3.65 \(\int \frac {(d+e x)^3 (A+B x+C x^2)}{(a+c x^2)^4} \, dx\)

Optimal. Leaf size=254 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (3 a e^2+5 c d^2\right )+a \left (a e^2 (B e+3 C d)+c d^2 (3 B e+C d)\right )\right )}{16 a^{7/2} c^{5/2}}-\frac {(d+e x) \left (a e (3 a B e+a C d+5 A c d)-x \left (3 c d (3 a B e+a C d+5 A c d)+4 a e^2 (2 a C+A c)\right )\right )}{48 a^3 c^2 \left (a+c x^2\right )}-\frac {(d+e x)^2 (2 a e (2 a C+A c)-c x (3 a B e+a C d+5 A c d))}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac {(d+e x)^3 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3} \]

[Out]

-1/6*(a*B-(A*c-C*a)*x)*(e*x+d)^3/a/c/(c*x^2+a)^3-1/24*(e*x+d)^2*(2*a*(A*c+2*C*a)*e-c*(5*A*c*d+3*B*a*e+C*a*d)*x
)/a^2/c^2/(c*x^2+a)^2-1/48*(e*x+d)*(a*e*(5*A*c*d+3*B*a*e+C*a*d)-(4*a*(A*c+2*C*a)*e^2+3*c*d*(5*A*c*d+3*B*a*e+C*
a*d))*x)/a^3/c^2/(c*x^2+a)+1/16*(A*c*d*(3*a*e^2+5*c*d^2)+a*(a*e^2*(B*e+3*C*d)+c*d^2*(3*B*e+C*d)))*arctan(x*c^(
1/2)/a^(1/2))/a^(7/2)/c^(5/2)

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Rubi [A]  time = 0.54, antiderivative size = 288, normalized size of antiderivative = 1.13, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {1645, 821, 778, 205} \[ -\frac {4 a e \left (A c \left (a e^2+5 c d^2\right )+a \left (2 a C e^2+c d (3 B e+C d)\right )\right )-c x \left (A c d \left (15 c d^2-a e^2\right )+a \left (a e^2 (7 C d-3 B e)+3 c d^2 (3 B e+C d)\right )\right )}{48 a^3 c^3 \left (a+c x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (3 a e^2+5 c d^2\right )+a \left (a e^2 (B e+3 C d)+c d^2 (3 B e+C d)\right )\right )}{16 a^{7/2} c^{5/2}}-\frac {(d+e x)^2 (2 a e (2 a C+A c)-c x (3 a B e+a C d+5 A c d))}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac {(d+e x)^3 (a B-x (A c-a C))}{6 a c \left (a+c x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^3*(A + B*x + C*x^2))/(a + c*x^2)^4,x]

[Out]

-((a*B - (A*c - a*C)*x)*(d + e*x)^3)/(6*a*c*(a + c*x^2)^3) - ((d + e*x)^2*(2*a*(A*c + 2*a*C)*e - c*(5*A*c*d +
a*C*d + 3*a*B*e)*x))/(24*a^2*c^2*(a + c*x^2)^2) - (4*a*e*(A*c*(5*c*d^2 + a*e^2) + a*(2*a*C*e^2 + c*d*(C*d + 3*
B*e))) - c*(A*c*d*(15*c*d^2 - a*e^2) + a*(a*e^2*(7*C*d - 3*B*e) + 3*c*d^2*(C*d + 3*B*e)))*x)/(48*a^3*c^3*(a +
c*x^2)) + ((A*c*d*(5*c*d^2 + 3*a*e^2) + a*(a*e^2*(3*C*d + B*e) + c*d^2*(C*d + 3*B*e)))*ArcTan[(Sqrt[c]*x)/Sqrt
[a]])/(16*a^(7/2)*c^(5/2))

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])

Rule 1645

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq,
a + c*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + c
*x^2, x], x, 1]}, 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)*ExpandToSum[2*a*c*(p + 1)*(d + e*x)*Q - 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}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && GtQ[m, 0] &&  !(IGtQ[m, 0] && RationalQ[a, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rubi steps

