3.520 \(\int \frac{(d+e x)^3}{(a+c x^2)^4} \, dx\)

Optimal. Leaf size=156 \[ -\frac{4 a e \left (a e^2+5 c d^2\right )-c d x \left (15 c d^2-a e^2\right )}{48 a^3 c^2 \left (a+c x^2\right )}+\frac{d \left (3 a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}}-\frac{(d+e x)^2 (2 a e-5 c d x)}{24 a^2 c \left (a+c x^2\right )^2}+\frac{x (d+e x)^3}{6 a \left (a+c x^2\right )^3} \]

[Out]

(x*(d + e*x)^3)/(6*a*(a + c*x^2)^3) - ((2*a*e - 5*c*d*x)*(d + e*x)^2)/(24*a^2*c*(a + c*x^2)^2) - (4*a*e*(5*c*d
^2 + a*e^2) - c*d*(15*c*d^2 - a*e^2)*x)/(48*a^3*c^2*(a + c*x^2)) + (d*(5*c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x)/S
qrt[a]])/(16*a^(7/2)*c^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.118263, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.235, Rules used = {737, 821, 778, 205} \[ -\frac{4 a e \left (a e^2+5 c d^2\right )-c d x \left (15 c d^2-a e^2\right )}{48 a^3 c^2 \left (a+c x^2\right )}+\frac{d \left (3 a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}}-\frac{(d+e x)^2 (2 a e-5 c d x)}{24 a^2 c \left (a+c x^2\right )^2}+\frac{x (d+e x)^3}{6 a \left (a+c x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + c*x^2)^4,x]

[Out]

(x*(d + e*x)^3)/(6*a*(a + c*x^2)^3) - ((2*a*e - 5*c*d*x)*(d + e*x)^2)/(24*a^2*c*(a + c*x^2)^2) - (4*a*e*(5*c*d
^2 + a*e^2) - c*d*(15*c*d^2 - a*e^2)*x)/(48*a^3*c^2*(a + c*x^2)) + (d*(5*c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x)/S
qrt[a]])/(16*a^(7/2)*c^(3/2))

Rule 737

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(x*(d + e*x)^m*(a + c*x^2)^(p + 1)
)/(2*a*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(d + e*x)^(m - 1)*(d*(2*p + 3) + e*(m + 2*p + 3)*x)*(a + c*x^2
)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (LtQ[m, 1]
|| (ILtQ[m + 2*p + 3, 0] && NeQ[m, 2])) && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 821

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

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

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]

Rubi steps

\begin{align*} \int \frac{(d+e x)^3}{\left (a+c x^2\right )^4} \, dx &=\frac{x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac{\int \frac{(-5 d-2 e x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx}{6 a}\\ &=\frac{x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac{(2 a e-5 c d x) (d+e x)^2}{24 a^2 c \left (a+c x^2\right )^2}-\frac{\int \frac{(d+e x) \left (-15 c d^2-4 a e^2-5 c d e x\right )}{\left (a+c x^2\right )^2} \, dx}{24 a^2 c}\\ &=\frac{x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac{(2 a e-5 c d x) (d+e x)^2}{24 a^2 c \left (a+c x^2\right )^2}-\frac{4 a e \left (5 c d^2+a e^2\right )-c d \left (15 c d^2-a e^2\right ) x}{48 a^3 c^2 \left (a+c x^2\right )}+\frac{\left (d \left (5 c d^2+3 a e^2\right )\right ) \int \frac{1}{a+c x^2} \, dx}{16 a^3 c}\\ &=\frac{x (d+e x)^3}{6 a \left (a+c x^2\right )^3}-\frac{(2 a e-5 c d x) (d+e x)^2}{24 a^2 c \left (a+c x^2\right )^2}-\frac{4 a e \left (5 c d^2+a e^2\right )-c d \left (15 c d^2-a e^2\right ) x}{48 a^3 c^2 \left (a+c x^2\right )}+\frac{d \left (5 c d^2+3 a e^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{16 a^{7/2} c^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.152602, size = 155, normalized size = 0.99 \[ \frac{\frac{\sqrt{a} \left (3 a^2 c^2 d x \left (11 d^2+8 e^2 x^2\right )-3 a^3 c e \left (8 d^2+3 d e x+4 e^2 x^2\right )-4 a^4 e^3+a c^3 d x^3 \left (40 d^2+9 e^2 x^2\right )+15 c^4 d^3 x^5\right )}{\left (a+c x^2\right )^3}+3 \sqrt{c} d \left (3 a e^2+5 c d^2\right ) \tan ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{48 a^{7/2} c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^3/(a + c*x^2)^4,x]

