3.479 \(\int \frac{(a+c x^2)^3}{(d+e x)^2} \, dx\)

Optimal. Leaf size=158 \[ \frac{c x \left (3 a^2 e^4+9 a c d^2 e^2+5 c^2 d^4\right )}{e^6}+\frac{c^2 x^3 \left (a e^2+c d^2\right )}{e^4}-\frac{c^2 d x^2 \left (3 a e^2+2 c d^2\right )}{e^5}-\frac{\left (a e^2+c d^2\right )^3}{e^7 (d+e x)}-\frac{6 c d \left (a e^2+c d^2\right )^2 \log (d+e x)}{e^7}-\frac{c^3 d x^4}{2 e^3}+\frac{c^3 x^5}{5 e^2} \]

[Out]

(c*(5*c^2*d^4 + 9*a*c*d^2*e^2 + 3*a^2*e^4)*x)/e^6 - (c^2*d*(2*c*d^2 + 3*a*e^2)*x^2)/e^5 + (c^2*(c*d^2 + a*e^2)
*x^3)/e^4 - (c^3*d*x^4)/(2*e^3) + (c^3*x^5)/(5*e^2) - (c*d^2 + a*e^2)^3/(e^7*(d + e*x)) - (6*c*d*(c*d^2 + a*e^
2)^2*Log[d + e*x])/e^7

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Rubi [A]  time = 0.151524, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {697} \[ \frac{c x \left (3 a^2 e^4+9 a c d^2 e^2+5 c^2 d^4\right )}{e^6}+\frac{c^2 x^3 \left (a e^2+c d^2\right )}{e^4}-\frac{c^2 d x^2 \left (3 a e^2+2 c d^2\right )}{e^5}-\frac{\left (a e^2+c d^2\right )^3}{e^7 (d+e x)}-\frac{6 c d \left (a e^2+c d^2\right )^2 \log (d+e x)}{e^7}-\frac{c^3 d x^4}{2 e^3}+\frac{c^3 x^5}{5 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^3/(d + e*x)^2,x]

[Out]

(c*(5*c^2*d^4 + 9*a*c*d^2*e^2 + 3*a^2*e^4)*x)/e^6 - (c^2*d*(2*c*d^2 + 3*a*e^2)*x^2)/e^5 + (c^2*(c*d^2 + a*e^2)
*x^3)/e^4 - (c^3*d*x^4)/(2*e^3) + (c^3*x^5)/(5*e^2) - (c*d^2 + a*e^2)^3/(e^7*(d + e*x)) - (6*c*d*(c*d^2 + a*e^
2)^2*Log[d + e*x])/e^7

Rule 697

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

Rubi steps

\begin{align*} \int \frac{\left (a+c x^2\right )^3}{(d+e x)^2} \, dx &=\int \left (\frac{c \left (5 c^2 d^4+9 a c d^2 e^2+3 a^2 e^4\right )}{e^6}-\frac{2 c^2 d \left (2 c d^2+3 a e^2\right ) x}{e^5}+\frac{3 c^2 \left (c d^2+a e^2\right ) x^2}{e^4}-\frac{2 c^3 d x^3}{e^3}+\frac{c^3 x^4}{e^2}+\frac{\left (c d^2+a e^2\right )^3}{e^6 (d+e x)^2}-\frac{6 c d \left (c d^2+a e^2\right )^2}{e^6 (d+e x)}\right ) \, dx\\ &=\frac{c \left (5 c^2 d^4+9 a c d^2 e^2+3 a^2 e^4\right ) x}{e^6}-\frac{c^2 d \left (2 c d^2+3 a e^2\right ) x^2}{e^5}+\frac{c^2 \left (c d^2+a e^2\right ) x^3}{e^4}-\frac{c^3 d x^4}{2 e^3}+\frac{c^3 x^5}{5 e^2}-\frac{\left (c d^2+a e^2\right )^3}{e^7 (d+e x)}-\frac{6 c d \left (c d^2+a e^2\right )^2 \log (d+e x)}{e^7}\\ \end{align*}

