3.1.22 \(\int \frac {(a+b x^2) (c+d x^2)^3}{(e+f x^2)^4} \, dx\)

Optimal. Leaf size=348 \[ \frac {d x \left (b e \left (-3 c^2 f^2-10 c d e f+105 d^2 e^2\right )-a f \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )}{48 e^3 f^4}-\frac {x \left (c+d x^2\right ) \left (b e \left (-3 c^2 f^2-8 c d e f+35 d^2 e^2\right )-a f \left (15 c^2 f^2+4 c d e f+5 d^2 e^2\right )\right )}{48 e^3 f^3 \left (e+f x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-c^3 f^3-3 c^2 d e f^2-15 c d^2 e^2 f+35 d^3 e^3\right )-a f \left (5 c^3 f^3+3 c^2 d e f^2+3 c d^2 e^2 f+5 d^3 e^3\right )\right )}{16 e^{7/2} f^{9/2}}-\frac {x \left (c+d x^2\right )^2 (b e (7 d e-c f)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.45, antiderivative size = 348, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {526, 388, 205} \begin {gather*} -\frac {x \left (c+d x^2\right ) \left (b e \left (-3 c^2 f^2-8 c d e f+35 d^2 e^2\right )-a f \left (15 c^2 f^2+4 c d e f+5 d^2 e^2\right )\right )}{48 e^3 f^3 \left (e+f x^2\right )}+\frac {d x \left (b e \left (-3 c^2 f^2-10 c d e f+105 d^2 e^2\right )-a f \left (15 c^2 f^2+14 c d e f+15 d^2 e^2\right )\right )}{48 e^3 f^4}-\frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-3 c^2 d e f^2-c^3 f^3-15 c d^2 e^2 f+35 d^3 e^3\right )-a f \left (3 c^2 d e f^2+5 c^3 f^3+3 c d^2 e^2 f+5 d^3 e^3\right )\right )}{16 e^{7/2} f^{9/2}}-\frac {x \left (c+d x^2\right )^2 (b e (7 d e-c f)-a f (5 c f+d e))}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{6 e f \left (e+f x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)*(c + d*x^2)^3)/(e + f*x^2)^4,x]

[Out]

(d*(b*e*(105*d^2*e^2 - 10*c*d*e*f - 3*c^2*f^2) - a*f*(15*d^2*e^2 + 14*c*d*e*f + 15*c^2*f^2))*x)/(48*e^3*f^4) -
((b*e - a*f)*x*(c + d*x^2)^3)/(6*e*f*(e + f*x^2)^3) - ((b*e*(7*d*e - c*f) - a*f*(d*e + 5*c*f))*x*(c + d*x^2)^
2)/(24*e^2*f^2*(e + f*x^2)^2) - ((b*e*(35*d^2*e^2 - 8*c*d*e*f - 3*c^2*f^2) - a*f*(5*d^2*e^2 + 4*c*d*e*f + 15*c
^2*f^2))*x*(c + d*x^2))/(48*e^3*f^3*(e + f*x^2)) - ((b*e*(35*d^3*e^3 - 15*c*d^2*e^2*f - 3*c^2*d*e*f^2 - c^3*f^
3) - a*f*(5*d^3*e^3 + 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 + 5*c^3*f^3))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(16*e^(7/2)*f^(
9/2))

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 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 526

