3.1337 \(\int \frac{A+B x}{(d+e x)^3 (a+c x^2)} \, dx\)

Optimal. Leaf size=251 \[ \frac{c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3}+\frac{B d-A e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac{-a B e^2-2 A c d e+B c d^2}{(d+e x) \left (a e^2+c d^2\right )^2}-\frac{c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{\left (a e^2+c d^2\right )^3}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a} \left (a e^2+c d^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.322575, antiderivative size = 251, 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} \[ \frac{c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3}+\frac{B d-A e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac{-a B e^2-2 A c d e+B c d^2}{(d+e x) \left (a e^2+c d^2\right )^2}-\frac{c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{\left (a e^2+c d^2\right )^3}+\frac{\sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a} \left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.405469, size = 223, normalized size = 0.89 \[ \frac{c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )+\frac{\left (a e^2+c d^2\right ) \left (B \left (c d^2 (3 d+2 e x)-a e^2 (d+2 e x)\right )-A e \left (a e^2+c d (5 d+4 e x)\right )\right )}{(d+e x)^2}-2 c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )+\frac{2 \sqrt{c} \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt{a}}}{2 \left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.011, size = 509, normalized size = 2. \begin{align*} -{\frac{Ae}{ \left ( 2\,a{e}^{2}+2\,c{d}^{2} \right ) \left ( ex+d \right ) ^{2}}}+{\frac{Bd}{ \left ( 2\,a{e}^{2}+2\,c{d}^{2} \right ) \left ( ex+d \right ) ^{2}}}-2\,{\frac{Acde}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}-{\frac{aB{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}+{\frac{Bc{d}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{2} \left ( ex+d \right ) }}-{\frac{c\ln \left ( ex+d \right ) aA{e}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+3\,{\frac{{c}^{2}\ln \left ( ex+d \right ) A{d}^{2}e}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+3\,{\frac{c\ln \left ( ex+d \right ) aBd{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-{\frac{{c}^{2}\ln \left ( ex+d \right ) B{d}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+{\frac{c\ln \left ( c{x}^{2}+a \right ) Aa{e}^{3}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-{\frac{3\,{c}^{2}\ln \left ( c{x}^{2}+a \right ) A{d}^{2}e}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-{\frac{3\,c\ln \left ( c{x}^{2}+a \right ) Bad{e}^{2}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}+{\frac{{c}^{2}\ln \left ( c{x}^{2}+a \right ) B{d}^{3}}{2\, \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}}-3\,{\frac{Ada{c}^{2}{e}^{2}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}\sqrt{ac}}\arctan \left ({\frac{cx}{\sqrt{ac}}} \right ) }+{\frac{A{c}^{3}{d}^{3}}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}-{\frac{B{e}^{3}{a}^{2}c}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+3\,{\frac{B{c}^{2}a{d}^{2}e}{ \left ( a{e}^{2}+c{d}^{2} \right ) ^{3}\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]

-1/2/(a*e^2+c*d^2)/(e*x+d)^2*A*e+1/2/(a*e^2+c*d^2)/(e*x+d)^2*B*d-2/(a*e^2+c*d^2)^2/(e*x+d)*A*c*d*e-1/(a*e^2+c*
d^2)^2/(e*x+d)*a*B*e^2+1/(a*e^2+c*d^2)^2/(e*x+d)*B*c*d^2-c/(a*e^2+c*d^2)^3*ln(e*x+d)*a*A*e^3+3*c^2/(a*e^2+c*d^
2)^3*ln(e*x+d)*A*d^2*e+3*c/(a*e^2+c*d^2)^3*ln(e*x+d)*a*B*d*e^2-c^2/(a*e^2+c*d^2)^3*ln(e*x+d)*B*d^3+1/2/(a*e^2+
c*d^2)^3*c*ln(c*x^2+a)*A*a*e^3-3/2/(a*e^2+c*d^2)^3*c^2*ln(c*x^2+a)*A*d^2*e-3/2/(a*e^2+c*d^2)^3*c*ln(c*x^2+a)*B
*a*d*e^2+1/2/(a*e^2+c*d^2)^3*c^2*ln(c*x^2+a)*B*d^3-3/(a*e^2+c*d^2)^3*c^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A
*a*d*e^2+1/(a*e^2+c*d^2)^3*c^3/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d^3-1/(a*e^2+c*d^2)^3*c/(a*c)^(1/2)*arcta
n(x*c/(a*c)^(1/2))*a^2*B*e^3+3/(a*e^2+c*d^2)^3*c^2/(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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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]

Timed out

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Giac [A]  time = 1.21858, size = 517, normalized size = 2.06 \begin{align*} \frac{{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A a c e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \,{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} - \frac{{\left (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} - 3 \, B a c d e^{3} + A a c e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{3} d^{6} e + 3 \, a c^{2} d^{4} e^{3} + 3 \, a^{2} c d^{2} e^{5} + a^{3} e^{7}} + \frac{{\left (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e - 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} \sqrt{a c}} + \frac{3 \, B c^{2} d^{5} - 5 \, A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} - B a^{2} d e^{4} - A a^{2} e^{5} + 2 \,{\left (B c^{2} d^{4} e - 2 \, A c^{2} d^{3} e^{2} - 2 \, A a c d e^{4} - B a^{2} e^{5}\right )} x}{2 \,{\left (c d^{2} + a e^{2}\right )}^{3}{\left (x e + d\right )}^{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, algorithm="giac")

[Out]

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