, algorithm="giac")

[Out]

-(B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3 + B*a^2*d*e^4 - A*a^2*e^5)*e^(-6)*log(abs(x*e +
d)) + 1/60*(12*B*c^2*x^5*e^4 - 15*B*c^2*d*x^4*e^3 + 20*B*c^2*d^2*x^3*e^2 - 30*B*c^2*d^3*x^2*e + 60*B*c^2*d^4*x
 + 15*A*c^2*x^4*e^4 - 20*A*c^2*d*x^3*e^3 + 30*A*c^2*d^2*x^2*e^2 - 60*A*c^2*d^3*x*e + 40*B*a*c*x^3*e^4 - 60*B*a
*c*d*x^2*e^3 + 120*B*a*c*d^2*x*e^2 + 60*A*a*c*x^2*e^4 - 120*A*a*c*d*x*e^3 + 60*B*a^2*x*e^4)*e^(-5)