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*b*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n
)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + b*e - a*f) + d*(b*e*n*(p + 1) + (b*e - a*f)*(n*q + 1))*x
^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && GtQ[q, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx &=-\frac {(b e-a f) x \left (c+d x^2\right )^3}{6 e f \left (e+f x^2\right )^3}-\frac {\int \frac {\left (c+d x^2\right )^2 \left (-c (b e+5 a f)-d (7 b e-a f) x^2\right )}{\left (e+f x^2\right )^3} \, dx}{6 e f}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^3}{6 e f \left (e+f x^2\right )^3}-\frac {(b e (7 d e-c f)-a f (d e+5 c f)) x \left (c+d x^2\right )^2}{24 e^2 f^2 \left (e+f x^2\right )^2}+\frac {\int \frac {\left (c+d x^2\right ) \left (c (d e (7 b e-a f)+3 c f (b e+5 a f))+d (b e (35 d e-c f)-5 a f (d e+c f)) x^2\right )}{\left (e+f x^2\right )^2} \, dx}{24 e^2 f^2}\\ &=-\frac {(b e-a f) x \left (c+d x^2\right )^3}{6 e f \left (e+f x^2\right )^3}-\frac {(b e (7 d e-c f)-a f (d e+5 c f)) x \left (c+d x^2\right )^2}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {\left (b e \left (35 d^2 e^2-8 c d e f-3 c^2 f^2\right )-a f \left (5 d^2 e^2+4 c d e f+15 c^2 f^2\right )\right ) x \left (c+d x^2\right )}{48 e^3 f^3 \left (e+f x^2\right )}-\frac {\int \frac {c \left (a f \left (5 d^2 e^2+6 c d e f-15 c^2 f^2\right )-b e \left (35 d^2 e^2+6 c d e f+3 c^2 f^2\right )\right )-d \left (b e \left (105 d^2 e^2-10 c d e f-3 c^2 f^2\right )-a f \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )\right ) x^2}{e+f x^2} \, dx}{48 e^3 f^3}\\ &=\frac {d \left (b e \left (105 d^2 e^2-10 c d e f-3 c^2 f^2\right )-a f \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )\right ) x}{48 e^3 f^4}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{6 e f \left (e+f x^2\right )^3}-\frac {(b e (7 d e-c f)-a f (d e+5 c f)) x \left (c+d x^2\right )^2}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {\left (b e \left (35 d^2 e^2-8 c d e f-3 c^2 f^2\right )-a f \left (5 d^2 e^2+4 c d e f+15 c^2 f^2\right )\right ) x \left (c+d x^2\right )}{48 e^3 f^3 \left (e+f x^2\right )}-\frac {\left (b e \left (35 d^3 e^3-15 c d^2 e^2 f-3 c^2 d e f^2-c^3 f^3\right )-a f \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )\right ) \int \frac {1}{e+f x^2} \, dx}{16 e^3 f^4}\\ &=\frac {d \left (b e \left (105 d^2 e^2-10 c d e f-3 c^2 f^2\right )-a f \left (15 d^2 e^2+14 c d e f+15 c^2 f^2\right )\right ) x}{48 e^3 f^4}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{6 e f \left (e+f x^2\right )^3}-\frac {(b e (7 d e-c f)-a f (d e+5 c f)) x \left (c+d x^2\right )^2}{24 e^2 f^2 \left (e+f x^2\right )^2}-\frac {\left (b e \left (35 d^2 e^2-8 c d e f-3 c^2 f^2\right )-a f \left (5 d^2 e^2+4 c d e f+15 c^2 f^2\right )\right ) x \left (c+d x^2\right )}{48 e^3 f^3 \left (e+f x^2\right )}-\frac {\left (b e \left (35 d^3 e^3-15 c d^2 e^2 f-3 c^2 d e f^2-c^3 f^3\right )-a f \left (5 d^3 e^3+3 c d^2 e^2 f+3 c^2 d e f^2+5 c^3 f^3\right )\right ) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{16 e^{7/2} f^{9/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 295, normalized size = 0.85 \begin {gather*} \frac {x (d e-c f) \left (b e \left (-c^2 f^2-4 c d e f+29 d^2 e^2\right )-a f \left (5 c^2 f^2+8 c d e f+11 d^2 e^2\right )\right )}{16 e^3 f^4 \left (e+f x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-c^3 f^3-3 c^2 d e f^2-15 c d^2 e^2 f+35 d^3 e^3\right )-a f \left (5 c^3 f^3+3 c^2 d e f^2+3 c d^2 e^2 f+5 d^3 e^3\right )\right )}{16 e^{7/2} f^{9/2}}-\frac {x (d e-c f)^2 (b e (19 d e-c f)-a f (5 c f+13 d e))}{24 e^2 f^4 \left (e+f x^2\right )^2}+\frac {x (b e-a f) (d e-c f)^3}{6 e f^4 \left (e+f x^2\right )^3}+\frac {b d^3 x}{f^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)*(c + d*x^2)^3)/(e + f*x^2)^4,x]

[Out]

