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

Optimal. Leaf size=167 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt {a} c^{5/2}}+\frac {e x \left (-a B e^2+3 A c d e+3 B c d^2\right )}{c^2}+\frac {\log \left (a+c x^2\right ) \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )}{2 c^2}+\frac {e^2 x^2 (A e+3 B d)}{2 c}+\frac {B e^3 x^3}{3 c} \]

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Rubi [A]  time = 0.18, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {801, 635, 205, 260} \begin {gather*} \frac {\log \left (a+c x^2\right ) \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )}{2 c^2}+\frac {e x \left (-a B e^2+3 A c d e+3 B c d^2\right )}{c^2}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )-a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt {a} c^{5/2}}+\frac {e^2 x^2 (A e+3 B d)}{2 c}+\frac {B e^3 x^3}{3 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(3*B*c*d^2 + 3*A*c*d*e - a*B*e^2)*x)/c^2 + (e^2*(3*B*d + A*e)*x^2)/(2*c) + (B*e^3*x^3)/(3*c) + ((A*c*d*(c*d
^2 - 3*a*e^2) - a*B*e*(3*c*d^2 - a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(5/2)) + ((B*c*d^3 + 3*A*c*d^
2*e - 3*a*B*d*e^2 - a*A*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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

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Mathematica [A]  time = 0.15, size = 151, normalized size = 0.90 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (a e^2-3 c d^2\right )\right )}{\sqrt {a} c^{5/2}}+\frac {e x \left (-6 a B e^2+3 A c e (6 d+e x)+B c \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+3 \log \left (a+c x^2\right ) \left (-a A e^3-3 a B d e^2+3 A c d^2 e+B c d^3\right )}{6 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*c*d*(c*d^2 - 3*a*e^2) + a*B*e*(-3*c*d^2 + a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(5/2)) + (e*x*(-
6*a*B*e^2 + 3*A*c*e*(6*d + e*x) + B*c*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 3*(B*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2
- a*A*e^3)*Log[a + c*x^2])/(6*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)^3}{a+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 398, normalized size = 2.38 \begin {gather*} \left [\frac {2 \, B a c^{2} e^{3} x^{3} + 3 \, {\left (3 \, B a c^{2} d e^{2} + A a c^{2} e^{3}\right )} x^{2} - 3 \, {\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 6 \, {\left (3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} x + 3 \, {\left (B a c^{2} d^{3} + 3 \, A a c^{2} d^{2} e - 3 \, B a^{2} c d e^{2} - A a^{2} c e^{3}\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}, \frac {2 \, B a c^{2} e^{3} x^{3} + 3 \, {\left (3 \, B a c^{2} d e^{2} + A a c^{2} e^{3}\right )} x^{2} + 6 \, {\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 6 \, {\left (3 \, B a c^{2} d^{2} e + 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\right )} x + 3 \, {\left (B a c^{2} d^{3} + 3 \, A a c^{2} d^{2} e - 3 \, B a^{2} c d e^{2} - A a^{2} c e^{3}\right )} \log \left (c x^{2} + a\right )}{6 \, a c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 165, normalized size = 0.99 \begin {gather*} \frac {{\left (B c d^{3} + 3 \, A c d^{2} e - 3 \, B a d e^{2} - A a e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {2 \, B c^{2} x^{3} e^{3} + 9 \, B c^{2} d x^{2} e^{2} + 18 \, B c^{2} d^{2} x e + 3 \, A c^{2} x^{2} e^{3} + 18 \, A c^{2} d x e^{2} - 6 \, B a c x e^{3}}{6 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*c*d^3 + 3*A*c*d^2*e - 3*B*a*d*e^2 - A*a*e^3)*log(c*x^2 + a)/c^2 + (A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*
d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/6*(2*B*c^2*x^3*e^3 + 9*B*c^2*d*x^2*e^2 + 18*B*c^2
*d^2*x*e + 3*A*c^2*x^2*e^3 + 18*A*c^2*d*x*e^2 - 6*B*a*c*x*e^3)/c^3

