3.1352 \(\int \frac{A+B x}{(d+e x) (a+c x^2)^3} \, dx\)

Optimal. Leaf size=307 \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (a B e \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )-A c d \left (15 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right )\right )}{8 a^{5/2} \sqrt{c} \left (a e^2+c d^2\right )^3}-\frac{4 a^2 e^2 (B d-A e)+x \left (a B e \left (c d^2-3 a e^2\right )-A c d \left (7 a e^2+3 c d^2\right )\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}-\frac{a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}+\frac{e^4 \log \left (a+c x^2\right ) (B d-A e)}{2 \left (a e^2+c d^2\right )^3}-\frac{e^4 (B d-A e) \log (d+e x)}{\left (a e^2+c d^2\right )^3} \]

[Out]

-(a*(B*d - A*e) - (A*c*d + a*B*e)*x)/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) - (4*a^2*e^2*(B*d - A*e) + (a*B*e*(c*
d^2 - 3*a*e^2) - A*c*d*(3*c*d^2 + 7*a*e^2))*x)/(8*a^2*(c*d^2 + a*e^2)^2*(a + c*x^2)) - ((a*B*e*(c^2*d^4 + 6*a*
c*d^2*e^2 - 3*a^2*e^4) - A*c*d*(3*c^2*d^4 + 10*a*c*d^2*e^2 + 15*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5
/2)*Sqrt[c]*(c*d^2 + a*e^2)^3) - (e^4*(B*d - A*e)*Log[d + e*x])/(c*d^2 + a*e^2)^3 + (e^4*(B*d - A*e)*Log[a + c
*x^2])/(2*(c*d^2 + a*e^2)^3)

________________________________________________________________________________________

Rubi [A]  time = 0.505059, antiderivative size = 307, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {823, 801, 635, 205, 260} \[ -\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (a B e \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )-A c d \left (15 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right )\right )}{8 a^{5/2} \sqrt{c} \left (a e^2+c d^2\right )^3}-\frac{4 a^2 e^2 (B d-A e)+x \left (a B e \left (c d^2-3 a e^2\right )-A c d \left (7 a e^2+3 c d^2\right )\right )}{8 a^2 \left (a+c x^2\right ) \left (a e^2+c d^2\right )^2}-\frac{a (B d-A e)-x (a B e+A c d)}{4 a \left (a+c x^2\right )^2 \left (a e^2+c d^2\right )}+\frac{e^4 \log \left (a+c x^2\right ) (B d-A e)}{2 \left (a e^2+c d^2\right )^3}-\frac{e^4 (B d-A e) \log (d+e x)}{\left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(B*d - A*e) - (A*c*d + a*B*e)*x)/(4*a*(c*d^2 + a*e^2)*(a + c*x^2)^2) - (4*a^2*e^2*(B*d - A*e) + (a*B*e*(c*
d^2 - 3*a*e^2) - A*c*d*(3*c*d^2 + 7*a*e^2))*x)/(8*a^2*(c*d^2 + a*e^2)^2*(a + c*x^2)) - ((a*B*e*(c^2*d^4 + 6*a*
c*d^2*e^2 - 3*a^2*e^4) - A*c*d*(3*c^2*d^4 + 10*a*c*d^2*e^2 + 15*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5
/2)*Sqrt[c]*(c*d^2 + a*e^2)^3) - (e^4*(B*d - A*e)*Log[d + e*x])/(c*d^2 + a*e^2)^3 + (e^4*(B*d - A*e)*Log[a + c
*x^2])/(2*(c*d^2 + a*e^2)^3)

