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

Optimal. Leaf size=142 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a A e^2+2 a B d e+3 A c d^2\right )}{8 a^{5/2} c^{3/2}}-\frac {2 a e (a B e+2 A c d)-c x \left (-a A e^2+2 a B d e+3 A c d^2\right )}{8 a^2 c^2 \left (a+c x^2\right )}-\frac {(d+e x)^2 (a B-A c x)}{4 a c \left (a+c x^2\right )^2} \]

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Rubi [A]  time = 0.13, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {821, 778, 205} \begin {gather*} -\frac {2 a e (a B e+2 A c d)-c x \left (-a A e^2+2 a B d e+3 A c d^2\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a A e^2+2 a B d e+3 A c d^2\right )}{8 a^{5/2} c^{3/2}}-\frac {(d+e x)^2 (a B-A c x)}{4 a c \left (a+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*B - A*c*x)*(d + e*x)^2)/(4*a*c*(a + c*x^2)^2) - (2*a*e*(2*A*c*d + a*B*e) - c*(3*A*c*d^2 + 2*a*B*d*e - a*A
*e^2)*x)/(8*a^2*c^2*(a + c*x^2)) + ((3*A*c*d^2 + 2*a*B*d*e + a*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 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 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])

Rubi steps

\begin {align*} \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx &=-\frac {(a B-A c x) (d+e x)^2}{4 a c \left (a+c x^2\right )^2}+\frac {\int \frac {(d+e x) (3 A c d+2 a B e+A c e x)}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac {(a B-A c x) (d+e x)^2}{4 a c \left (a+c x^2\right )^2}-\frac {2 a e (2 A c d+a B e)-c \left (3 A c d^2+2 a B d e-a A e^2\right ) x}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (3 A c d^2+2 a B d e+a A e^2\right ) \int \frac {1}{a+c x^2} \, dx}{8 a^2 c}\\ &=-\frac {(a B-A c x) (d+e x)^2}{4 a c \left (a+c x^2\right )^2}-\frac {2 a e (2 A c d+a B e)-c \left (3 A c d^2+2 a B d e-a A e^2\right ) x}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (3 A c d^2+2 a B d e+a 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 = 0.11, size = 158, normalized size = 1.11 \begin {gather*} \frac {\frac {\sqrt {a} \left (-4 a^2 B e^2+a c e x (A e+2 B d)+3 A c^2 d^2 x\right )}{a+c x^2}+\frac {2 a^{3/2} \left (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\right )}{\left (a+c x^2\right )^2}+\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a A e^2+2 a B d e+3 A c d^2\right )}{8 a^{5/2} c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((Sqrt[a]*(-4*a^2*B*e^2 + 3*A*c^2*d^2*x + a*c*e*(2*B*d + A*e)*x))/(a + c*x^2) + (2*a^(3/2)*(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 + c*x^2)^2 + Sqrt[c]*(3*A*c*d^2 + 2*a*B*d*e + a*A*e^2)*Ar
cTan[(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 {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 537, normalized size = 3.78 \begin {gather*} \left [-\frac {8 \, B a^{3} c e^{2} x^{2} + 4 \, B a^{3} c d^{2} + 8 \, A a^{3} c d e + 4 \, B a^{4} e^{2} - 2 \, {\left (3 \, A a c^{3} d^{2} + 2 \, B a^{2} c^{2} d e + A a^{2} c^{2} e^{2}\right )} x^{3} + {\left (3 \, A a^{2} c d^{2} + 2 \, B a^{3} d e + A a^{3} e^{2} + {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d^{2} + 2 \, B a^{2} c d e + A a^{2} 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 (5 \, A a^{2} c^{2} d^{2} - 2 \, B a^{3} c d e - A a^{3} c 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 \, B a^{3} c e^{2} x^{2} + 2 \, B a^{3} c d^{2} + 4 \, A a^{3} c d e + 2 \, B a^{4} e^{2} - {\left (3 \, A a c^{3} d^{2} + 2 \, B a^{2} c^{2} d e + A a^{2} c^{2} e^{2}\right )} x^{3} - {\left (3 \, A a^{2} c d^{2} + 2 \, B a^{3} d e + A a^{3} e^{2} + {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d^{2} + 2 \, B a^{2} c d e + A a^{2} c e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, A a^{2} c^{2} d^{2} - 2 \, B a^{3} c d e - A a^{3} c 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((B*x+A)*(e*x+d)^2/(c*x^2+a)^3,x, algorithm="fricas")

[Out]