(b*d^3*x)/f^4 + ((b*e - a*f)*(d*e - c*f)^3*x)/(6*e*f^4*(e + f*x^2)^3) - ((d*e - c*f)^2*(b*e*(19*d*e - c*f) - a
*f*(13*d*e + 5*c*f))*x)/(24*e^2*f^4*(e + f*x^2)^2) + ((d*e - c*f)*(b*e*(29*d^2*e^2 - 4*c*d*e*f - c^2*f^2) - a*
f*(11*d^2*e^2 + 8*c*d*e*f + 5*c^2*f^2))*x)/(16*e^3*f^4*(e + f*x^2)) - ((b*e*(35*d^3*e^3 - 15*c*d^2*e^2*f - 3*c
^2*d*e*f^2 - c^3*f^3) - a*f*(5*d^3*e^3 + 3*c*d^2*e^2*f + 3*c^2*d*e*f^2 + 5*c^3*f^3))*ArcTan[(Sqrt[f]*x)/Sqrt[e
]])/(16*e^(7/2)*f^(9/2))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a+b x^2\right ) \left (c+d x^2\right )^3}{\left (e+f x^2\right )^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((a + b*x^2)*(c + d*x^2)^3)/(e + f*x^2)^4,x]

[Out]

IntegrateAlgebraic[((a + b*x^2)*(c + d*x^2)^3)/(e + f*x^2)^4, x]

________________________________________________________________________________________

fricas [B]  time = 1.38, size = 1422, normalized size = 4.09

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/96*(96*b*d^3*e^4*f^4*x^7 + 6*(77*b*d^3*e^5*f^3 + 5*a*c^3*e*f^7 - 11*(3*b*c*d^2 + a*d^3)*e^4*f^4 + 3*(b*c^2*
d + a*c*d^2)*e^3*f^5 + (b*c^3 + 3*a*c^2*d)*e^2*f^6)*x^5 + 16*(35*b*d^3*e^6*f^2 + 5*a*c^3*e^2*f^6 - 5*(3*b*c*d^
2 + a*d^3)*e^5*f^3 - 3*(b*c^2*d + a*c*d^2)*e^4*f^4 + (b*c^3 + 3*a*c^2*d)*e^3*f^5)*x^3 + 3*(35*b*d^3*e^7 - 5*a*
c^3*e^3*f^4 - 5*(3*b*c*d^2 + a*d^3)*e^6*f - 3*(b*c^2*d + a*c*d^2)*e^5*f^2 - (b*c^3 + 3*a*c^2*d)*e^4*f^3 + (35*
b*d^3*e^4*f^3 - 5*a*c^3*f^7 - 5*(3*b*c*d^2 + a*d^3)*e^3*f^4 - 3*(b*c^2*d + a*c*d^2)*e^2*f^5 - (b*c^3 + 3*a*c^2
*d)*e*f^6)*x^6 + 3*(35*b*d^3*e^5*f^2 - 5*a*c^3*e*f^6 - 5*(3*b*c*d^2 + a*d^3)*e^4*f^3 - 3*(b*c^2*d + a*c*d^2)*e
^3*f^4 - (b*c^3 + 3*a*c^2*d)*e^2*f^5)*x^4 + 3*(35*b*d^3*e^6*f - 5*a*c^3*e^2*f^5 - 5*(3*b*c*d^2 + a*d^3)*e^5*f^
2 - 3*(b*c^2*d + a*c*d^2)*e^4*f^3 - (b*c^3 + 3*a*c^2*d)*e^3*f^4)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x -
e)/(f*x^2 + e)) + 6*(35*b*d^3*e^7*f + 11*a*c^3*e^3*f^5 - 5*(3*b*c*d^2 + a*d^3)*e^6*f^2 - 3*(b*c^2*d + a*c*d^2
)*e^5*f^3 - (b*c^3 + 3*a*c^2*d)*e^4*f^4)*x)/(e^4*f^8*x^6 + 3*e^5*f^7*x^4 + 3*e^6*f^6*x^2 + e^7*f^5), 1/48*(48*
b*d^3*e^4*f^4*x^7 + 3*(77*b*d^3*e^5*f^3 + 5*a*c^3*e*f^7 - 11*(3*b*c*d^2 + a*d^3)*e^4*f^4 + 3*(b*c^2*d + a*c*d^
2)*e^3*f^5 + (b*c^3 + 3*a*c^2*d)*e^2*f^6)*x^5 + 8*(35*b*d^3*e^6*f^2 + 5*a*c^3*e^2*f^6 - 5*(3*b*c*d^2 + a*d^3)*
e^5*f^3 - 3*(b*c^2*d + a*c*d^2)*e^4*f^4 + (b*c^3 + 3*a*c^2*d)*e^3*f^5)*x^3 - 3*(35*b*d^3*e^7 - 5*a*c^3*e^3*f^4
- 5*(3*b*c*d^2 + a*d^3)*e^6*f - 3*(b*c^2*d + a*c*d^2)*e^5*f^2 - (b*c^3 + 3*a*c^2*d)*e^4*f^3 + (35*b*d^3*e^4*f
^3 - 5*a*c^3*f^7 - 5*(3*b*c*d^2 + a*d^3)*e^3*f^4 - 3*(b*c^2*d + a*c*d^2)*e^2*f^5 - (b*c^3 + 3*a*c^2*d)*e*f^6)*
x^6 + 3*(35*b*d^3*e^5*f^2 - 5*a*c^3*e*f^6 - 5*(3*b*c*d^2 + a*d^3)*e^4*f^3 - 3*(b*c^2*d + a*c*d^2)*e^3*f^4 - (b
*c^3 + 3*a*c^2*d)*e^2*f^5)*x^4 + 3*(35*b*d^3*e^6*f - 5*a*c^3*e^2*f^5 - 5*(3*b*c*d^2 + a*d^3)*e^5*f^2 - 3*(b*c^
2*d + a*c*d^2)*e^4*f^3 - (b*c^3 + 3*a*c^2*d)*e^3*f^4)*x^2)*sqrt(e*f)*arctan(sqrt(e*f)*x/e) + 3*(35*b*d^3*e^7*f
+ 11*a*c^3*e^3*f^5 - 5*(3*b*c*d^2 + a*d^3)*e^6*f^2 - 3*(b*c^2*d + a*c*d^2)*e^5*f^3 - (b*c^3 + 3*a*c^2*d)*e^4*
f^4)*x)/(e^4*f^8*x^6 + 3*e^5*f^7*x^4 + 3*e^6*f^6*x^2 + e^7*f^5)]