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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) \left (a+c x^2\right )^3} \, dx &=-\frac{a (B d-A e)-(A c d+a B e) x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac{\int \frac{-c \left (3 A c d^2-a B d e+4 a A e^2\right )-3 c e (A c d+a B e) x}{(d+e x) \left (a+c x^2\right )^2} \, dx}{4 a c \left (c d^2+a e^2\right )}\\ &=-\frac{a (B d-A e)-(A c d+a B e) x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac{4 a^2 e^2 (B d-A e)+\left (a B e \left (c d^2-3 a e^2\right )-A c d \left (3 c d^2+7 a e^2\right )\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{\int \frac{-c^2 \left (a B d e \left (c d^2+5 a e^2\right )-A \left (3 c^2 d^4+7 a c d^2 e^2+8 a^2 e^4\right )\right )-c^2 e \left (a B e \left (c d^2-3 a e^2\right )-A c d \left (3 c d^2+7 a e^2\right )\right ) x}{(d+e x) \left (a+c x^2\right )} \, dx}{8 a^2 c^2 \left (c d^2+a e^2\right )^2}\\ &=-\frac{a (B d-A e)-(A c d+a B e) x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac{4 a^2 e^2 (B d-A e)+\left (a B e \left (c d^2-3 a e^2\right )-A c d \left (3 c d^2+7 a e^2\right )\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}+\frac{\int \left (\frac{8 a^2 c^2 e^5 (-B d+A e)}{\left (c d^2+a e^2\right ) (d+e x)}+\frac{c^2 \left (-a B e \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+A c d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right )+8 a^2 c e^4 (B d-A e) x\right )}{\left (c d^2+a e^2\right ) \left (a+c x^2\right )}\right ) \, dx}{8 a^2 c^2 \left (c d^2+a e^2\right )^2}\\ &=-\frac{a (B d-A e)-(A c d+a B e) x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac{4 a^2 e^2 (B d-A e)+\left (a B e \left (c d^2-3 a e^2\right )-A c d \left (3 c d^2+7 a e^2\right )\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac{e^4 (B d-A e) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac{\int \frac{-a B e \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )+A c d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right )+8 a^2 c e^4 (B d-A e) x}{a+c x^2} \, dx}{8 a^2 \left (c d^2+a e^2\right )^3}\\ &=-\frac{a (B d-A e)-(A c d+a B e) x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac{4 a^2 e^2 (B d-A e)+\left (a B e \left (c d^2-3 a e^2\right )-A c d \left (3 c d^2+7 a e^2\right )\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac{e^4 (B d-A e) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac{\left (c e^4 (B d-A e)\right ) \int \frac{x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}-\frac{\left (a B e \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-A c d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right )\right ) \int \frac{1}{a+c x^2} \, dx}{8 a^2 \left (c d^2+a e^2\right )^3}\\ &=-\frac{a (B d-A e)-(A c d+a B e) x}{4 a \left (c d^2+a e^2\right ) \left (a+c x^2\right )^2}-\frac{4 a^2 e^2 (B d-A e)+\left (a B e \left (c d^2-3 a e^2\right )-A c d \left (3 c d^2+7 a e^2\right )\right ) x}{8 a^2 \left (c d^2+a e^2\right )^2 \left (a+c x^2\right )}-\frac{\left (a B e \left (c^2 d^4+6 a c d^2 e^2-3 a^2 e^4\right )-A c d \left (3 c^2 d^4+10 a c d^2 e^2+15 a^2 e^4\right )\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{8 a^{5/2} \sqrt{c} \left (c d^2+a e^2\right )^3}-\frac{e^4 (B d-A e) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac{e^4 (B d-A e) \log \left (a+c x^2\right )}{2 \left (c d^2+a e^2\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.316742, size = 263, normalized size = 0.86 \[ \frac{\frac{\left (a e^2+c d^2\right ) \left (a^2 e^2 (4 A e-4 B d+3 B e x)+a c d e x (7 A e-B d)+3 A c^2 d^3 x\right )}{a^2 \left (a+c x^2\right )}+\frac{\tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right ) \left (A c d \left (15 a^2 e^4+10 a c d^2 e^2+3 c^2 d^4\right )+a B e \left (3 a^2 e^4-6 a c d^2 e^2-c^2 d^4\right )\right )}{a^{5/2} \sqrt{c}}+\frac{2 \left (a e^2+c d^2\right )^2 (a (A e-B d+B e x)+A c d x)}{a \left (a+c x^2\right )^2}+4 e^4 \log \left (a+c x^2\right ) (B d-A e)+8 e^4 (A e-B d) \log (d+e x)}{8 \left (a e^2+c d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(c*d^2 + a*e^2)^2*(A*c*d*x + a*(-(B*d) + A*e + B*e*x)))/(a*(a + c*x^2)^2) + ((c*d^2 + a*e^2)*(3*A*c^2*d^3*
x + a*c*d*e*(-(B*d) + 7*A*e)*x + a^2*e^2*(-4*B*d + 4*A*e + 3*B*e*x)))/(a^2*(a + c*x^2)) + ((a*B*e*(-(c^2*d^4)
- 6*a*c*d^2*e^2 + 3*a^2*e^4) + A*c*d*(3*c^2*d^4 + 10*a*c*d^2*e^2 + 15*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(
a^(5/2)*Sqrt[c]) + 8*e^4*(-(B*d) + A*e)*Log[d + e*x] + 4*e^4*(B*d - A*e)*Log[a + c*x^2])/(8*(c*d^2 + a*e^2)^3)

