3.6.15 \(\int \frac {(d+e x)^2}{(a+c x^2)^3} \, dx\) [515]

Optimal. Leaf size=113 \[ -\frac {(a e-c d x) (d+e x)}{4 a c \left (a+c x^2\right )^2}-\frac {2 a d e-\left (3 c d^2+a e^2\right ) x}{8 a^2 c \left (a+c x^2\right )}+\frac {\left (3 c d^2+a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 113, 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 = {753, 653, 211} \begin {gather*} \frac {\text {ArcTan}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a e^2+3 c d^2\right )}{8 a^{5/2} c^{3/2}}-\frac {2 a d e-x \left (a e^2+3 c d^2\right )}{8 a^2 c \left (a+c x^2\right )}-\frac {(d+e x) (a e-c d x)}{4 a c \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a + c*x^2)^3,x]

[Out]

-1/4*((a*e - c*d*x)*(d + e*x))/(a*c*(a + c*x^2)^2) - (2*a*d*e - (3*c*d^2 + a*e^2)*x)/(8*a^2*c*(a + c*x^2)) + (
(3*c*d^2 + a*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 653

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

Rule 753

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]

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{\left (a+c x^2\right )^3} \, dx &=-\frac {(a e-c d x) (d+e x)}{4 a c \left (a+c x^2\right )^2}+\frac {\int \frac {3 c d^2+a e^2+2 c d e x}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac {(a e-c d x) (d+e x)}{4 a c \left (a+c x^2\right )^2}-\frac {2 a d e-\left (3 c d^2+a e^2\right ) x}{8 a^2 c \left (a+c x^2\right )}+\frac {\left (3 c d^2+a e^2\right ) \int \frac {1}{a+c x^2} \, dx}{8 a^2 c}\\ &=-\frac {(a e-c d x) (d+e x)}{4 a c \left (a+c x^2\right )^2}-\frac {2 a d e-\left (3 c d^2+a e^2\right ) x}{8 a^2 c \left (a+c x^2\right )}+\frac {\left (3 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 = 0.05, size = 101, normalized size = 0.89 \begin {gather*} \frac {3 c^2 d^2 x^3-a^2 e (4 d+e x)+a c x \left (5 d^2+e^2 x^2\right )}{8 a^2 c \left (a+c x^2\right )^2}+\frac {\left (3 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 {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.50, size = 95, normalized size = 0.84

method	result	size
default	\(\frac {\frac {\left (e^{2} a +3 c \,d^{2}\right ) x^{3}}{8 a^{2}}-\frac {\left (e^{2} a -5 c \,d^{2}\right ) x}{8 a c}-\frac {d e}{2 c}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\left (e^{2} a +3 c \,d^{2}\right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 a^{2} c \sqrt {a c}}\)	\(95\)
risch	\(\frac {\frac {\left (e^{2} a +3 c \,d^{2}\right ) x^{3}}{8 a^{2}}-\frac {\left (e^{2} a -5 c \,d^{2}\right ) x}{8 a c}-\frac {d e}{2 c}}{\left (c \,x^{2}+a \right )^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) e^{2}}{16 \sqrt {-a c}\, c a}-\frac {3 \ln \left (c x +\sqrt {-a c}\right ) d^{2}}{16 \sqrt {-a c}\, a^{2}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) e^{2}}{16 \sqrt {-a c}\, c a}+\frac {3 \ln \left (-c x +\sqrt {-a c}\right ) d^{2}}{16 \sqrt {-a c}\, a^{2}}\)	\(169\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.53, size = 109, normalized size = 0.96 \begin {gather*} \frac {{\left (3 \, c^{2} d^{2} + a c e^{2}\right )} x^{3} - 4 \, a^{2} d e + {\left (5 \, a c d^{2} - a^{2} e^{2}\right )} x}{8 \, {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} + \frac {{\left (3 \, c d^{2} + a 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)^2/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.69, size = 346, normalized size = 3.06 \begin {gather*} \left [\frac {6 \, a c^{3} d^{2} x^{3} + 10 \, a^{2} c^{2} d^{2} x - 8 \, a^{3} c d e - {\left (3 \, c^{3} d^{2} x^{4} + 6 \, a c^{2} d^{2} x^{2} + 3 \, a^{2} c d^{2} + {\left (a c^{2} x^{4} + 2 \, a^{2} c x^{2} + a^{3}\right )} e^{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^{2} c^{2} x^{3} - a^{3} c x\right )} e^{2}}{16 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, \frac {3 \, a c^{3} d^{2} x^{3} + 5 \, a^{2} c^{2} d^{2} x - 4 \, a^{3} c d e + {\left (3 \, c^{3} d^{2} x^{4} + 6 \, a c^{2} d^{2} x^{2} + 3 \, a^{2} c d^{2} + {\left (a c^{2} x^{4} + 2 \, a^{2} c x^{2} + a^{3}\right )} e^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (a^{2} c^{2} x^{3} - a^{3} c x\right )} e^{2}}{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)^2/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[1/16*(6*a*c^3*d^2*x^3 + 10*a^2*c^2*d^2*x - 8*a^3*c*d*e - (3*c^3*d^2*x^4 + 6*a*c^2*d^2*x^2 + 3*a^2*c*d^2 + (a*
c^2*x^4 + 2*a^2*c*x^2 + a^3)*e^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(a^2*c^2*x^3 -
a^3*c*x)*e^2)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), 1/8*(3*a*c^3*d^2*x^3 + 5*a^2*c^2*d^2*x - 4*a^3*c*d*e +
(3*c^3*d^2*x^4 + 6*a*c^2*d^2*x^2 + 3*a^2*c*d^2 + (a*c^2*x^4 + 2*a^2*c*x^2 + a^3)*e^2)*sqrt(a*c)*arctan(sqrt(a*
c)*x/a) + (a^2*c^2*x^3 - a^3*c*x)*e^2)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2)]

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Sympy [A]
time = 0.41, size = 172, normalized size = 1.52 \begin {gather*} - \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \left (a e^{2} + 3 c d^{2}\right ) \log {\left (- a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \left (a e^{2} + 3 c d^{2}\right ) \log {\left (a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {- 4 a^{2} d e + x^{3} \left (a c e^{2} + 3 c^{2} d^{2}\right ) + x \left (- a^{2} e^{2} + 5 a c d^{2}\right )}{8 a^{4} c + 16 a^{3} c^{2} x^{2} + 8 a^{2} c^{3} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.22, size = 95, normalized size = 0.84 \begin {gather*} \frac {{\left (3 \, c d^{2} + a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} + \frac {3 \, c^{2} d^{2} x^{3} + a c x^{3} e^{2} + 5 \, a c d^{2} x - a^{2} x e^{2} - 4 \, a^{2} d e}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.35, size = 101, normalized size = 0.89 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,c\,d^2+a\,e^2\right )}{8\,a^{5/2}\,c^{3/2}}-\frac {\frac {d\,e}{2\,c}-\frac {x^3\,\left (3\,c\,d^2+a\,e^2\right )}{8\,a^2}+\frac {x\,\left (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)^2/(a + c*x^2)^3,x)

[Out]

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