\begin {align*} \int \frac {(d+e x)^3 \left (A+B x+C x^2\right )}{\left (a+c x^2\right )^4} \, dx &=-\frac {(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac {\int \frac {(d+e x)^2 (-5 A c d-a C d-3 a B e-2 (A c+2 a C) e x)}{\left (a+c x^2\right )^3} \, dx}{6 a c}\\ &=-\frac {(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac {(d+e x)^2 (2 a (A c+2 a C) e-c (5 A c d+a C d+3 a B e) x)}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac {\int \frac {(d+e x) \left (-4 a (A c+2 a C) e^2-3 c d (5 A c d+a C d+3 a B e)-c e (5 A c d+a C d+3 a B e) x\right )}{\left (a+c x^2\right )^2} \, dx}{24 a^2 c^2}\\ &=-\frac {(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac {(d+e x)^2 (2 a (A c+2 a C) e-c (5 A c d+a C d+3 a B e) x)}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac {4 a e \left (A c \left (5 c d^2+a e^2\right )+a \left (2 a C e^2+c d (C d+3 B e)\right )\right )-c \left (A c d \left (15 c d^2-a e^2\right )+a \left (a e^2 (7 C d-3 B e)+3 c d^2 (C d+3 B e)\right )\right ) x}{48 a^3 c^3 \left (a+c x^2\right )}+\frac {\left (A c d \left (5 c d^2+3 a e^2\right )+a \left (a e^2 (3 C d+B e)+c d^2 (C d+3 B e)\right )\right ) \int \frac {1}{a+c x^2} \, dx}{16 a^3 c^2}\\ &=-\frac {(a B-(A c-a C) x) (d+e x)^3}{6 a c \left (a+c x^2\right )^3}-\frac {(d+e x)^2 (2 a (A c+2 a C) e-c (5 A c d+a C d+3 a B e) x)}{24 a^2 c^2 \left (a+c x^2\right )^2}-\frac {4 a e \left (A c \left (5 c d^2+a e^2\right )+a \left (2 a C e^2+c d (C d+3 B e)\right )\right )-c \left (A c d \left (15 c d^2-a e^2\right )+a \left (a e^2 (7 C d-3 B e)+3 c d^2 (C d+3 B e)\right )\right ) x}{48 a^3 c^3 \left (a+c x^2\right )}+\frac {\left (A c d \left (5 c d^2+3 a e^2\right )+a \left (a e^2 (3 C d+B e)+c d^2 (C d+3 B e)\right )\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{16 a^{7/2} c^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.30, size = 350, normalized size = 1.38 \[ \frac {-\frac {3 \sqrt {a} \left (8 a^3 C e^3-a^2 c e^2 x (B e+3 C d)-a c^2 d x \left (3 e (A e+B d)+C d^2\right )-5 A c^3 d^3 x\right )}{a+c x^2}-\frac {8 a^{5/2} \left (a^3 C e^3-a^2 c e (e (A e+3 B d+B e x)+3 C d (d+e x))+a c^2 d \left (3 A e (d+e x)+B d (d+3 e x)+C d^2 x\right )-A c^3 d^3 x\right )}{\left (a+c x^2\right )^3}+\frac {2 a^{3/2} \left (12 a^3 C e^3-a^2 c e (e (6 A e+18 B d+7 B e x)+3 C d (6 d+7 e x))+a c^2 d x \left (3 e (A e+B d)+C d^2\right )+5 A c^3 d^3 x\right )}{\left (a+c x^2\right )^2}+3 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (3 a e^2+5 c d^2\right )+a \left (a e^2 (B e+3 C d)+c d^2 (3 B e+C d)\right )\right )}{48 a^{7/2} c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x)^3*(A + B*x + C*x^2))/(a + c*x^2)^4,x]

[Out]