________________________________________________________________________________________

giac [A]  time = 0.37, size = 447, normalized size = 1.28 \begin {gather*} \frac {b d^{3} x}{f^{4}} + \frac {{\left (5 \, a c^{3} f^{4} + b c^{3} f^{3} e + 3 \, a c^{2} d f^{3} e + 3 \, b c^{2} d f^{2} e^{2} + 3 \, a c d^{2} f^{2} e^{2} + 15 \, b c d^{2} f e^{3} + 5 \, a d^{3} f e^{3} - 35 \, b d^{3} e^{4}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {7}{2}\right )}}{16 \, f^{\frac {9}{2}}} + \frac {{\left (15 \, a c^{3} f^{6} x^{5} + 3 \, b c^{3} f^{5} x^{5} e + 9 \, a c^{2} d f^{5} x^{5} e + 9 \, b c^{2} d f^{4} x^{5} e^{2} + 9 \, a c d^{2} f^{4} x^{5} e^{2} - 99 \, b c d^{2} f^{3} x^{5} e^{3} - 33 \, a d^{3} f^{3} x^{5} e^{3} + 40 \, a c^{3} f^{5} x^{3} e + 87 \, b d^{3} f^{2} x^{5} e^{4} + 8 \, b c^{3} f^{4} x^{3} e^{2} + 24 \, a c^{2} d f^{4} x^{3} e^{2} - 24 \, b c^{2} d f^{3} x^{3} e^{3} - 24 \, a c d^{2} f^{3} x^{3} e^{3} - 120 \, b c d^{2} f^{2} x^{3} e^{4} - 40 \, a d^{3} f^{2} x^{3} e^{4} + 33 \, a c^{3} f^{4} x e^{2} + 136 \, b d^{3} f x^{3} e^{5} - 3 \, b c^{3} f^{3} x e^{3} - 9 \, a c^{2} d f^{3} x e^{3} - 9 \, b c^{2} d f^{2} x e^{4} - 9 \, a c d^{2} f^{2} x e^{4} - 45 \, b c d^{2} f x e^{5} - 15 \, a d^{3} f x e^{5} + 57 \, b d^{3} x e^{6}\right )} e^{\left (-3\right )}}{48 \, {\left (f x^{2} + e\right )}^{3} f^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d^3*x/f^4 + 1/16*(5*a*c^3*f^4 + b*c^3*f^3*e + 3*a*c^2*d*f^3*e + 3*b*c^2*d*f^2*e^2 + 3*a*c*d^2*f^2*e^2 + 15*b
*c*d^2*f*e^3 + 5*a*d^3*f*e^3 - 35*b*d^3*e^4)*arctan(sqrt(f)*x*e^(-1/2))*e^(-7/2)/f^(9/2) + 1/48*(15*a*c^3*f^6*
x^5 + 3*b*c^3*f^5*x^5*e + 9*a*c^2*d*f^5*x^5*e + 9*b*c^2*d*f^4*x^5*e^2 + 9*a*c*d^2*f^4*x^5*e^2 - 99*b*c*d^2*f^3
*x^5*e^3 - 33*a*d^3*f^3*x^5*e^3 + 40*a*c^3*f^5*x^3*e + 87*b*d^3*f^2*x^5*e^4 + 8*b*c^3*f^4*x^3*e^2 + 24*a*c^2*d
*f^4*x^3*e^2 - 24*b*c^2*d*f^3*x^3*e^3 - 24*a*c*d^2*f^3*x^3*e^3 - 120*b*c*d^2*f^2*x^3*e^4 - 40*a*d^3*f^2*x^3*e^
4 + 33*a*c^3*f^4*x*e^2 + 136*b*d^3*f*x^3*e^5 - 3*b*c^3*f^3*x*e^3 - 9*a*c^2*d*f^3*x*e^3 - 9*b*c^2*d*f^2*x*e^4 -
9*a*c*d^2*f^2*x*e^4 - 45*b*c*d^2*f*x*e^5 - 15*a*d^3*f*x*e^5 + 57*b*d^3*x*e^6)*e^(-3)/((f*x^2 + e)^3*f^4)

