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

Optimal. Leaf size=112 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a e (A e+2 B d)+A c d^2\right )}{2 a^{3/2} c^{3/2}}-\frac {(d+e x) (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )}+\frac {B e^2 \log \left (a+c x^2\right )}{2 c^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {819, 635, 205, 260} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a e (A e+2 B d)+A c d^2\right )}{2 a^{3/2} c^{3/2}}-\frac {(d+e x) (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )}+\frac {B e^2 \log \left (a+c x^2\right )}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((d + e*x)*(a*(B*d + A*e) - (A*c*d - a*B*e)*x))/(2*a*c*(a + c*x^2)) + ((A*c*d^2 + a*e*(2*B*d + A*e))*ArcTan[(
Sqrt[c]*x)/Sqrt[a]])/(2*a^(3/2)*c^(3/2)) + (B*e^2*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 819

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m - 1)*(a + c*x^2)^(p + 1)*(a*(e*f + d*g) - (c*d*f - a*e*g)*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*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, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 119, normalized size = 1.06 \begin {gather*} \frac {\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a A e^2+2 a B d e+A c d^2\right )}{a^{3/2}}+\frac {a^2 B e^2-a c (A e (2 d+e x)+B d (d+2 e x))+A c^2 d^2 x}{a \left (a+c x^2\right )}+B e^2 \log \left (a+c x^2\right )}{2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 384, normalized size = 3.43 \begin {gather*} \left [-\frac {2 \, B a^{2} c d^{2} + 4 \, A a^{2} c d e - 2 \, B a^{3} e^{2} + {\left (A a c d^{2} + 2 \, B a^{2} d e + A a^{2} e^{2} + {\left (A c^{2} d^{2} + 2 \, B a c d e + A a c 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 a c^{2} d^{2} - 2 \, B a^{2} c d e - A a^{2} c e^{2}\right )} x - 2 \, {\left (B a^{2} c e^{2} x^{2} + B a^{3} e^{2}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{3} x^{2} + a^{3} c^{2}\right )}}, -\frac {B a^{2} c d^{2} + 2 \, A a^{2} c d e - B a^{3} e^{2} - {\left (A a c d^{2} + 2 \, B a^{2} d e + A a^{2} e^{2} + {\left (A c^{2} d^{2} + 2 \, B a c d e + A a c e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (A a c^{2} d^{2} - 2 \, B a^{2} c d e - A a^{2} c e^{2}\right )} x - {\left (B a^{2} c e^{2} x^{2} + B a^{3} e^{2}\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((B*x+A)*(e*x+d)^2/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 127, normalized size = 1.13 \begin {gather*} \frac {B e^{2} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (A c d^{2} + 2 \, B a d e + A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c} + \frac {{\left (A c d^{2} - 2 \, B a d e - A a e^{2}\right )} x - \frac {B a c d^{2} + 2 \, A a c d e - B a^{2} e^{2}}{c}}{2 \, {\left (c x^{2} + a\right )} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.06, size = 151, normalized size = 1.35 \begin {gather*} \frac {A \,d^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}+\frac {A \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c}+\frac {B d e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {B \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {-\frac {\left (A a \,e^{2}-A c \,d^{2}+2 a B d e \right ) x}{2 a c}-\frac {2 A c d e -B a \,e^{2}+B c \,d^{2}}{2 c^{2}}}{c \,x^{2}+a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.28, size = 130, normalized size = 1.16 \begin {gather*} \frac {B e^{2} \log \left (c x^{2} + a\right )}{2 \, c^{2}} - \frac {B a c d^{2} + 2 \, A a c d e - B a^{2} e^{2} - {\left (A c^{2} d^{2} - 2 \, B a c d e - A a c e^{2}\right )} x}{2 \, {\left (a c^{3} x^{2} + a^{2} c^{2}\right )}} + \frac {{\left (A c d^{2} + 2 \, B a d e + A a 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((B*x+A)*(e*x+d)^2/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.80, size = 203, normalized size = 1.81 \begin {gather*} \frac {B\,a\,e^2}{2\,\left (c^3\,x^2+a\,c^2\right )}-\frac {B\,d^2}{2\,\left (c^2\,x^2+a\,c\right )}-\frac {A\,d\,e}{c^2\,x^2+a\,c}+\frac {A\,d^2\,x}{2\,\left (a^2+c\,a\,x^2\right )}-\frac {A\,e^2\,x}{2\,\left (c^2\,x^2+a\,c\right )}+\frac {B\,e^2\,\ln \left (c\,x^2+a\right )}{2\,c^2}+\frac {A\,d^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,a^{3/2}\,\sqrt {c}}+\frac {A\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,\sqrt {a}\,c^{3/2}}-\frac {B\,d\,e\,x}{c^2\,x^2+a\,c}+\frac {B\,d\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{\sqrt {a}\,c^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 2.96, size = 382, normalized size = 3.41 \begin {gather*} \left (\frac {B e^{2}}{2 c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right ) \log {\left (x + \frac {- 2 B a^{2} e^{2} + 4 a^{2} c^{2} \left (\frac {B e^{2}}{2 c^{2}} - \frac {\sqrt {- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right )}{A a c e^{2} + A c^{2} d^{2} + 2 B a c d e} \right )} + \left (\frac {B e^{2}}{2 c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right ) \log {\left (x + \frac {- 2 B a^{2} e^{2} + 4 a^{2} c^{2} \left (\frac {B e^{2}}{2 c^{2}} + \frac {\sqrt {- a^{3} c^{5}} \left (A a e^{2} + A c d^{2} + 2 B a d e\right )}{4 a^{3} c^{4}}\right )}{A a c e^{2} + A c^{2} d^{2} + 2 B a c d e} \right )} + \frac {- 2 A a c d e + B a^{2} e^{2} - B a c d^{2} + x \left (- A a c e^{2} + A c^{2} d^{2} - 2 B a c d e\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((B*x+A)*(e*x+d)**2/(c*x**2+a)**2,x)

[Out]

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

________________________________________________________________________________________