((-3*Sqrt[a]*(8*a^3*C*e^3 - 5*A*c^3*d^3*x - a^2*c*e^2*(3*C*d + B*e)*x - a*c^2*d*(C*d^2 + 3*e*(B*d + A*e))*x))/
(a + c*x^2) - (8*a^(5/2)*(a^3*C*e^3 - A*c^3*d^3*x + a*c^2*d*(C*d^2*x + 3*A*e*(d + e*x) + B*d*(d + 3*e*x)) - a^
2*c*e*(3*C*d*(d + e*x) + e*(3*B*d + A*e + B*e*x))))/(a + c*x^2)^3 + (2*a^(3/2)*(12*a^3*C*e^3 + 5*A*c^3*d^3*x +
a*c^2*d*(C*d^2 + 3*e*(B*d + A*e))*x - a^2*c*e*(3*C*d*(6*d + 7*e*x) + e*(18*B*d + 6*A*e + 7*B*e*x))))/(a + c*x
^2)^2 + 3*Sqrt[c]*(A*c*d*(5*c*d^2 + 3*a*e^2) + a*(a*e^2*(3*C*d + B*e) + c*d^2*(C*d + 3*B*e)))*ArcTan[(Sqrt[c]*
x)/Sqrt[a]])/(48*a^(7/2)*c^3)

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fricas [B]  time = 0.83, size = 1378, normalized size = 5.43 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(C*x^2+B*x+A)/(c*x^2+a)^4,x, algorithm="fricas")

[Out]

[-1/96*(48*C*a^4*c^2*e^3*x^4 + 16*B*a^4*c^2*d^3 + 24*B*a^5*c*d*e^2 - 6*(3*B*a^2*c^4*d^2*e + B*a^3*c^3*e^3 + (C
*a^2*c^4 + 5*A*a*c^5)*d^3 + 3*(C*a^3*c^3 + A*a^2*c^4)*d*e^2)*x^5 + 24*(C*a^5*c + 2*A*a^4*c^2)*d^2*e + 8*(2*C*a
^6 + A*a^5*c)*e^3 - 16*(3*B*a^3*c^3*d^2*e - B*a^4*c^2*e^3 + (C*a^3*c^3 + 5*A*a^2*c^4)*d^3 - 3*(C*a^4*c^2 - A*a
^3*c^3)*d*e^2)*x^3 + 24*(3*C*a^4*c^2*d^2*e + 3*B*a^4*c^2*d*e^2 + (2*C*a^5*c + A*a^4*c^2)*e^3)*x^2 + 3*(3*B*a^4
*c*d^2*e + B*a^5*e^3 + (3*B*a*c^4*d^2*e + B*a^2*c^3*e^3 + (C*a*c^4 + 5*A*c^5)*d^3 + 3*(C*a^2*c^3 + A*a*c^4)*d*
e^2)*x^6 + 3*(3*B*a^2*c^3*d^2*e + B*a^3*c^2*e^3 + (C*a^2*c^3 + 5*A*a*c^4)*d^3 + 3*(C*a^3*c^2 + A*a^2*c^3)*d*e^
2)*x^4 + (C*a^4*c + 5*A*a^3*c^2)*d^3 + 3*(C*a^5 + A*a^4*c)*d*e^2 + 3*(3*B*a^3*c^2*d^2*e + B*a^4*c*e^3 + (C*a^3
*c^2 + 5*A*a^2*c^3)*d^3 + 3*(C*a^4*c + A*a^3*c^2)*d*e^2)*x^2)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x
^2 + a)) + 6*(3*B*a^4*c^2*d^2*e + B*a^5*c*e^3 + (C*a^4*c^2 - 11*A*a^3*c^3)*d^3 + 3*(C*a^5*c + A*a^4*c^2)*d*e^2
)*x)/(a^4*c^6*x^6 + 3*a^5*c^5*x^4 + 3*a^6*c^4*x^2 + a^7*c^3), -1/48*(24*C*a^4*c^2*e^3*x^4 + 8*B*a^4*c^2*d^3 +
12*B*a^5*c*d*e^2 - 3*(3*B*a^2*c^4*d^2*e + B*a^3*c^3*e^3 + (C*a^2*c^4 + 5*A*a*c^5)*d^3 + 3*(C*a^3*c^3 + A*a^2*c
^4)*d*e^2)*x^5 + 12*(C*a^5*c + 2*A*a^4*c^2)*d^2*e + 4*(2*C*a^6 + A*a^5*c)*e^3 - 8*(3*B*a^3*c^3*d^2*e - B*a^4*c
^2*e^3 + (C*a^3*c^3 + 5*A*a^2*c^4)*d^3 - 3*(C*a^4*c^2 - A*a^3*c^3)*d*e^2)*x^3 + 12*(3*C*a^4*c^2*d^2*e + 3*B*a^
4*c^2*d*e^2 + (2*C*a^5*c + A*a^4*c^2)*e^3)*x^2 - 3*(3*B*a^4*c*d^2*e + B*a^5*e^3 + (3*B*a*c^4*d^2*e + B*a^2*c^3
*e^3 + (C*a*c^4 + 5*A*c^5)*d^3 + 3*(C*a^2*c^3 + A*a*c^4)*d*e^2)*x^6 + 3*(3*B*a^2*c^3*d^2*e + B*a^3*c^2*e^3 + (
C*a^2*c^3 + 5*A*a*c^4)*d^3 + 3*(C*a^3*c^2 + A*a^2*c^3)*d*e^2)*x^4 + (C*a^4*c + 5*A*a^3*c^2)*d^3 + 3*(C*a^5 + A
*a^4*c)*d*e^2 + 3*(3*B*a^3*c^2*d^2*e + B*a^4*c*e^3 + (C*a^3*c^2 + 5*A*a^2*c^3)*d^3 + 3*(C*a^4*c + A*a^3*c^2)*d
*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + 3*(3*B*a^4*c^2*d^2*e + B*a^5*c*e^3 + (C*a^4*c^2 - 11*A*a^3*c^3)*d
^3 + 3*(C*a^5*c + A*a^4*c^2)*d*e^2)*x)/(a^4*c^6*x^6 + 3*a^5*c^5*x^4 + 3*a^6*c^4*x^2 + a^7*c^3)]

