3.1304 \(\int \frac{(A+B x) (a+c x^2)^2}{(d+e x)^2} \, dx\)

Optimal. Leaf size=180 \[ \frac{c x^2 \left (2 a B e^2-2 A c d e+3 B c d^2\right )}{2 e^4}-\frac{c x \left (-2 a A e^3+4 a B d e^2-3 A c d^2 e+4 B c d^3\right )}{e^5}+\frac{\left (a e^2+c d^2\right )^2 (B d-A e)}{e^6 (d+e x)}+\frac{\left (a e^2+c d^2\right ) \log (d+e x) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^6}-\frac{c^2 x^3 (2 B d-A e)}{3 e^3}+\frac{B c^2 x^4}{4 e^2} \]

[Out]

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

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Rubi [A]  time = 0.215351, antiderivative size = 180, 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} \[ \frac{c x^2 \left (2 a B e^2-2 A c d e+3 B c d^2\right )}{2 e^4}-\frac{c x \left (-2 a A e^3+4 a B d e^2-3 A c d^2 e+4 B c d^3\right )}{e^5}+\frac{\left (a e^2+c d^2\right )^2 (B d-A e)}{e^6 (d+e x)}+\frac{\left (a e^2+c d^2\right ) \log (d+e x) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^6}-\frac{c^2 x^3 (2 B d-A e)}{3 e^3}+\frac{B c^2 x^4}{4 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.150424, size = 175, normalized size = 0.97 \[ \frac{6 c e^2 x^2 \left (2 a B e^2-2 A c d e+3 B c d^2\right )+12 c e x \left (A e \left (2 a e^2+3 c d^2\right )-4 B \left (a d e^2+c d^3\right )\right )+\frac{12 \left (a e^2+c d^2\right )^2 (B d-A e)}{d+e x}+12 \left (a e^2+c d^2\right ) \log (d+e x) \left (a B e^2-4 A c d e+5 B c d^2\right )+4 c^2 e^3 x^3 (A e-2 B d)+3 B c^2 e^4 x^4}{12 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.01, size = 309, normalized size = 1.7 \begin{align*}{\frac{B{c}^{2}{x}^{4}}{4\,{e}^{2}}}+{\frac{A{c}^{2}{x}^{3}}{3\,{e}^{2}}}-{\frac{2\,B{c}^{2}{x}^{3}d}{3\,{e}^{3}}}-{\frac{A{c}^{2}{x}^{2}d}{{e}^{3}}}+{\frac{aBc{x}^{2}}{{e}^{2}}}+{\frac{3\,B{c}^{2}{x}^{2}{d}^{2}}{2\,{e}^{4}}}+2\,{\frac{aAcx}{{e}^{2}}}+3\,{\frac{A{c}^{2}{d}^{2}x}{{e}^{4}}}-4\,{\frac{Bacdx}{{e}^{3}}}-4\,{\frac{B{c}^{2}{d}^{3}x}{{e}^{5}}}-{\frac{A{a}^{2}}{e \left ( ex+d \right ) }}-2\,{\frac{A{d}^{2}ac}{{e}^{3} \left ( ex+d \right ) }}-{\frac{A{d}^{4}{c}^{2}}{{e}^{5} \left ( ex+d \right ) }}+{\frac{Bd{a}^{2}}{{e}^{2} \left ( ex+d \right ) }}+2\,{\frac{aBc{d}^{3}}{{e}^{4} \left ( ex+d \right ) }}+{\frac{B{c}^{2}{d}^{5}}{{e}^{6} \left ( ex+d \right ) }}-4\,{\frac{\ln \left ( ex+d \right ) Aacd}{{e}^{3}}}-4\,{\frac{{d}^{3}\ln \left ( ex+d \right ) A{c}^{2}}{{e}^{5}}}+{\frac{\ln \left ( ex+d \right ) B{a}^{2}}{{e}^{2}}}+6\,{\frac{\ln \left ( ex+d \right ) Bac{d}^{2}}{{e}^{4}}}+5\,{\frac{{d}^{4}\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)^2,x)

[Out]