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maple [A]  time = 0.05, size = 238, normalized size = 1.43 \begin {gather*} \frac {B \,e^{3} x^{3}}{3 c}-\frac {3 A a d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {A \,e^{3} x^{2}}{2 c}+\frac {A \,d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}+\frac {B \,a^{2} e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}-\frac {3 B a \,d^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {3 B d \,e^{2} x^{2}}{2 c}-\frac {A a \,e^{3} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}+\frac {3 A \,d^{2} e \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {3 A d \,e^{2} x}{c}-\frac {3 B a d \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 c^{2}}-\frac {B a \,e^{3} x}{c^{2}}+\frac {B \,d^{3} \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {3 B \,d^{2} e x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B/c*e^3*x^3+1/2/c*e^3*A*x^2+3/2/c*e^2*B*x^2*d+3/c*e^2*A*x*d-1/c^2*e^3*B*x*a+3/c*e*B*x*d^2-1/2/c^2*ln(c*x^2
+a)*A*a*e^3+3/2/c*ln(c*x^2+a)*A*d^2*e-3/2/c^2*ln(c*x^2+a)*B*a*d*e^2+1/2/c*ln(c*x^2+a)*B*d^3-3/c/(a*c)^(1/2)*ar
ctan(1/(a*c)^(1/2)*c*x)*A*a*d*e^2+1/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*d^3+1/c^2/(a*c)^(1/2)*arctan(1/(a*
c)^(1/2)*c*x)*B*a^2*e^3-3/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*a*d^2*e

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maxima [A]  time = 1.16, size = 160, normalized size = 0.96 \begin {gather*} \frac {{\left (B c d^{3} + 3 \, A c d^{2} e - 3 \, B a d e^{2} - A a e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{2}} + \frac {{\left (A c^{2} d^{3} - 3 \, B a c d^{2} e - 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {2 \, B c e^{3} x^{3} + 3 \, {\left (3 \, B c d e^{2} + A c e^{3}\right )} x^{2} + 6 \, {\left (3 \, B c d^{2} e + 3 \, A c d e^{2} - B a e^{3}\right )} x}{6 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*c*d^3 + 3*A*c*d^2*e - 3*B*a*d*e^2 - A*a*e^3)*log(c*x^2 + a)/c^2 + (A*c^2*d^3 - 3*B*a*c*d^2*e - 3*A*a*c*
d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2) + 1/6*(2*B*c*e^3*x^3 + 3*(3*B*c*d*e^2 + A*c*e^3)*x^2
+ 6*(3*B*c*d^2*e + 3*A*c*d*e^2 - B*a*e^3)*x)/c^2

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mupad [B]  time = 1.82, size = 175, normalized size = 1.05 \begin {gather*} x\,\left (\frac {3\,d\,e\,\left (A\,e+B\,d\right )}{c}-\frac {B\,a\,e^3}{c^2}\right )+\frac {x^2\,\left (A\,e^3+3\,B\,d\,e^2\right )}{2\,c}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (B\,a^2\,e^3-3\,B\,a\,c\,d^2\,e-3\,A\,a\,c\,d\,e^2+A\,c^2\,d^3\right )}{\sqrt {a}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (-12\,B\,a^2\,c^3\,d\,e^2-4\,A\,a^2\,c^3\,e^3+4\,B\,a\,c^4\,d^3+12\,A\,a\,c^4\,d^2\,e\right )}{8\,a\,c^5}+\frac {B\,e^3\,x^3}{3\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((3*d*e*(A*e + B*d))/c - (B*a*e^3)/c^2) + (x^2*(A*e^3 + 3*B*d*e^2))/(2*c) + (atan((c^(1/2)*x)/a^(1/2))*(A*c^
2*d^3 + B*a^2*e^3 - 3*A*a*c*d*e^2 - 3*B*a*c*d^2*e))/(a^(1/2)*c^(5/2)) + (log(a + c*x^2)*(4*B*a*c^4*d^3 - 4*A*a
^2*c^3*e^3 - 12*B*a^2*c^3*d*e^2 + 12*A*a*c^4*d^2*e))/(8*a*c^5) + (B*e^3*x^3)/(3*c)

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sympy [B]  time = 2.15, size = 641, normalized size = 3.84 \begin {gather*} \frac {B e^{3} x^{3}}{3 c} + x^{2} \left (\frac {A e^{3}}{2 c} + \frac {3 B d e^{2}}{2 c}\right ) + x \left (\frac {3 A d e^{2}}{c} - \frac {B a e^{3}}{c^{2}} + \frac {3 B d^{2} e}{c}\right ) + \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right ) \log {\left (x + \frac {A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + 2 a c^{2} \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} - \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right )}{- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e} \right )} + \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right ) \log {\left (x + \frac {A a^{2} e^{3} - 3 A a c d^{2} e + 3 B a^{2} d e^{2} - B a c d^{3} + 2 a c^{2} \left (- \frac {A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} - B c d^{3}}{2 c^{2}} + \frac {\sqrt {- a c^{5}} \left (- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e\right )}{2 a c^{5}}\right )}{- 3 A a c d e^{2} + A c^{2} d^{3} + B a^{2} e^{3} - 3 B a c d^{2} e} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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