Mathematica [A]  time = 0.0583553, size = 193, normalized size = 1.22 \[ \frac{30 a^2 c e^4 \left (-d^2+d e x+e^2 x^2\right )-10 a^3 e^6+10 a c^2 e^2 \left (6 d^2 e^2 x^2+9 d^3 e x-3 d^4-2 d e^3 x^3+e^4 x^4\right )-60 c d (d+e x) \left (a e^2+c d^2\right )^2 \log (d+e x)+c^3 \left (30 d^4 e^2 x^2-10 d^3 e^3 x^3+5 d^2 e^4 x^4+50 d^5 e x-10 d^6-3 d e^5 x^5+2 e^6 x^6\right )}{10 e^7 (d+e x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^2)^3/(d + e*x)^2,x]

[Out]

(-10*a^3*e^6 + 30*a^2*c*e^4*(-d^2 + d*e*x + e^2*x^2) + 10*a*c^2*e^2*(-3*d^4 + 9*d^3*e*x + 6*d^2*e^2*x^2 - 2*d*
e^3*x^3 + e^4*x^4) + c^3*(-10*d^6 + 50*d^5*e*x + 30*d^4*e^2*x^2 - 10*d^3*e^3*x^3 + 5*d^2*e^4*x^4 - 3*d*e^5*x^5
+ 2*e^6*x^6) - 60*c*d*(c*d^2 + a*e^2)^2*(d + e*x)*Log[d + e*x])/(10*e^7*(d + e*x))

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Maple [A]  time = 0.05, size = 233, normalized size = 1.5 \begin{align*}{\frac{{c}^{3}{x}^{5}}{5\,{e}^{2}}}-{\frac{{c}^{3}d{x}^{4}}{2\,{e}^{3}}}+{\frac{{c}^{2}{x}^{3}a}{{e}^{2}}}+{\frac{{x}^{3}{c}^{3}{d}^{2}}{{e}^{4}}}-3\,{\frac{{c}^{2}{x}^{2}ad}{{e}^{3}}}-2\,{\frac{{c}^{3}{x}^{2}{d}^{3}}{{e}^{5}}}+3\,{\frac{{a}^{2}cx}{{e}^{2}}}+9\,{\frac{a{c}^{2}{d}^{2}x}{{e}^{4}}}+5\,{\frac{{c}^{3}{d}^{4}x}{{e}^{6}}}-6\,{\frac{cd\ln \left ( ex+d \right ){a}^{2}}{{e}^{3}}}-12\,{\frac{{c}^{2}{d}^{3}\ln \left ( ex+d \right ) a}{{e}^{5}}}-6\,{\frac{{c}^{3}{d}^{5}\ln \left ( ex+d \right ) }{{e}^{7}}}-{\frac{{a}^{3}}{e \left ( ex+d \right ) }}-3\,{\frac{{a}^{2}c{d}^{2}}{{e}^{3} \left ( ex+d \right ) }}-3\,{\frac{{d}^{4}a{c}^{2}}{{e}^{5} \left ( ex+d \right ) }}-{\frac{{d}^{6}{c}^{3}}{{e}^{7} \left ( ex+d \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+a)^3/(e*x+d)^2,x)

[Out]

1/5*c^3*x^5/e^2-1/2*c^3*d*x^4/e^3+c^2/e^2*x^3*a+c^3/e^4*x^3*d^2-3*c^2/e^3*x^2*a*d-2*c^3/e^5*x^2*d^3+3*c/e^2*a^
2*x+9*c^2/e^4*a*d^2*x+5*c^3/e^6*d^4*x-6*c*d/e^3*ln(e*x+d)*a^2-12*c^2*d^3/e^5*ln(e*x+d)*a-6*c^3*d^5/e^7*ln(e*x+
d)-1/e/(e*x+d)*a^3-3/e^3/(e*x+d)*a^2*c*d^2-3/e^5/(e*x+d)*d^4*a*c^2-1/e^7/(e*x+d)*d^6*c^3