[Out]

((Sqrt[a]*(-4*a^4*e^3 + 15*c^4*d^3*x^5 - 3*a^3*c*e*(8*d^2 + 3*d*e*x + 4*e^2*x^2) + 3*a^2*c^2*d*x*(11*d^2 + 8*e
^2*x^2) + a*c^3*d*x^3*(40*d^2 + 9*e^2*x^2)))/(a + c*x^2)^3 + 3*Sqrt[c]*d*(5*c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x
)/Sqrt[a]])/(48*a^(7/2)*c^2)

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 158, normalized size = 1. \begin{align*}{\frac{1}{ \left ( c{x}^{2}+a \right ) ^{3}} \left ({\frac{d \left ( 3\,a{e}^{2}+5\,c{d}^{2} \right ) c{x}^{5}}{16\,{a}^{3}}}+{\frac{d \left ( 3\,a{e}^{2}+5\,c{d}^{2} \right ){x}^{3}}{6\,{a}^{2}}}-{\frac{{e}^{3}{x}^{2}}{4\,c}}-{\frac{d \left ( 3\,a{e}^{2}-11\,c{d}^{2} \right ) x}{16\,ac}}-{\frac{e \left ( a{e}^{2}+6\,c{d}^{2} \right ) }{12\,{c}^{2}}} \right ) }+{\frac{3\,d{e}^{2}}{16\,{a}^{2}c}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}}+{\frac{5\,{d}^{3}}{16\,{a}^{3}}\arctan \left ({cx{\frac{1}{\sqrt{ac}}}} \right ){\frac{1}{\sqrt{ac}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/(c*x^2+a)^4,x)

[Out]

(1/16*d*(3*a*e^2+5*c*d^2)/a^3*c*x^5+1/6/a^2*d*(3*a*e^2+5*c*d^2)*x^3-1/4*e^3*x^2/c-1/16*d*(3*a*e^2-11*c*d^2)/a/
c*x-1/12*e*(a*e^2+6*c*d^2)/c^2)/(c*x^2+a)^3+3/16*d/a^2/c/(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))*e^2+5/16*d^3/a^3/
(a*c)^(1/2)*arctan(x*c/(a*c)^(1/2))

