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

Optimal. Leaf size=109 \[ \frac {d \left (3 a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{3/2}}+\frac {e^3 \log \left (a+c x^2\right )}{2 c^2}-\frac {d e^2 x}{2 a c}-\frac {(d+e x)^2 (a e-c d x)}{2 a c \left (a+c x^2\right )} \]

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Rubi [A]  time = 0.08, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {739, 774, 635, 205, 260} \begin {gather*} \frac {d \left (3 a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{3/2}}+\frac {e^3 \log \left (a+c x^2\right )}{2 c^2}-\frac {d e^2 x}{2 a c}-\frac {(d+e x)^2 (a e-c d x)}{2 a c \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^3/(a + c*x^2)^2,x]

[Out]

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

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 739

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[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, 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, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 774

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

Rubi steps

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

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Mathematica [A]  time = 0.08, size = 107, normalized size = 0.98 \begin {gather*} \frac {\frac {\sqrt {a} \left (a^2 e^3-3 a c d e (d+e x)+a e^3 \left (a+c x^2\right ) \log \left (a+c x^2\right )+c^2 d^3 x\right )}{a+c x^2}+\sqrt {c} d \left (3 a e^2+c d^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*d*(c*d^2 + 3*a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]] + (Sqrt[a]*(a^2*e^3 + c^2*d^3*x - 3*a*c*d*e*(d + e*x)
+ a*e^3*(a + c*x^2)*Log[a + c*x^2]))/(a + c*x^2))/(2*a^(3/2)*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 )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 311, normalized size = 2.85 \begin {gather*} \left [-\frac {6 \, a^{2} c d^{2} e - 2 \, a^{3} e^{3} + {\left (a c d^{3} + 3 \, a^{2} d e^{2} + {\left (c^{2} d^{3} + 3 \, a 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 (a c^{2} d^{3} - 3 \, a^{2} c d e^{2}\right )} x - 2 \, {\left (a^{2} c e^{3} x^{2} + a^{3} e^{3}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{3} x^{2} + a^{3} c^{2}\right )}}, -\frac {3 \, a^{2} c d^{2} e - a^{3} e^{3} - {\left (a c d^{3} + 3 \, a^{2} d e^{2} + {\left (c^{2} d^{3} + 3 \, a c d e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (a c^{2} d^{3} - 3 \, a^{2} c d e^{2}\right )} x - {\left (a^{2} c e^{3} x^{2} + a^{3} e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a^{2} c^{3} x^{2} + a^{3} 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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 104, normalized size = 0.95 \begin {gather*} \frac {e^{3} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (c d^{3} + 3 \, a d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c} + \frac {{\left (c d^{3} - 3 \, a d e^{2}\right )} x - \frac {3 \, a c d^{2} e - a^{2} e^{3}}{c}}{2 \, {\left (c x^{2} + a\right )} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^3*log(c*x^2 + a)/c^2 + 1/2*(c*d^3 + 3*a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c) + 1/2*((c*d^3 - 3*a
*d*e^2)*x - (3*a*c*d^2*e - a^2*e^3)/c)/((c*x^2 + a)*a*c)

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maple [A]  time = 0.06, size = 115, normalized size = 1.06 \begin {gather*} \frac {d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}+\frac {3 d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c}+\frac {e^{3} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {-\frac {\left (3 a \,e^{2}-c \,d^{2}\right ) d x}{2 a c}+\frac {\left (a \,e^{2}-3 c \,d^{2}\right ) e}{2 c^{2}}}{c \,x^{2}+a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/2*d*(3*a*e^2-c*d^2)/a/c*x+1/2*e*(a*e^2-3*c*d^2)/c^2)/(c*x^2+a)+1/2*e^3*ln(c*x^2+a)/c^2+3/2/c/(a*c)^(1/2)*a
rctan(1/(a*c)^(1/2)*c*x)*d*e^2+1/2/a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^3

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maxima [A]  time = 2.98, size = 108, normalized size = 0.99 \begin {gather*} \frac {e^{3} \log \left (c x^{2} + a\right )}{2 \, c^{2}} - \frac {3 \, a c d^{2} e - a^{2} e^{3} - {\left (c^{2} d^{3} - 3 \, a c d e^{2}\right )} x}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}} + \frac {{\left (c d^{3} + 3 \, a d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^3*log(c*x^2 + a)/c^2 - 1/2*(3*a*c*d^2*e - a^2*e^3 - (c^2*d^3 - 3*a*c*d*e^2)*x)/(a*c^3*x^2 + a^2*c^2) + 1
/2*(c*d^3 + 3*a*d*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c)

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mupad [B]  time = 0.33, size = 143, normalized size = 1.31 \begin {gather*} \frac {d^3\,x}{2\,\left (a^2+c\,a\,x^2\right )}-\frac {3\,d^2\,e}{2\,\left (c^2\,x^2+a\,c\right )}+\frac {e^3\,\ln \left (c\,x^2+a\right )}{2\,c^2}+\frac {a\,e^3}{2\,\left (c^3\,x^2+a\,c^2\right )}+\frac {d^3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,a^{3/2}\,\sqrt {c}}-\frac {3\,d\,e^2\,x}{2\,\left (c^2\,x^2+a\,c\right )}+\frac {3\,d\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,\sqrt {a}\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^3/(a + c*x^2)^2,x)

[Out]

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

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sympy [B]  time = 1.01, size = 298, normalized size = 2.73 \begin {gather*} \left (\frac {e^{3}}{2 c^{2}} - \frac {d \sqrt {- a^{3} c^{5}} \left (3 a e^{2} + c d^{2}\right )}{4 a^{3} c^{4}}\right ) \log {\left (x + \frac {4 a^{2} c^{2} \left (\frac {e^{3}}{2 c^{2}} - \frac {d \sqrt {- a^{3} c^{5}} \left (3 a e^{2} + c d^{2}\right )}{4 a^{3} c^{4}}\right ) - 2 a^{2} e^{3}}{3 a c d e^{2} + c^{2} d^{3}} \right )} + \left (\frac {e^{3}}{2 c^{2}} + \frac {d \sqrt {- a^{3} c^{5}} \left (3 a e^{2} + c d^{2}\right )}{4 a^{3} c^{4}}\right ) \log {\left (x + \frac {4 a^{2} c^{2} \left (\frac {e^{3}}{2 c^{2}} + \frac {d \sqrt {- a^{3} c^{5}} \left (3 a e^{2} + c d^{2}\right )}{4 a^{3} c^{4}}\right ) - 2 a^{2} e^{3}}{3 a c d e^{2} + c^{2} d^{3}} \right )} + \frac {a^{2} e^{3} - 3 a c d^{2} e + x \left (- 3 a c d e^{2} + c^{2} d^{3}\right )}{2 a^{2} c^{2} + 2 a c^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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