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

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Maxima [A]  time = 1.04848, size = 336, normalized size = 1.87 \begin{align*} \frac{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^{7} x + d e^{6}} + \frac{3 \, B c^{2} e^{3} x^{4} - 4 \,{\left (2 \, B c^{2} d e^{2} - A c^{2} e^{3}\right )} x^{3} + 6 \,{\left (3 \, B c^{2} d^{2} e - 2 \, A c^{2} d e^{2} + 2 \, B a c e^{3}\right )} x^{2} - 12 \,{\left (4 \, B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 4 \, B a c d e^{2} - 2 \, A a c e^{3}\right )} x}{12 \, e^{5}} + \frac{{\left (5 \, B c^{2} d^{4} - 4 \, A c^{2} d^{3} e + 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} + B a^{2} e^{4}\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)^2,x, algorithm="maxima")

[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^7*x + d*e^6) + 1/12
*(3*B*c^2*e^3*x^4 - 4*(2*B*c^2*d*e^2 - A*c^2*e^3)*x^3 + 6*(3*B*c^2*d^2*e - 2*A*c^2*d*e^2 + 2*B*a*c*e^3)*x^2 -
12*(4*B*c^2*d^3 - 3*A*c^2*d^2*e + 4*B*a*c*d*e^2 - 2*A*a*c*e^3)*x)/e^5 + (5*B*c^2*d^4 - 4*A*c^2*d^3*e + 6*B*a*c
*d^2*e^2 - 4*A*a*c*d*e^3 + B*a^2*e^4)*log(e*x + d)/e^6

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Fricas [B]  time = 1.70791, size = 763, normalized size = 4.24 \begin{align*} \frac{3 \, B c^{2} e^{5} x^{5} + 12 \, B c^{2} d^{5} - 12 \, A c^{2} d^{4} e + 24 \, B a c d^{3} e^{2} - 24 \, A a c d^{2} e^{3} + 12 \, B a^{2} d e^{4} - 12 \, A a^{2} e^{5} -{\left (5 \, B c^{2} d e^{4} - 4 \, A c^{2} e^{5}\right )} x^{4} + 2 \,{\left (5 \, B c^{2} d^{2} e^{3} - 4 \, A c^{2} d e^{4} + 6 \, B a c e^{5}\right )} x^{3} - 6 \,{\left (5 \, B c^{2} d^{3} e^{2} - 4 \, A c^{2} d^{2} e^{3} + 6 \, B a c d e^{4} - 4 \, A a c e^{5}\right )} x^{2} - 12 \,{\left (4 \, B c^{2} d^{4} e - 3 \, A c^{2} d^{3} e^{2} + 4 \, B a c d^{2} e^{3} - 2 \, A a c d e^{4}\right )} x + 12 \,{\left (5 \, B c^{2} d^{5} - 4 \, A c^{2} d^{4} e + 6 \, B a c d^{3} e^{2} - 4 \, A a c d^{2} e^{3} + B a^{2} d e^{4} +{\left (5 \, B c^{2} d^{4} e - 4 \, A c^{2} d^{3} e^{2} + 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} + B a^{2} e^{5}\right )} x\right )} \log \left (e x + d\right )}{12 \,{\left (e^{7} x + d e^{6}\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)^2,x, algorithm="fricas")

[Out]

1/12*(3*B*c^2*e^5*x^5 + 12*B*c^2*d^5 - 12*A*c^2*d^4*e + 24*B*a*c*d^3*e^2 - 24*A*a*c*d^2*e^3 + 12*B*a^2*d*e^4 -
12*A*a^2*e^5 - (5*B*c^2*d*e^4 - 4*A*c^2*e^5)*x^4 + 2*(5*B*c^2*d^2*e^3 - 4*A*c^2*d*e^4 + 6*B*a*c*e^5)*x^3 - 6*
(5*B*c^2*d^3*e^2 - 4*A*c^2*d^2*e^3 + 6*B*a*c*d*e^4 - 4*A*a*c*e^5)*x^2 - 12*(4*B*c^2*d^4*e - 3*A*c^2*d^3*e^2 +
4*B*a*c*d^2*e^3 - 2*A*a*c*d*e^4)*x + 12*(5*B*c^2*d^5 - 4*A*c^2*d^4*e + 6*B*a*c*d^3*e^2 - 4*A*a*c*d^2*e^3 + B*a
^2*d*e^4 + (5*B*c^2*d^4*e - 4*A*c^2*d^3*e^2 + 6*B*a*c*d^2*e^3 - 4*A*a*c*d*e^4 + B*a^2*e^5)*x)*log(e*x + d))/(e
^7*x + d*e^6)