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Maple [B]  time = 0.02, size = 1087, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/(a*e^2+c*d^2)^3/(c*x^2+a)^2*x^2*A*a*c*e^5+1/2/(a*e^2+c*d^2)^3/(c*x^2+a)^2*x^2*A*c^2*d^2*e^3-1/2/(a*e^2+c*d
^2)^3/(c*x^2+a)^2*x^2*B*c^2*d^3*e^2+7/4/(a*e^2+c*d^2)^3/(c*x^2+a)^2*x*A*c^2*d^3*e^2-1/(a*e^2+c*d^2)^3/(c*x^2+a
)^2*a*B*c*d^3*e^2+3/8/(a*e^2+c*d^2)^3/(c*x^2+a)^2*c^4/a^2*x^3*A*d^5+3/8/(a*e^2+c*d^2)^3/(c*x^2+a)^2*c*a*x^3*B*
e^5+7/8/(a*e^2+c*d^2)^3/(c*x^2+a)^2*c^2*x^3*A*d*e^4+15/8/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A
*c*d*e^4+1/8/(a*e^2+c*d^2)^3/(c*x^2+a)^2*x*B*c^2*d^4*e+1/(a*e^2+c*d^2)^3/(c*x^2+a)^2*A*d^2*a*c*e^3+5/8/(a*e^2+
c*d^2)^3/(c*x^2+a)^2*a^2*x*B*e^5+1/4/(a*e^2+c*d^2)^3/(c*x^2+a)^2*A*d^4*c^2*e-3/4/(a*e^2+c*d^2)^3/(c*x^2+a)^2*B
*a^2*d*e^4+3/8/(a*e^2+c*d^2)^3*a/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*e^5+e^5/(a*e^2+c*d^2)^3*ln(e*x+d)*A-1/2
/(a*e^2+c*d^2)^3*ln(c*x^2+a)*A*e^5+1/2/(a*e^2+c*d^2)^3*ln(c*x^2+a)*B*d*e^4-e^4/(a*e^2+c*d^2)^3*ln(e*x+d)*B*d-1
/4/(a*e^2+c*d^2)^3/(c*x^2+a)^2*B*c^2*d^5+3/4/(a*e^2+c*d^2)^3/(c*x^2+a)^2*A*a^2*e^5+1/4/(a*e^2+c*d^2)^3/(c*x^2+
a)^2*c^2*x^3*B*d^2*e^3-3/4/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*B*c*d^2*e^3+5/4/(a*e^2+c*d^2)^3
/(c*x^2+a)^2*c^3/a*x^3*A*d^3*e^2-1/8/(a*e^2+c*d^2)^3/(c*x^2+a)^2*c^3/a*x^3*B*d^4*e-1/2/(a*e^2+c*d^2)^3/(c*x^2+
a)^2*x^2*B*a*c*d*e^4+9/8/(a*e^2+c*d^2)^3/(c*x^2+a)^2*a*x*A*c*d*e^4+3/4/(a*e^2+c*d^2)^3/(c*x^2+a)^2*a*x*B*c*d^2
*e^3+5/4/(a*e^2+c*d^2)^3/a/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*c^2*d^3*e^2-1/8/(a*e^2+c*d^2)^3/a/(a*c)^(1/2)
*arctan(x*c/(a*c)^(1/2))*B*c^2*d^4*e+3/8/(a*e^2+c*d^2)^3/a^2/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*A*d^5*c^3+5/8
/(a*e^2+c*d^2)^3/(c*x^2+a)^2/a*x*A*d^5*c^3

