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

Optimal. Leaf size=98 \[ \frac {3 d \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}}-\frac {3 d (d+e x) (a e-c d x)}{8 a^2 c \left (a+c x^2\right )}+\frac {x (d+e x)^3}{4 a \left (a+c x^2\right )^2} \]

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Rubi [A]  time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {729, 723, 205} \begin {gather*} \frac {3 d \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}}-\frac {3 d (d+e x) (a e-c d x)}{8 a^2 c \left (a+c x^2\right )}+\frac {x (d+e x)^3}{4 a \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + c*x^2)^3,x]

[Out]

(x*(d + e*x)^3)/(4*a*(a + c*x^2)^2) - (3*d*(a*e - c*d*x)*(d + e*x))/(8*a^2*c*(a + c*x^2)) + (3*d*(c*d^2 + a*e^
2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

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 723

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

Rule 729

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^m*(2*c*x)*(a + c*x^2)^(
p + 1))/(4*a*c*(p + 1)), x] - Dist[(m*(2*c*d))/(4*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x],
 x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 0] && LtQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 127, normalized size = 1.30 \begin {gather*} \frac {\frac {\sqrt {a} \left (-2 a^3 e^3-a^2 c e \left (6 d^2+3 d e x+4 e^2 x^2\right )+a c^2 d x \left (5 d^2+3 e^2 x^2\right )+3 c^3 d^3 x^3\right )}{\left (a+c x^2\right )^2}+3 \sqrt {c} d \left (a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*(-2*a^3*e^3 + 3*c^3*d^3*x^3 + a*c^2*d*x*(5*d^2 + 3*e^2*x^2) - a^2*c*e*(6*d^2 + 3*d*e*x + 4*e^2*x^2))
)/(a + c*x^2)^2 + 3*Sqrt[c]*d*(c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^2)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^3}{\left (a+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.41, size = 406, normalized size = 4.14 \begin {gather*} \left [-\frac {8 \, a^{3} c e^{3} x^{2} + 12 \, a^{3} c d^{2} e + 4 \, a^{4} e^{3} - 6 \, {\left (a c^{3} d^{3} + a^{2} c^{2} d e^{2}\right )} x^{3} + 3 \, {\left (a^{2} c d^{3} + a^{3} d e^{2} + {\left (c^{3} d^{3} + a c^{2} d e^{2}\right )} x^{4} + 2 \, {\left (a c^{2} d^{3} + a^{2} 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 ) - 2 \, {\left (5 \, a^{2} c^{2} d^{3} - 3 \, a^{3} c d e^{2}\right )} x}{16 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac {4 \, a^{3} c e^{3} x^{2} + 6 \, a^{3} c d^{2} e + 2 \, a^{4} e^{3} - 3 \, {\left (a c^{3} d^{3} + a^{2} c^{2} d e^{2}\right )} x^{3} - 3 \, {\left (a^{2} c d^{3} + a^{3} d e^{2} + {\left (c^{3} d^{3} + a c^{2} d e^{2}\right )} x^{4} + 2 \, {\left (a c^{2} d^{3} + a^{2} c d e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, a^{2} c^{2} d^{3} - 3 \, a^{3} c d e^{2}\right )} x}{8 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 124, normalized size = 1.27 \begin {gather*} \frac {3 \, {\left (c d^{3} + a d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} + \frac {3 \, c^{3} d^{3} x^{3} + 3 \, a c^{2} d x^{3} e^{2} + 5 \, a c^{2} d^{3} x - 4 \, a^{2} c x^{2} e^{3} - 3 \, a^{2} c d x e^{2} - 6 \, a^{2} c d^{2} e - 2 \, a^{3} e^{3}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(c*d^3 + a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c) + 1/8*(3*c^3*d^3*x^3 + 3*a*c^2*d*x^3*e^2 + 5*a*c
^2*d^3*x - 4*a^2*c*x^2*e^3 - 3*a^2*c*d*x*e^2 - 6*a^2*c*d^2*e - 2*a^3*e^3)/((c*x^2 + a)^2*a^2*c^2)

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maple [A]  time = 0.05, size = 133, normalized size = 1.36 \begin {gather*} \frac {3 d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a c}+\frac {3 d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a^{2}}+\frac {-\frac {e^{3} x^{2}}{2 c}+\frac {3 \left (a \,e^{2}+c \,d^{2}\right ) d \,x^{3}}{8 a^{2}}-\frac {\left (3 a \,e^{2}-5 c \,d^{2}\right ) d x}{8 a c}-\frac {\left (a \,e^{2}+3 c \,d^{2}\right ) e}{4 c^{2}}}{\left (c \,x^{2}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3/8*d*(a*e^2+c*d^2)/a^2*x^3-1/2/c*e^3*x^2-1/8*d*(3*a*e^2-5*c*d^2)/a/c*x-1/4*e*(a*e^2+3*c*d^2)/c^2)/(c*x^2+a)^
2+3/8*d/a/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*e^2+3/8*d^3/a^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)