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Maxima [A]  time = 1.15595, size = 278, normalized size = 1.76 \begin{align*} -\frac{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}}{e^{8} x + d e^{7}} + \frac{2 \, c^{3} e^{4} x^{5} - 5 \, c^{3} d e^{3} x^{4} + 10 \,{\left (c^{3} d^{2} e^{2} + a c^{2} e^{4}\right )} x^{3} - 10 \,{\left (2 \, c^{3} d^{3} e + 3 \, a c^{2} d e^{3}\right )} x^{2} + 10 \,{\left (5 \, c^{3} d^{4} + 9 \, a c^{2} d^{2} e^{2} + 3 \, a^{2} c e^{4}\right )} x}{10 \, e^{6}} - \frac{6 \,{\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} \log \left (e x + d\right )}{e^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^3/(e*x+d)^2,x, algorithm="maxima")

[Out]

-(c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 + a^3*e^6)/(e^8*x + d*e^7) + 1/10*(2*c^3*e^4*x^5 - 5*c^3*d*e^3*x
^4 + 10*(c^3*d^2*e^2 + a*c^2*e^4)*x^3 - 10*(2*c^3*d^3*e + 3*a*c^2*d*e^3)*x^2 + 10*(5*c^3*d^4 + 9*a*c^2*d^2*e^2
+ 3*a^2*c*e^4)*x)/e^6 - 6*(c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*log(e*x + d)/e^7

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Fricas [A]  time = 2.18339, size = 558, normalized size = 3.53 \begin{align*} \frac{2 \, c^{3} e^{6} x^{6} - 3 \, c^{3} d e^{5} x^{5} - 10 \, c^{3} d^{6} - 30 \, a c^{2} d^{4} e^{2} - 30 \, a^{2} c d^{2} e^{4} - 10 \, a^{3} e^{6} + 5 \,{\left (c^{3} d^{2} e^{4} + 2 \, a c^{2} e^{6}\right )} x^{4} - 10 \,{\left (c^{3} d^{3} e^{3} + 2 \, a c^{2} d e^{5}\right )} x^{3} + 30 \,{\left (c^{3} d^{4} e^{2} + 2 \, a c^{2} d^{2} e^{4} + a^{2} c e^{6}\right )} x^{2} + 10 \,{\left (5 \, c^{3} d^{5} e + 9 \, a c^{2} d^{3} e^{3} + 3 \, a^{2} c d e^{5}\right )} x - 60 \,{\left (c^{3} d^{6} + 2 \, a c^{2} d^{4} e^{2} + a^{2} c d^{2} e^{4} +{\left (c^{3} d^{5} e + 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x\right )} \log \left (e x + d\right )}{10 \,{\left (e^{8} x + d e^{7}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^3/(e*x+d)^2,x, algorithm="fricas")

[Out]

1/10*(2*c^3*e^6*x^6 - 3*c^3*d*e^5*x^5 - 10*c^3*d^6 - 30*a*c^2*d^4*e^2 - 30*a^2*c*d^2*e^4 - 10*a^3*e^6 + 5*(c^3
*d^2*e^4 + 2*a*c^2*e^6)*x^4 - 10*(c^3*d^3*e^3 + 2*a*c^2*d*e^5)*x^3 + 30*(c^3*d^4*e^2 + 2*a*c^2*d^2*e^4 + a^2*c
*e^6)*x^2 + 10*(5*c^3*d^5*e + 9*a*c^2*d^3*e^3 + 3*a^2*c*d*e^5)*x - 60*(c^3*d^6 + 2*a*c^2*d^4*e^2 + a^2*c*d^2*e
^4 + (c^3*d^5*e + 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x)*log(e*x + d))/(e^8*x + d*e^7)