________________________________________________________________________________________

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((e*x+d)^3/(c*x^2+a)^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.20226, size = 1148, normalized size = 7.36 \begin{align*} \left [-\frac{24 \, a^{4} c e^{3} x^{2} + 48 \, a^{4} c d^{2} e + 8 \, a^{5} e^{3} - 6 \,{\left (5 \, a c^{4} d^{3} + 3 \, a^{2} c^{3} d e^{2}\right )} x^{5} - 16 \,{\left (5 \, a^{2} c^{3} d^{3} + 3 \, a^{3} c^{2} d e^{2}\right )} x^{3} + 3 \,{\left (5 \, a^{3} c d^{3} + 3 \, a^{4} d e^{2} +{\left (5 \, c^{4} d^{3} + 3 \, a c^{3} d e^{2}\right )} x^{6} + 3 \,{\left (5 \, a c^{3} d^{3} + 3 \, a^{2} c^{2} d e^{2}\right )} x^{4} + 3 \,{\left (5 \, a^{2} c^{2} d^{3} + 3 \, a^{3} c d e^{2}\right )} x^{2}\right )} \sqrt{-a c} \log \left (\frac{c x^{2} - 2 \, \sqrt{-a c} x - a}{c x^{2} + a}\right ) - 6 \,{\left (11 \, a^{3} c^{2} d^{3} - 3 \, a^{4} c d e^{2}\right )} x}{96 \,{\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}, -\frac{12 \, a^{4} c e^{3} x^{2} + 24 \, a^{4} c d^{2} e + 4 \, a^{5} e^{3} - 3 \,{\left (5 \, a c^{4} d^{3} + 3 \, a^{2} c^{3} d e^{2}\right )} x^{5} - 8 \,{\left (5 \, a^{2} c^{3} d^{3} + 3 \, a^{3} c^{2} d e^{2}\right )} x^{3} - 3 \,{\left (5 \, a^{3} c d^{3} + 3 \, a^{4} d e^{2} +{\left (5 \, c^{4} d^{3} + 3 \, a c^{3} d e^{2}\right )} x^{6} + 3 \,{\left (5 \, a c^{3} d^{3} + 3 \, a^{2} c^{2} d e^{2}\right )} x^{4} + 3 \,{\left (5 \, a^{2} c^{2} d^{3} + 3 \, a^{3} c d e^{2}\right )} x^{2}\right )} \sqrt{a c} \arctan \left (\frac{\sqrt{a c} x}{a}\right ) - 3 \,{\left (11 \, a^{3} c^{2} d^{3} - 3 \, a^{4} c d e^{2}\right )} x}{48 \,{\left (a^{4} c^{5} x^{6} + 3 \, a^{5} c^{4} x^{4} + 3 \, a^{6} c^{3} x^{2} + a^{7} c^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a)^4,x, algorithm="fricas")

[Out]

[-1/96*(24*a^4*c*e^3*x^2 + 48*a^4*c*d^2*e + 8*a^5*e^3 - 6*(5*a*c^4*d^3 + 3*a^2*c^3*d*e^2)*x^5 - 16*(5*a^2*c^3*
d^3 + 3*a^3*c^2*d*e^2)*x^3 + 3*(5*a^3*c*d^3 + 3*a^4*d*e^2 + (5*c^4*d^3 + 3*a*c^3*d*e^2)*x^6 + 3*(5*a*c^3*d^3 +
3*a^2*c^2*d*e^2)*x^4 + 3*(5*a^2*c^2*d^3 + 3*a^3*c*d*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*
x^2 + a)) - 6*(11*a^3*c^2*d^3 - 3*a^4*c*d*e^2)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2), -1/
48*(12*a^4*c*e^3*x^2 + 24*a^4*c*d^2*e + 4*a^5*e^3 - 3*(5*a*c^4*d^3 + 3*a^2*c^3*d*e^2)*x^5 - 8*(5*a^2*c^3*d^3 +
3*a^3*c^2*d*e^2)*x^3 - 3*(5*a^3*c*d^3 + 3*a^4*d*e^2 + (5*c^4*d^3 + 3*a*c^3*d*e^2)*x^6 + 3*(5*a*c^3*d^3 + 3*a^
2*c^2*d*e^2)*x^4 + 3*(5*a^2*c^2*d^3 + 3*a^3*c*d*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - 3*(11*a^3*c^2*d^3
- 3*a^4*c*d*e^2)*x)/(a^4*c^5*x^6 + 3*a^5*c^4*x^4 + 3*a^6*c^3*x^2 + a^7*c^2)]