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maxima [A]  time = 2.95, size = 144, normalized size = 1.47 \begin {gather*} -\frac {4 \, a^{2} c e^{3} x^{2} + 6 \, a^{2} c d^{2} e + 2 \, a^{3} e^{3} - 3 \, {\left (c^{3} d^{3} + a c^{2} d e^{2}\right )} x^{3} - {\left (5 \, a c^{2} d^{3} - 3 \, a^{2} c d e^{2}\right )} x}{8 \, {\left (a^{2} c^{4} x^{4} + 2 \, a^{3} c^{3} x^{2} + a^{4} c^{2}\right )}} + \frac {3 \, {\left (c d^{3} + a d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(4*a^2*c*e^3*x^2 + 6*a^2*c*d^2*e + 2*a^3*e^3 - 3*(c^3*d^3 + a*c^2*d*e^2)*x^3 - (5*a*c^2*d^3 - 3*a^2*c*d*e
^2)*x)/(a^2*c^4*x^4 + 2*a^3*c^3*x^2 + a^4*c^2) + 3/8*(c*d^3 + a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c)

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mupad [B]  time = 0.13, size = 125, normalized size = 1.28 \begin {gather*} \frac {3\,d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (c\,d^2+a\,e^2\right )}{8\,a^{5/2}\,c^{3/2}}-\frac {\frac {e^3\,x^2}{2\,c}+\frac {e\,\left (3\,c\,d^2+a\,e^2\right )}{4\,c^2}-\frac {3\,d\,x^3\,\left (c\,d^2+a\,e^2\right )}{8\,a^2}+\frac {d\,x\,\left (3\,a\,e^2-5\,c\,d^2\right )}{8\,a\,c}}{a^2+2\,a\,c\,x^2+c^2\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^2)^3,x)

[Out]

(3*d*atan((c^(1/2)*x)/a^(1/2))*(a*e^2 + c*d^2))/(8*a^(5/2)*c^(3/2)) - ((e^3*x^2)/(2*c) + (e*(a*e^2 + 3*c*d^2))
/(4*c^2) - (3*d*x^3*(a*e^2 + c*d^2))/(8*a^2) + (d*x*(3*a*e^2 - 5*c*d^2))/(8*a*c))/(a^2 + c^2*x^4 + 2*a*c*x^2)

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sympy [B]  time = 1.27, size = 272, normalized size = 2.78 \begin {gather*} - \frac {3 d \sqrt {- \frac {1}{a^{5} c^{3}}} \left (a e^{2} + c d^{2}\right ) \log {\left (- \frac {3 a^{3} c d \sqrt {- \frac {1}{a^{5} c^{3}}} \left (a e^{2} + c d^{2}\right )}{3 a d e^{2} + 3 c d^{3}} + x \right )}}{16} + \frac {3 d \sqrt {- \frac {1}{a^{5} c^{3}}} \left (a e^{2} + c d^{2}\right ) \log {\left (\frac {3 a^{3} c d \sqrt {- \frac {1}{a^{5} c^{3}}} \left (a e^{2} + c d^{2}\right )}{3 a d e^{2} + 3 c d^{3}} + x \right )}}{16} + \frac {- 2 a^{3} e^{3} - 6 a^{2} c d^{2} e - 4 a^{2} c e^{3} x^{2} + x^{3} \left (3 a c^{2} d e^{2} + 3 c^{3} d^{3}\right ) + x \left (- 3 a^{2} c d e^{2} + 5 a c^{2} d^{3}\right )}{8 a^{4} c^{2} + 16 a^{3} c^{3} x^{2} + 8 a^{2} c^{4} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*d*sqrt(-1/(a**5*c**3))*(a*e**2 + c*d**2)*log(-3*a**3*c*d*sqrt(-1/(a**5*c**3))*(a*e**2 + c*d**2)/(3*a*d*e**2
 + 3*c*d**3) + x)/16 + 3*d*sqrt(-1/(a**5*c**3))*(a*e**2 + c*d**2)*log(3*a**3*c*d*sqrt(-1/(a**5*c**3))*(a*e**2
+ c*d**2)/(3*a*d*e**2 + 3*c*d**3) + x)/16 + (-2*a**3*e**3 - 6*a**2*c*d**2*e - 4*a**2*c*e**3*x**2 + x**3*(3*a*c
**2*d*e**2 + 3*c**3*d**3) + x*(-3*a**2*c*d*e**2 + 5*a*c**2*d**3))/(8*a**4*c**2 + 16*a**3*c**3*x**2 + 8*a**2*c*
*4*x**4)

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