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Sympy [A]  time = 1.0412, size = 189, normalized size = 1.2 \begin{align*} - \frac{c^{3} d x^{4}}{2 e^{3}} + \frac{c^{3} x^{5}}{5 e^{2}} - \frac{6 c d \left (a e^{2} + c d^{2}\right )^{2} \log{\left (d + e x \right )}}{e^{7}} - \frac{a^{3} e^{6} + 3 a^{2} c d^{2} e^{4} + 3 a c^{2} d^{4} e^{2} + c^{3} d^{6}}{d e^{7} + e^{8} x} + \frac{x^{3} \left (a c^{2} e^{2} + c^{3} d^{2}\right )}{e^{4}} - \frac{x^{2} \left (3 a c^{2} d e^{2} + 2 c^{3} d^{3}\right )}{e^{5}} + \frac{x \left (3 a^{2} c e^{4} + 9 a c^{2} d^{2} e^{2} + 5 c^{3} d^{4}\right )}{e^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+a)**3/(e*x+d)**2,x)

[Out]

-c**3*d*x**4/(2*e**3) + c**3*x**5/(5*e**2) - 6*c*d*(a*e**2 + c*d**2)**2*log(d + e*x)/e**7 - (a**3*e**6 + 3*a**
2*c*d**2*e**4 + 3*a*c**2*d**4*e**2 + c**3*d**6)/(d*e**7 + e**8*x) + x**3*(a*c**2*e**2 + c**3*d**2)/e**4 - x**2
*(3*a*c**2*d*e**2 + 2*c**3*d**3)/e**5 + x*(3*a**2*c*e**4 + 9*a*c**2*d**2*e**2 + 5*c**3*d**4)/e**6

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Giac [A]  time = 1.2953, size = 351, normalized size = 2.22 \begin{align*} \frac{1}{10} \,{\left (2 \, c^{3} - \frac{15 \, c^{3} d}{x e + d} + \frac{10 \,{\left (5 \, c^{3} d^{2} e^{2} + a c^{2} e^{4}\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}} - \frac{20 \,{\left (5 \, c^{3} d^{3} e^{3} + 3 \, a c^{2} d e^{5}\right )} e^{\left (-3\right )}}{{\left (x e + d\right )}^{3}} + \frac{30 \,{\left (5 \, c^{3} d^{4} e^{4} + 6 \, a c^{2} d^{2} e^{6} + a^{2} c e^{8}\right )} e^{\left (-4\right )}}{{\left (x e + d\right )}^{4}}\right )}{\left (x e + d\right )}^{5} e^{\left (-7\right )} + 6 \,{\left (c^{3} d^{5} + 2 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4}\right )} e^{\left (-7\right )} \log \left (\frac{{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) -{\left (\frac{c^{3} d^{6} e^{5}}{x e + d} + \frac{3 \, a c^{2} d^{4} e^{7}}{x e + d} + \frac{3 \, a^{2} c d^{2} e^{9}}{x e + d} + \frac{a^{3} e^{11}}{x e + d}\right )} e^{\left (-12\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+a)^3/(e*x+d)^2,x, algorithm="giac")

[Out]

1/10*(2*c^3 - 15*c^3*d/(x*e + d) + 10*(5*c^3*d^2*e^2 + a*c^2*e^4)*e^(-2)/(x*e + d)^2 - 20*(5*c^3*d^3*e^3 + 3*a
*c^2*d*e^5)*e^(-3)/(x*e + d)^3 + 30*(5*c^3*d^4*e^4 + 6*a*c^2*d^2*e^6 + a^2*c*e^8)*e^(-4)/(x*e + d)^4)*(x*e + d
)^5*e^(-7) + 6*(c^3*d^5 + 2*a*c^2*d^3*e^2 + a^2*c*d*e^4)*e^(-7)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) - (c^3*d^
6*e^5/(x*e + d) + 3*a*c^2*d^4*e^7/(x*e + d) + 3*a^2*c*d^2*e^9/(x*e + d) + a^3*e^11/(x*e + d))*e^(-12)