3.1331 \(\int \frac{(A+B x) (d+e x)^3}{a+c x^2} \, dx\)

Optimal. Leaf size=167 \[ \frac{\log \left (a+c x^2\right ) \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )}{2 c^2}+\frac{e x \left (-a B e^2+3 A c d e+3 B c d^2\right )}{c^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a} c^{5/2}}+\frac{e^2 x^2 (A e+3 B d)}{2 c}+\frac{B e^3 x^3}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.179365, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {801, 635, 205, 260} \[ \frac{\log \left (a+c x^2\right ) \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )}{2 c^2}+\frac{e x \left (-a B e^2+3 A c d e+3 B c d^2\right )}{c^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a} c^{5/2}}+\frac{e^2 x^2 (A e+3 B d)}{2 c}+\frac{B e^3 x^3}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 801

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

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{(A+B x) (d+e x)^3}{a+c x^2} \, dx &=\int \left (\frac{e \left (3 B c d^2+3 A c d e-a B e^2\right )}{c^2}+\frac{e^2 (3 B d+A e) x}{c}+\frac{B e^3 x^2}{c}+\frac{A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x}{c^2 \left (a+c x^2\right )}\right ) \, dx\\ &=\frac{e \left (3 B c d^2+3 A c d e-a B e^2\right ) x}{c^2}+\frac{e^2 (3 B d+A e) x^2}{2 c}+\frac{B e^3 x^3}{3 c}+\frac{\int \frac{A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) x}{a+c x^2} \, dx}{c^2}\\ &=\frac{e \left (3 B c d^2+3 A c d e-a B e^2\right ) x}{c^2}+\frac{e^2 (3 B d+A e) x^2}{2 c}+\frac{B e^3 x^3}{3 c}+\frac{\left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) \int \frac{x}{a+c x^2} \, dx}{c}+\frac{\left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right ) \int \frac{1}{a+c x^2} \, dx}{c^2}\\ &=\frac{e \left (3 B c d^2+3 A c d e-a B e^2\right ) x}{c^2}+\frac{e^2 (3 B d+A e) x^2}{2 c}+\frac{B e^3 x^3}{3 c}+\frac{\left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a} c^{5/2}}+\frac{\left (B c d^3+3 A c d^2 e-3 a B d e^2-a A e^3\right ) \log \left (a+c x^2\right )}{2 c^2}\\ \end{align*}

Mathematica [A]  time = 0.156008, size = 151, normalized size = 0.9 \[ \frac{e x \left (-6 a B e^2+3 A c e (6 d+e x)+B c \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+3 \log \left (a+c x^2\right ) \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )}{6 c^2}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (a e^2-3 c d^2\right )\right )}{\sqrt{a} c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*c*d*(c*d^2 - 3*a*e^2) + a*B*e*(-3*c*d^2 + a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(5/2)) + (e*x*(-
6*a*B*e^2 + 3*A*c*e*(6*d + e*x) + B*c*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 3*(B*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2
- a*A*e^3)*Log[a + c*x^2])/(6*c^2)

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Maple [A]  time = 0.006, size = 238, normalized size = 1.4 \begin{align*}{\frac{B{e}^{3}{x}^{3}}{3\,c}}+{\frac{A{e}^{3}{x}^{2}}{2\,c}}+{\frac{3\,{e}^{2}B{x}^{2}d}{2\,c}}+3\,{\frac{Ad{e}^{2}x}{c}}-{\frac{B{e}^{3}ax}{{c}^{2}}}+3\,{\frac{Be{d}^{2}x}{c}}-{\frac{\ln \left ( c{x}^{2}+a \right ) Aa{e}^{3}}{2\,{c}^{2}}}+{\frac{3\,\ln \left ( c{x}^{2}+a \right ) A{d}^{2}e}{2\,c}}-{\frac{3\,\ln \left ( c{x}^{2}+a \right ) Bad{e}^{2}}{2\,{c}^{2}}}+{\frac{\ln \left ( c{x}^{2}+a \right ) B{d}^{3}}{2\,c}}-3\,{\frac{aAd{e}^{2}}{c\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) }+{A{d}^{3}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{{a}^{2}B{e}^{3}}{{c}^{2}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-3\,{\frac{aB{d}^{2}e}{c\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*e^3*x^3/c+1/2*e^3/c*A*x^2+3/2*e^2/c*B*x^2*d+3*e^2/c*A*d*x-e^3/c^2*a*B*x+3*e/c*B*d^2*x-1/2/c^2*ln(c*x^2+a
)*A*a*e^3+3/2/c*ln(c*x^2+a)*A*d^2*e-3/2/c^2*ln(c*x^2+a)*B*a*d*e^2+1/2/c*ln(c*x^2+a)*B*d^3-3/c/(a*c)^(1/2)*arct
an(x*c/(a*c)^(1/2))*A*a*d*e^2+1/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d^3+1/c^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(
1/2))*a^2*B*e^3-3/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*a*d^2*e

