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

Optimal. Leaf size=94 \[ \frac{c x^2 \left (2 a e^2+c d^2\right )}{2 e^3}-\frac{c d x \left (2 a e^2+c d^2\right )}{e^4}+\frac{\left (a e^2+c d^2\right )^2 \log (d+e x)}{e^5}-\frac{c^2 d x^3}{3 e^2}+\frac{c^2 x^4}{4 e} \]

[Out]

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

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Rubi [A]  time = 0.0753735, antiderivative size = 94, 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} \[ \frac{c x^2 \left (2 a e^2+c d^2\right )}{2 e^3}-\frac{c d x \left (2 a e^2+c d^2\right )}{e^4}+\frac{\left (a e^2+c d^2\right )^2 \log (d+e x)}{e^5}-\frac{c^2 d x^3}{3 e^2}+\frac{c^2 x^4}{4 e} \]

Antiderivative was successfully verified.

[In]

Int[(a + c*x^2)^2/(d + e*x),x]

[Out]

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

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

Mathematica [A]  time = 0.0301811, size = 79, normalized size = 0.84 \[ \frac{c e x \left (12 a e^2 (e x-2 d)+c \left (6 d^2 e x-12 d^3-4 d e^2 x^2+3 e^3 x^3\right )\right )+12 \left (a e^2+c d^2\right )^2 \log (d+e x)}{12 e^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + c*x^2)^2/(d + e*x),x]

[Out]

(c*e*x*(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)) + 12*(c*d^2 + a*e^2)^2*Log[
d + e*x])/(12*e^5)

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Maple [A]  time = 0.045, size = 114, normalized size = 1.2 \begin{align*}{\frac{{c}^{2}{x}^{4}}{4\,e}}-{\frac{{c}^{2}d{x}^{3}}{3\,{e}^{2}}}+{\frac{c{x}^{2}a}{e}}+{\frac{{c}^{2}{x}^{2}{d}^{2}}{2\,{e}^{3}}}-2\,{\frac{acdx}{{e}^{2}}}-{\frac{{c}^{2}{d}^{3}x}{{e}^{4}}}+{\frac{\ln \left ( ex+d \right ){a}^{2}}{e}}+2\,{\frac{\ln \left ( ex+d \right ) ac{d}^{2}}{{e}^{3}}}+{\frac{{d}^{4}\ln \left ( ex+d \right ){c}^{2}}{{e}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.18506, size = 142, normalized size = 1.51 \begin{align*} \frac{3 \, c^{2} e^{3} x^{4} - 4 \, c^{2} d e^{2} x^{3} + 6 \,{\left (c^{2} d^{2} e + 2 \, a c e^{3}\right )} x^{2} - 12 \,{\left (c^{2} d^{3} + 2 \, a c d e^{2}\right )} x}{12 \, e^{4}} + \frac{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (e x + d\right )}{e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.81788, size = 223, normalized size = 2.37 \begin{align*} \frac{3 \, c^{2} e^{4} x^{4} - 4 \, c^{2} d e^{3} x^{3} + 6 \,{\left (c^{2} d^{2} e^{2} + 2 \, a c e^{4}\right )} x^{2} - 12 \,{\left (c^{2} d^{3} e + 2 \, a c d e^{3}\right )} x + 12 \,{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \log \left (e x + d\right )}{12 \, e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.579513, size = 90, normalized size = 0.96 \begin{align*} - \frac{c^{2} d x^{3}}{3 e^{2}} + \frac{c^{2} x^{4}}{4 e} + \frac{x^{2} \left (2 a c e^{2} + c^{2} d^{2}\right )}{2 e^{3}} - \frac{x \left (2 a c d e^{2} + c^{2} d^{3}\right )}{e^{4}} + \frac{\left (a e^{2} + c d^{2}\right )^{2} \log{\left (d + e x \right )}}{e^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.2773, size = 135, normalized size = 1.44 \begin{align*}{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac{1}{12} \,{\left (3 \, c^{2} x^{4} e^{3} - 4 \, c^{2} d x^{3} e^{2} + 6 \, c^{2} d^{2} x^{2} e - 12 \, c^{2} d^{3} x + 12 \, a c x^{2} e^{3} - 24 \, a c d x e^{2}\right )} e^{\left (-4\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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