________________________________________________________________________________________

Sympy [B]  time = 2.35928, size = 320, normalized size = 2.05 \begin{align*} - \frac{d \sqrt{- \frac{1}{a^{7} c^{3}}} \left (3 a e^{2} + 5 c d^{2}\right ) \log{\left (- \frac{a^{4} c d \sqrt{- \frac{1}{a^{7} c^{3}}} \left (3 a e^{2} + 5 c d^{2}\right )}{3 a d e^{2} + 5 c d^{3}} + x \right )}}{32} + \frac{d \sqrt{- \frac{1}{a^{7} c^{3}}} \left (3 a e^{2} + 5 c d^{2}\right ) \log{\left (\frac{a^{4} c d \sqrt{- \frac{1}{a^{7} c^{3}}} \left (3 a e^{2} + 5 c d^{2}\right )}{3 a d e^{2} + 5 c d^{3}} + x \right )}}{32} + \frac{- 4 a^{4} e^{3} - 24 a^{3} c d^{2} e - 12 a^{3} c e^{3} x^{2} + x^{5} \left (9 a c^{3} d e^{2} + 15 c^{4} d^{3}\right ) + x^{3} \left (24 a^{2} c^{2} d e^{2} + 40 a c^{3} d^{3}\right ) + x \left (- 9 a^{3} c d e^{2} + 33 a^{2} c^{2} d^{3}\right )}{48 a^{6} c^{2} + 144 a^{5} c^{3} x^{2} + 144 a^{4} c^{4} x^{4} + 48 a^{3} c^{5} x^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/(c*x**2+a)**4,x)

[Out]

-d*sqrt(-1/(a**7*c**3))*(3*a*e**2 + 5*c*d**2)*log(-a**4*c*d*sqrt(-1/(a**7*c**3))*(3*a*e**2 + 5*c*d**2)/(3*a*d*
e**2 + 5*c*d**3) + x)/32 + d*sqrt(-1/(a**7*c**3))*(3*a*e**2 + 5*c*d**2)*log(a**4*c*d*sqrt(-1/(a**7*c**3))*(3*a
*e**2 + 5*c*d**2)/(3*a*d*e**2 + 5*c*d**3) + x)/32 + (-4*a**4*e**3 - 24*a**3*c*d**2*e - 12*a**3*c*e**3*x**2 + x
**5*(9*a*c**3*d*e**2 + 15*c**4*d**3) + x**3*(24*a**2*c**2*d*e**2 + 40*a*c**3*d**3) + x*(-9*a**3*c*d*e**2 + 33*
a**2*c**2*d**3))/(48*a**6*c**2 + 144*a**5*c**3*x**2 + 144*a**4*c**4*x**4 + 48*a**3*c**5*x**6)

________________________________________________________________________________________

Giac [A]  time = 1.18849, size = 208, normalized size = 1.33 \begin{align*} \frac{{\left (5 \, c d^{3} + 3 \, a d e^{2}\right )} \arctan \left (\frac{c x}{\sqrt{a c}}\right )}{16 \, \sqrt{a c} a^{3} c} + \frac{15 \, c^{4} d^{3} x^{5} + 9 \, a c^{3} d x^{5} e^{2} + 40 \, a c^{3} d^{3} x^{3} + 24 \, a^{2} c^{2} d x^{3} e^{2} + 33 \, a^{2} c^{2} d^{3} x - 12 \, a^{3} c x^{2} e^{3} - 9 \, a^{3} c d x e^{2} - 24 \, a^{3} c d^{2} e - 4 \, a^{4} e^{3}}{48 \,{\left (c x^{2} + a\right )}^{3} a^{3} c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/(c*x^2+a)^4,x, algorithm="giac")

[Out]

1/16*(5*c*d^3 + 3*a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c) + 1/48*(15*c^4*d^3*x^5 + 9*a*c^3*d*x^5*e^2
+ 40*a*c^3*d^3*x^3 + 24*a^2*c^2*d*x^3*e^2 + 33*a^2*c^2*d^3*x - 12*a^3*c*x^2*e^3 - 9*a^3*c*d*x*e^2 - 24*a^3*c*d
^2*e - 4*a^4*e^3)/((c*x^2 + a)^3*a^3*c^2)