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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)/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

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)/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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)/(c*x**2+a)**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.15664, size = 721, normalized size = 2.35 \begin{align*} \frac{{\left (B d e^{4} - A e^{5}\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 d e^{5} - A e^{6}\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 (3 \, A c^{3} d^{5} - B a c^{2} d^{4} e + 10 \, A a c^{2} d^{3} e^{2} - 6 \, B a^{2} c d^{2} e^{3} + 15 \, A a^{2} c d e^{4} + 3 \, B a^{3} e^{5}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{8 \,{\left (a^{2} c^{3} d^{6} + 3 \, a^{3} c^{2} d^{4} e^{2} + 3 \, a^{4} c d^{2} e^{4} + a^{5} e^{6}\right )} \sqrt{a c}} - \frac{2 \, B a^{2} c^{2} d^{5} - 2 \, A a^{2} c^{2} d^{4} e + 8 \, B a^{3} c d^{3} e^{2} - 8 \, A a^{3} c d^{2} e^{3} + 6 \, B a^{4} d e^{4} - 6 \, A a^{4} e^{5} -{\left (3 \, A c^{4} d^{5} - B a c^{3} d^{4} e + 10 \, A a c^{3} d^{3} e^{2} + 2 \, B a^{2} c^{2} d^{2} e^{3} + 7 \, A a^{2} c^{2} d e^{4} + 3 \, B a^{3} c e^{5}\right )} x^{3} + 4 \,{\left (B a^{2} c^{2} d^{3} e^{2} - A a^{2} c^{2} d^{2} e^{3} + B a^{3} c d e^{4} - A a^{3} c e^{5}\right )} x^{2} -{\left (5 \, A a c^{3} d^{5} + B a^{2} c^{2} d^{4} e + 14 \, A a^{2} c^{2} d^{3} e^{2} + 6 \, B a^{3} c d^{2} e^{3} + 9 \, A a^{3} c d e^{4} + 5 \, B a^{4} e^{5}\right )} x}{8 \,{\left (c d^{2} + a e^{2}\right )}^{3}{\left (c x^{2} + a\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*d*e^4 - A*e^5)*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*d*e^5 - A*e^
6)*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) + 1/8*(3*A*c^3*d^5 - B*a*c^2*d^
4*e + 10*A*a*c^2*d^3*e^2 - 6*B*a^2*c*d^2*e^3 + 15*A*a^2*c*d*e^4 + 3*B*a^3*e^5)*arctan(c*x/sqrt(a*c))/((a^2*c^3
*d^6 + 3*a^3*c^2*d^4*e^2 + 3*a^4*c*d^2*e^4 + a^5*e^6)*sqrt(a*c)) - 1/8*(2*B*a^2*c^2*d^5 - 2*A*a^2*c^2*d^4*e +
8*B*a^3*c*d^3*e^2 - 8*A*a^3*c*d^2*e^3 + 6*B*a^4*d*e^4 - 6*A*a^4*e^5 - (3*A*c^4*d^5 - B*a*c^3*d^4*e + 10*A*a*c^
3*d^3*e^2 + 2*B*a^2*c^2*d^2*e^3 + 7*A*a^2*c^2*d*e^4 + 3*B*a^3*c*e^5)*x^3 + 4*(B*a^2*c^2*d^3*e^2 - A*a^2*c^2*d^
2*e^3 + B*a^3*c*d*e^4 - A*a^3*c*e^5)*x^2 - (5*A*a*c^3*d^5 + B*a^2*c^2*d^4*e + 14*A*a^2*c^2*d^3*e^2 + 6*B*a^3*c
*d^2*e^3 + 9*A*a^3*c*d*e^4 + 5*B*a^4*e^5)*x)/((c*d^2 + a*e^2)^3*(c*x^2 + a)^2*a^2)