________________________________________________________________________________________

maple [B]  time = 0.01, size = 735, normalized size = 2.11 \begin {gather*} \frac {5 a \,c^{3} f^{2} x^{5}}{16 \left (f \,x^{2}+e \right )^{3} e^{3}}+\frac {3 a \,c^{2} d f \,x^{5}}{16 \left (f \,x^{2}+e \right )^{3} e^{2}}+\frac {3 a c \,d^{2} x^{5}}{16 \left (f \,x^{2}+e \right )^{3} e}-\frac {11 a \,d^{3} x^{5}}{16 \left (f \,x^{2}+e \right )^{3} f}+\frac {b \,c^{3} f \,x^{5}}{16 \left (f \,x^{2}+e \right )^{3} e^{2}}+\frac {3 b \,c^{2} d \,x^{5}}{16 \left (f \,x^{2}+e \right )^{3} e}-\frac {33 b c \,d^{2} x^{5}}{16 \left (f \,x^{2}+e \right )^{3} f}+\frac {29 b \,d^{3} e \,x^{5}}{16 \left (f \,x^{2}+e \right )^{3} f^{2}}+\frac {5 a \,c^{3} f \,x^{3}}{6 \left (f \,x^{2}+e \right )^{3} e^{2}}+\frac {a \,c^{2} d \,x^{3}}{2 \left (f \,x^{2}+e \right )^{3} e}-\frac {a c \,d^{2} x^{3}}{2 \left (f \,x^{2}+e \right )^{3} f}-\frac {5 a \,d^{3} e \,x^{3}}{6 \left (f \,x^{2}+e \right )^{3} f^{2}}+\frac {b \,c^{3} x^{3}}{6 \left (f \,x^{2}+e \right )^{3} e}-\frac {b \,c^{2} d \,x^{3}}{2 \left (f \,x^{2}+e \right )^{3} f}-\frac {5 b c \,d^{2} e \,x^{3}}{2 \left (f \,x^{2}+e \right )^{3} f^{2}}+\frac {17 b \,d^{3} e^{2} x^{3}}{6 \left (f \,x^{2}+e \right )^{3} f^{3}}+\frac {11 a \,c^{3} x}{16 \left (f \,x^{2}+e \right )^{3} e}-\frac {3 a \,c^{2} d x}{16 \left (f \,x^{2}+e \right )^{3} f}-\frac {3 a c \,d^{2} e x}{16 \left (f \,x^{2}+e \right )^{3} f^{2}}-\frac {5 a \,d^{3} e^{2} x}{16 \left (f \,x^{2}+e \right )^{3} f^{3}}-\frac {b \,c^{3} x}{16 \left (f \,x^{2}+e \right )^{3} f}-\frac {3 b \,c^{2} d e x}{16 \left (f \,x^{2}+e \right )^{3} f^{2}}-\frac {15 b c \,d^{2} e^{2} x}{16 \left (f \,x^{2}+e \right )^{3} f^{3}}+\frac {19 b \,d^{3} e^{3} x}{16 \left (f \,x^{2}+e \right )^{3} f^{4}}+\frac {5 a \,c^{3} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e^{3}}+\frac {3 a \,c^{2} d \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e^{2} f}+\frac {3 a c \,d^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e \,f^{2}}+\frac {5 a \,d^{3} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, f^{3}}+\frac {b \,c^{3} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e^{2} f}+\frac {3 b \,c^{2} d \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, e \,f^{2}}+\frac {15 b c \,d^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, f^{3}}-\frac {35 b \,d^{3} e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \sqrt {e f}\, f^{4}}+\frac {b \,d^{3} x}{f^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)*(d*x^2+c)^3/(f*x^2+e)^4,x)