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giac [B]  time = 0.17, size = 475, normalized size = 1.87 \[ \frac {{\left (C a c d^{3} + 5 \, A c^{2} d^{3} + 3 \, B a c d^{2} e + 3 \, C a^{2} d e^{2} + 3 \, A a c d e^{2} + B a^{2} e^{3}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} + \frac {3 \, C a c^{4} d^{3} x^{5} + 15 \, A c^{5} d^{3} x^{5} + 9 \, B a c^{4} d^{2} x^{5} e + 9 \, C a^{2} c^{3} d x^{5} e^{2} + 9 \, A a c^{4} d x^{5} e^{2} + 8 \, C a^{2} c^{3} d^{3} x^{3} + 40 \, A a c^{4} d^{3} x^{3} + 3 \, B a^{2} c^{3} x^{5} e^{3} + 24 \, B a^{2} c^{3} d^{2} x^{3} e - 24 \, C a^{3} c^{2} x^{4} e^{3} - 24 \, C a^{3} c^{2} d x^{3} e^{2} + 24 \, A a^{2} c^{3} d x^{3} e^{2} - 36 \, C a^{3} c^{2} d^{2} x^{2} e - 3 \, C a^{3} c^{2} d^{3} x + 33 \, A a^{2} c^{3} d^{3} x - 8 \, B a^{3} c^{2} x^{3} e^{3} - 36 \, B a^{3} c^{2} d x^{2} e^{2} - 9 \, B a^{3} c^{2} d^{2} x e - 8 \, B a^{3} c^{2} d^{3} - 24 \, C a^{4} c x^{2} e^{3} - 12 \, A a^{3} c^{2} x^{2} e^{3} - 9 \, C a^{4} c d x e^{2} - 9 \, A a^{3} c^{2} d x e^{2} - 12 \, C a^{4} c d^{2} e - 24 \, A a^{3} c^{2} d^{2} e - 3 \, B a^{4} c x e^{3} - 12 \, B a^{4} c d e^{2} - 8 \, C a^{5} e^{3} - 4 \, A a^{4} c e^{3}}{48 \, {\left (c x^{2} + a\right )}^{3} a^{3} c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(C*x^2+B*x+A)/(c*x^2+a)^4,x, algorithm="giac")

[Out]