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89556, size = 860, normalized size = 5.15 \begin{align*} \left [\frac{2 \, B a c^{2} e^{3} x^{3} + 3 \,{\left (3 \, B a c^{2} d e^{2} + A a c^{2} e^{3}\right )} x^{2} - 3 \,{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \sqrt{-a c} \log \left (\frac{c x^{2} - 2 \, \sqrt{-a c} x - a}{c x^{2} + a}\right ) + 6 \,{\left (3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} x + 3 \,{\left (B a c^{2} d^{3} + 3 \, A a c^{2} d^{2} e - 3 \, B a^{2} c d e^{2} - A a^{2} c e^{3}\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}, \frac{2 \, B a c^{2} e^{3} x^{3} + 3 \,{\left (3 \, B a c^{2} d e^{2} + A a c^{2} e^{3}\right )} x^{2} + 6 \,{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) + 6 \,{\left (3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} x + 3 \,{\left (B a c^{2} d^{3} + 3 \, A a c^{2} d^{2} e - 3 \, B a^{2} c d e^{2} - A a^{2} c e^{3}\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(2*B*a*c^2*e^3*x^3 + 3*(3*B*a*c^2*d*e^2 + A*a*c^2*e^3)*x^2 - 3*(A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*d*e^2
+ B*a^2*e^3)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 6*(3*B*a*c^2*d^2*e + 3*A*a*c^2*d*e^2
- B*a^2*c*e^3)*x + 3*(B*a*c^2*d^3 + 3*A*a*c^2*d^2*e - 3*B*a^2*c*d*e^2 - A*a^2*c*e^3)*log(c*x^2 + a))/(a*c^3),
1/6*(2*B*a*c^2*e^3*x^3 + 3*(3*B*a*c^2*d*e^2 + A*a*c^2*e^3)*x^2 + 6*(A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*d*e^2
+ B*a^2*e^3)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 6*(3*B*a*c^2*d^2*e + 3*A*a*c^2*d*e^2 - B*a^2*c*e^3)*x + 3*(B*a*
c^2*d^3 + 3*A*a*c^2*d^2*e - 3*B*a^2*c*d*e^2 - A*a^2*c*e^3)*log(c*x^2 + a))/(a*c^3)]

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Sympy [B]  time = 5.01014, size = 638, normalized size = 3.82 \begin{align*} \frac{B e^{3} x^{3}}{3 c} + \left (- \frac{A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} - \frac{\sqrt{- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right ) \log{\left (x + \frac{A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + 2 a c^{2} \left (- \frac{A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} - \frac{\sqrt{- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right )}{- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \left (- \frac{A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} + \frac{\sqrt{- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right ) \log{\left (x + \frac{A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + 2 a c^{2} \left (- \frac{A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} + \frac{\sqrt{- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right )}{- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \frac{x^{2} \left (A e^{3} + 3 B d e^{2}\right )}{2 c} - \frac{x \left (- 3 A c d e^{2} + B a e^{3} - 3 B c d^{2} e\right )}{c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.19687, size = 223, normalized size = 1.34 \begin{align*} \frac{{\left (B c d^{3} + 3 \, A c d^{2} e - 3 \, B a d e^{2} - A a e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac{{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{\sqrt{a c} c^{2}} + \frac{2 \, B c^{2} x^{3} e^{3} + 9 \, B c^{2} d x^{2} e^{2} + 18 \, B c^{2} d^{2} x e + 3 \, A c^{2} x^{2} e^{3} + 18 \, A c^{2} d x e^{2} - 6 \, B a c x e^{3}}{6 \, c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*c*d^3 + 3*A*c*d^2*e - 3*B*a*d*e^2 - A*a*e^3)*log(c*x^2 + a)/c^2 + (A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*
d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/6*(2*B*c^2*x^3*e^3 + 9*B*c^2*d*x^2*e^2 + 18*B*c^2
*d^2*x*e + 3*A*c^2*x^2*e^3 + 18*A*c^2*d*x*e^2 - 6*B*a*c*x*e^3)/c^3