[-1/16*(8*B*a^3*c*e^2*x^2 + 4*B*a^3*c*d^2 + 8*A*a^3*c*d*e + 4*B*a^4*e^2 - 2*(3*A*a*c^3*d^2 + 2*B*a^2*c^2*d*e +
A*a^2*c^2*e^2)*x^3 + (3*A*a^2*c*d^2 + 2*B*a^3*d*e + A*a^3*e^2 + (3*A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x
^4 + 2*(3*A*a*c^2*d^2 + 2*B*a^2*c*d*e + A*a^2*c*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 +
a)) - 2*(5*A*a^2*c^2*d^2 - 2*B*a^3*c*d*e - A*a^3*c*e^2)*x)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/8*(4*B
*a^3*c*e^2*x^2 + 2*B*a^3*c*d^2 + 4*A*a^3*c*d*e + 2*B*a^4*e^2 - (3*A*a*c^3*d^2 + 2*B*a^2*c^2*d*e + A*a^2*c^2*e^
2)*x^3 - (3*A*a^2*c*d^2 + 2*B*a^3*d*e + A*a^3*e^2 + (3*A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x^4 + 2*(3*A*a
*c^2*d^2 + 2*B*a^2*c*d*e + A*a^2*c*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (5*A*a^2*c^2*d^2 - 2*B*a^3*c*d*
e - A*a^3*c*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 = 0.16, size = 169, normalized size = 1.19 \begin {gather*} \frac {{\left (3 \, A c d^{2} + 2 \, B a d e + A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} + \frac {3 \, A c^{3} d^{2} x^{3} + 2 \, B a c^{2} d x^{3} e + A a c^{2} x^{3} e^{2} + 5 \, A a c^{2} d^{2} x - 4 \, B a^{2} c x^{2} e^{2} - 2 \, B a^{2} c d x e - 2 \, B a^{2} c d^{2} - A a^{2} c x e^{2} - 4 \, A a^{2} c d e - 2 \, B a^{3} e^{2}}{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((B*x+A)*(e*x+d)^2/(c*x^2+a)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 180, normalized size = 1.27 \begin {gather*} \frac {A \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a c}+\frac {3 A \,d^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a^{2}}+\frac {B d e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{4 \sqrt {a c}\, a c}+\frac {-\frac {B \,e^{2} x^{2}}{2 c}+\frac {\left (A a \,e^{2}+3 A c \,d^{2}+2 a B d e \right ) x^{3}}{8 a^{2}}-\frac {\left (A a \,e^{2}-5 A c \,d^{2}+2 a B d e \right ) x}{8 a c}-\frac {2 A c d e +B a \,e^{2}+B c \,d^{2}}{4 c^{2}}}{\left (c \,x^{2}+a \right )^{2}} \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)^3,x)

[Out]

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

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maxima [A]  time = 1.18, size = 184, normalized size = 1.30 \begin {gather*} -\frac {4 \, B a^{2} c e^{2} x^{2} + 2 \, B a^{2} c d^{2} + 4 \, A a^{2} c d e + 2 \, B a^{3} e^{2} - {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{3} - {\left (5 \, A a c^{2} d^{2} - 2 \, B a^{2} c d e - A a^{2} c 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 {{\left (3 \, A c d^{2} + 2 \, B a d e + A 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((B*x+A)*(e*x+d)^2/(c*x^2+a)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.81, size = 154, normalized size = 1.08 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a^{5/2}\,c^{3/2}}-\frac {\frac {B\,c\,d^2+2\,A\,c\,d\,e+B\,a\,e^2}{4\,c^2}-\frac {x^3\,\left (3\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a^2}+\frac {x\,\left (-5\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a\,c}+\frac {B\,e^2\,x^2}{2\,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(((A + B*x)*(d + e*x)^2)/(a + c*x^2)^3,x)

[Out]

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

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sympy [A]  time = 7.84, size = 274, normalized size = 1.93 \begin {gather*} - \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \left (A a e^{2} + 3 A c d^{2} + 2 B a d e\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 a e^{2} + 3 A c d^{2} + 2 B a d e\right ) \log {\left (a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {- 4 A a^{2} c d e - 2 B a^{3} e^{2} - 2 B a^{2} c d^{2} - 4 B a^{2} c e^{2} x^{2} + x^{3} \left (A a c^{2} e^{2} + 3 A c^{3} d^{2} + 2 B a c^{2} d e\right ) + x \left (- A a^{2} c e^{2} + 5 A a c^{2} d^{2} - 2 B a^{2} c d e\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((B*x+A)*(e*x+d)**2/(c*x**2+a)**3,x)

[Out]

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

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