1/16*(C*a*c*d^3 + 5*A*c^2*d^3 + 3*B*a*c*d^2*e + 3*C*a^2*d*e^2 + 3*A*a*c*d*e^2 + B*a^2*e^3)*arctan(c*x/sqrt(a*c
))/(sqrt(a*c)*a^3*c^2) + 1/48*(3*C*a*c^4*d^3*x^5 + 15*A*c^5*d^3*x^5 + 9*B*a*c^4*d^2*x^5*e + 9*C*a^2*c^3*d*x^5*
e^2 + 9*A*a*c^4*d*x^5*e^2 + 8*C*a^2*c^3*d^3*x^3 + 40*A*a*c^4*d^3*x^3 + 3*B*a^2*c^3*x^5*e^3 + 24*B*a^2*c^3*d^2*
x^3*e - 24*C*a^3*c^2*x^4*e^3 - 24*C*a^3*c^2*d*x^3*e^2 + 24*A*a^2*c^3*d*x^3*e^2 - 36*C*a^3*c^2*d^2*x^2*e - 3*C*
a^3*c^2*d^3*x + 33*A*a^2*c^3*d^3*x - 8*B*a^3*c^2*x^3*e^3 - 36*B*a^3*c^2*d*x^2*e^2 - 9*B*a^3*c^2*d^2*x*e - 8*B*
a^3*c^2*d^3 - 24*C*a^4*c*x^2*e^3 - 12*A*a^3*c^2*x^2*e^3 - 9*C*a^4*c*d*x*e^2 - 9*A*a^3*c^2*d*x*e^2 - 12*C*a^4*c
*d^2*e - 24*A*a^3*c^2*d^2*e - 3*B*a^4*c*x*e^3 - 12*B*a^4*c*d*e^2 - 8*C*a^5*e^3 - 4*A*a^4*c*e^3)/((c*x^2 + a)^3
*a^3*c^3)

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maple [A]  time = 0.01, size = 464, normalized size = 1.83 \[ \frac {3 A d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a^{2} c}+\frac {5 A \,d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a^{3}}+\frac {B \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a \,c^{2}}+\frac {3 B \,d^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a^{2} c}+\frac {3 C d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a \,c^{2}}+\frac {C \,d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \sqrt {a c}\, a^{2} c}+\frac {-\frac {C \,e^{3} x^{4}}{2 c}+\frac {\left (3 A a c d \,e^{2}+5 A \,c^{2} d^{3}+B \,a^{2} e^{3}+3 B a c \,d^{2} e +3 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) x^{5}}{16 a^{3}}-\frac {\left (A c \,e^{2}+3 B c d e +2 a C \,e^{2}+3 C c \,d^{2}\right ) e \,x^{2}}{4 c^{2}}+\frac {\left (3 A a c d \,e^{2}+5 A \,c^{2} d^{3}-B \,a^{2} e^{3}+3 B a c \,d^{2} e -3 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) x^{3}}{6 a^{2} c}-\frac {\left (3 A a c d \,e^{2}-11 A \,c^{2} d^{3}+B \,a^{2} e^{3}+3 B a c \,d^{2} e +3 C \,a^{2} d \,e^{2}+C a c \,d^{3}\right ) x}{16 a \,c^{2}}-\frac {A c \,e^{3} a +6 A \,c^{2} d^{2} e +3 B c d \,e^{2} a +2 d^{3} c^{2} B +2 C \,a^{2} e^{3}+3 C a c \,d^{2} e}{12 c^{3}}}{\left (c \,x^{2}+a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3*(C*x^2+B*x+A)/(c*x^2+a)^4,x)

[Out]

(1/16*(3*A*a*c*d*e^2+5*A*c^2*d^3+B*a^2*e^3+3*B*a*c*d^2*e+3*C*a^2*d*e^2+C*a*c*d^3)/a^3*x^5-1/2*C/c*e^3*x^4+1/6*
(3*A*a*c*d*e^2+5*A*c^2*d^3-B*a^2*e^3+3*B*a*c*d^2*e-3*C*a^2*d*e^2+C*a*c*d^3)/a^2/c*x^3-1/4*e*(A*c*e^2+3*B*c*d*e
+2*C*a*e^2+3*C*c*d^2)/c^2*x^2-1/16*(3*A*a*c*d*e^2-11*A*c^2*d^3+B*a^2*e^3+3*B*a*c*d^2*e+3*C*a^2*d*e^2+C*a*c*d^3
)/a/c^2*x-1/12*(A*a*c*e^3+6*A*c^2*d^2*e+3*B*a*c*d*e^2+2*B*c^2*d^3+2*C*a^2*e^3+3*C*a*c*d^2*e)/c^3)/(c*x^2+a)^3+
3/16/a^2/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*d*e^2+5/16/a^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*d^3+
1/16/a/c^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*e^3+3/16/a^2/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*d^2*
e+3/16/a/c^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*C*d*e^2+1/16/a^2/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*C*
d^3