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Sympy [A]  time = 3.03884, size = 243, normalized size = 1.35 \begin{align*} \frac{B c^{2} x^{4}}{4 e^{2}} + \frac{- A a^{2} e^{5} - 2 A a c d^{2} e^{3} - A c^{2} d^{4} e + B a^{2} d e^{4} + 2 B a c d^{3} e^{2} + B c^{2} d^{5}}{d e^{6} + e^{7} x} - \frac{x^{3} \left (- A c^{2} e + 2 B c^{2} d\right )}{3 e^{3}} + \frac{x^{2} \left (- 2 A c^{2} d e + 2 B a c e^{2} + 3 B c^{2} d^{2}\right )}{2 e^{4}} - \frac{x \left (- 2 A a c e^{3} - 3 A c^{2} d^{2} e + 4 B a c d e^{2} + 4 B c^{2} d^{3}\right )}{e^{5}} + \frac{\left (a e^{2} + c d^{2}\right ) \left (- 4 A c d e + B a e^{2} + 5 B c d^{2}\right ) \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)**2,x)

[Out]

B*c**2*x**4/(4*e**2) + (-A*a**2*e**5 - 2*A*a*c*d**2*e**3 - A*c**2*d**4*e + B*a**2*d*e**4 + 2*B*a*c*d**3*e**2 +
B*c**2*d**5)/(d*e**6 + e**7*x) - x**3*(-A*c**2*e + 2*B*c**2*d)/(3*e**3) + x**2*(-2*A*c**2*d*e + 2*B*a*c*e**2
+ 3*B*c**2*d**2)/(2*e**4) - x*(-2*A*a*c*e**3 - 3*A*c**2*d**2*e + 4*B*a*c*d*e**2 + 4*B*c**2*d**3)/e**5 + (a*e**
2 + c*d**2)*(-4*A*c*d*e + B*a*e**2 + 5*B*c*d**2)*log(d + e*x)/e**6

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Giac [A]  time = 1.25382, size = 428, normalized size = 2.38 \begin{align*} \frac{1}{12} \,{\left (3 \, B c^{2} - \frac{4 \,{\left (5 \, B c^{2} d e - A c^{2} e^{2}\right )} e^{\left (-1\right )}}{x e + d} + \frac{12 \,{\left (5 \, B c^{2} d^{2} e^{2} - 2 \, A c^{2} d e^{3} + B a c e^{4}\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}} - \frac{24 \,{\left (5 \, B c^{2} d^{3} e^{3} - 3 \, A c^{2} d^{2} e^{4} + 3 \, B a c d e^{5} - A a c e^{6}\right )} e^{\left (-3\right )}}{{\left (x e + d\right )}^{3}}\right )}{\left (x e + d\right )}^{4} e^{\left (-6\right )} -{\left (5 \, B c^{2} d^{4} - 4 \, A c^{2} d^{3} e + 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} + B a^{2} e^{4}\right )} e^{\left (-6\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) +{\left (\frac{B c^{2} d^{5} e^{4}}{x e + d} - \frac{A c^{2} d^{4} e^{5}}{x e + d} + \frac{2 \, B a c d^{3} e^{6}}{x e + d} - \frac{2 \, A a c d^{2} e^{7}}{x e + d} + \frac{B a^{2} d e^{8}}{x e + d} - \frac{A a^{2} e^{9}}{x e + d}\right )} e^{\left (-10\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)^2,x, algorithm="giac")

[Out]

1/12*(3*B*c^2 - 4*(5*B*c^2*d*e - A*c^2*e^2)*e^(-1)/(x*e + d) + 12*(5*B*c^2*d^2*e^2 - 2*A*c^2*d*e^3 + B*a*c*e^4
)*e^(-2)/(x*e + d)^2 - 24*(5*B*c^2*d^3*e^3 - 3*A*c^2*d^2*e^4 + 3*B*a*c*d*e^5 - A*a*c*e^6)*e^(-3)/(x*e + d)^3)*
(x*e + d)^4*e^(-6) - (5*B*c^2*d^4 - 4*A*c^2*d^3*e + 6*B*a*c*d^2*e^2 - 4*A*a*c*d*e^3 + B*a^2*e^4)*e^(-6)*log(ab
s(x*e + d)*e^(-1)/(x*e + d)^2) + (B*c^2*d^5*e^4/(x*e + d) - A*c^2*d^4*e^5/(x*e + d) + 2*B*a*c*d^3*e^6/(x*e + d
) - 2*A*a*c*d^2*e^7/(x*e + d) + B*a^2*d*e^8/(x*e + d) - A*a^2*e^9/(x*e + d))*e^(-10)