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maxima [A]  time = 1.02, size = 457, normalized size = 1.80 \[ -\frac {24 \, C a^{3} c^{2} e^{3} x^{4} + 8 \, B a^{3} c^{2} d^{3} + 12 \, B a^{4} c d e^{2} - 3 \, {\left (3 \, B a c^{4} d^{2} e + B a^{2} c^{3} e^{3} + {\left (C a c^{4} + 5 \, A c^{5}\right )} d^{3} + 3 \, {\left (C a^{2} c^{3} + A a c^{4}\right )} d e^{2}\right )} x^{5} + 12 \, {\left (C a^{4} c + 2 \, A a^{3} c^{2}\right )} d^{2} e + 4 \, {\left (2 \, C a^{5} + A a^{4} c\right )} e^{3} - 8 \, {\left (3 \, B a^{2} c^{3} d^{2} e - B a^{3} c^{2} e^{3} + {\left (C a^{2} c^{3} + 5 \, A a c^{4}\right )} d^{3} - 3 \, {\left (C a^{3} c^{2} - A a^{2} c^{3}\right )} d e^{2}\right )} x^{3} + 12 \, {\left (3 \, C a^{3} c^{2} d^{2} e + 3 \, B a^{3} c^{2} d e^{2} + {\left (2 \, C a^{4} c + A a^{3} c^{2}\right )} e^{3}\right )} x^{2} + 3 \, {\left (3 \, B a^{3} c^{2} d^{2} e + B a^{4} c e^{3} + {\left (C a^{3} c^{2} - 11 \, A a^{2} c^{3}\right )} d^{3} + 3 \, {\left (C a^{4} c + A a^{3} c^{2}\right )} d e^{2}\right )} x}{48 \, {\left (a^{3} c^{6} x^{6} + 3 \, a^{4} c^{5} x^{4} + 3 \, a^{5} c^{4} x^{2} + a^{6} c^{3}\right )}} + \frac {{\left (3 \, B a c d^{2} e + B a^{2} e^{3} + {\left (C a c + 5 \, A c^{2}\right )} d^{3} + 3 \, {\left (C a^{2} + A a c\right )} d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{16 \, \sqrt {a c} a^{3} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3*(C*x^2+B*x+A)/(c*x^2+a)^4,x, algorithm="maxima")

[Out]

-1/48*(24*C*a^3*c^2*e^3*x^4 + 8*B*a^3*c^2*d^3 + 12*B*a^4*c*d*e^2 - 3*(3*B*a*c^4*d^2*e + B*a^2*c^3*e^3 + (C*a*c
^4 + 5*A*c^5)*d^3 + 3*(C*a^2*c^3 + A*a*c^4)*d*e^2)*x^5 + 12*(C*a^4*c + 2*A*a^3*c^2)*d^2*e + 4*(2*C*a^5 + A*a^4
*c)*e^3 - 8*(3*B*a^2*c^3*d^2*e - B*a^3*c^2*e^3 + (C*a^2*c^3 + 5*A*a*c^4)*d^3 - 3*(C*a^3*c^2 - A*a^2*c^3)*d*e^2
)*x^3 + 12*(3*C*a^3*c^2*d^2*e + 3*B*a^3*c^2*d*e^2 + (2*C*a^4*c + A*a^3*c^2)*e^3)*x^2 + 3*(3*B*a^3*c^2*d^2*e +
B*a^4*c*e^3 + (C*a^3*c^2 - 11*A*a^2*c^3)*d^3 + 3*(C*a^4*c + A*a^3*c^2)*d*e^2)*x)/(a^3*c^6*x^6 + 3*a^4*c^5*x^4
+ 3*a^5*c^4*x^2 + a^6*c^3) + 1/16*(3*B*a*c*d^2*e + B*a^2*e^3 + (C*a*c + 5*A*c^2)*d^3 + 3*(C*a^2 + A*a*c)*d*e^2
)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^3*c^2)

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mupad [B]  time = 4.07, size = 402, normalized size = 1.58 \[ \frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,C\,a^2\,d\,e^2+B\,a^2\,e^3+C\,a\,c\,d^3+3\,B\,a\,c\,d^2\,e+3\,A\,a\,c\,d\,e^2+5\,A\,c^2\,d^3\right )}{16\,a^{7/2}\,c^{5/2}}-\frac {\frac {2\,C\,a^2\,e^3+3\,C\,a\,c\,d^2\,e+3\,B\,a\,c\,d\,e^2+A\,a\,c\,e^3+2\,B\,c^2\,d^3+6\,A\,c^2\,d^2\,e}{12\,c^3}+\frac {x^2\,\left (A\,c\,e^3+2\,C\,a\,e^3+3\,B\,c\,d\,e^2+3\,C\,c\,d^2\,e\right )}{4\,c^2}-\frac {x^5\,\left (3\,C\,a^2\,d\,e^2+B\,a^2\,e^3+C\,a\,c\,d^3+3\,B\,a\,c\,d^2\,e+3\,A\,a\,c\,d\,e^2+5\,A\,c^2\,d^3\right )}{16\,a^3}+\frac {C\,e^3\,x^4}{2\,c}-\frac {x^3\,\left (-3\,C\,a^2\,d\,e^2-B\,a^2\,e^3+C\,a\,c\,d^3+3\,B\,a\,c\,d^2\,e+3\,A\,a\,c\,d\,e^2+5\,A\,c^2\,d^3\right )}{6\,a^2\,c}+\frac {x\,\left (3\,C\,a^2\,d\,e^2+B\,a^2\,e^3+C\,a\,c\,d^3+3\,B\,a\,c\,d^2\,e+3\,A\,a\,c\,d\,e^2-11\,A\,c^2\,d^3\right )}{16\,a\,c^2}}{a^3+3\,a^2\,c\,x^2+3\,a\,c^2\,x^4+c^3\,x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((d + e*x)^3*(A + B*x + C*x^2))/(a + c*x^2)^4,x)

[Out]

(atan((c^(1/2)*x)/a^(1/2))*(5*A*c^2*d^3 + B*a^2*e^3 + C*a*c*d^3 + 3*C*a^2*d*e^2 + 3*A*a*c*d*e^2 + 3*B*a*c*d^2*
e))/(16*a^(7/2)*c^(5/2)) - ((2*B*c^2*d^3 + 2*C*a^2*e^3 + A*a*c*e^3 + 6*A*c^2*d^2*e + 3*B*a*c*d*e^2 + 3*C*a*c*d
^2*e)/(12*c^3) + (x^2*(A*c*e^3 + 2*C*a*e^3 + 3*B*c*d*e^2 + 3*C*c*d^2*e))/(4*c^2) - (x^5*(5*A*c^2*d^3 + B*a^2*e
^3 + C*a*c*d^3 + 3*C*a^2*d*e^2 + 3*A*a*c*d*e^2 + 3*B*a*c*d^2*e))/(16*a^3) + (C*e^3*x^4)/(2*c) - (x^3*(5*A*c^2*
d^3 - B*a^2*e^3 + C*a*c*d^3 - 3*C*a^2*d*e^2 + 3*A*a*c*d*e^2 + 3*B*a*c*d^2*e))/(6*a^2*c) + (x*(B*a^2*e^3 - 11*A
*c^2*d^3 + C*a*c*d^3 + 3*C*a^2*d*e^2 + 3*A*a*c*d*e^2 + 3*B*a*c*d^2*e))/(16*a*c^2))/(a^3 + c^3*x^6 + 3*a^2*c*x^
2 + 3*a*c^2*x^4)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3*(C*x**2+B*x+A)/(c*x**2+a)**4,x)

[Out]

Timed out

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