[Out]

-15/16/f^3/(f*x^2+e)^3*b*c*d^2*e^2*x+3/16/f/e^2/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*a*c^2*d+3/16/f^2/e/(e*f)
^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*a*c*d^2+3/16/f^2/e/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*b*c^2*d-5/2/f^2/(f*x
^2+e)^3*x^3*b*c*d^2*e-3/16/f^2/(f*x^2+e)^3*a*c*d^2*e*x-3/16/f^2/(f*x^2+e)^3*b*c^2*d*e*x+3/16*f/(f*x^2+e)^3/e^2
*x^5*a*c^2*d+29/16/f^2/(f*x^2+e)^3*x^5*b*d^3*e+5/6*f/(f*x^2+e)^3/e^2*x^3*a*c^3-1/2/f/(f*x^2+e)^3*x^3*a*c*d^2-5
/6/f^2/(f*x^2+e)^3*x^3*a*d^3*e+3/16/(f*x^2+e)^3/e*x^5*b*c^2*d+1/2/(f*x^2+e)^3/e*x^3*a*c^2*d+3/16/(f*x^2+e)^3/e
*x^5*a*c*d^2+19/16/f^4/(f*x^2+e)^3*b*d^3*e^3*x+15/16/f^3/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*b*c*d^2-35/16/f
^4*e/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*b*d^3+1/16/f/e^2/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*b*c^3-1/2/f/
(f*x^2+e)^3*x^3*b*c^2*d+17/6/f^3/(f*x^2+e)^3*x^3*b*d^3*e^2-3/16/f/(f*x^2+e)^3*a*c^2*d*x-5/16/f^3/(f*x^2+e)^3*a
*d^3*e^2*x+5/16*f^2/(f*x^2+e)^3/e^3*x^5*a*c^3+1/16*f/(f*x^2+e)^3/e^2*x^5*b*c^3-33/16/f/(f*x^2+e)^3*x^5*b*c*d^2
+b*d^3/f^4*x+5/16/e^3/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*a*c^3+5/16/f^3/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*
x)*a*d^3-11/16/f/(f*x^2+e)^3*x^5*a*d^3-1/16/f/(f*x^2+e)^3*b*c^3*x+1/6/(f*x^2+e)^3/e*x^3*b*c^3+11/16/(f*x^2+e)^
3/e*x*a*c^3

________________________________________________________________________________________

maxima [A]  time = 2.39, size = 416, normalized size = 1.20 \begin {gather*} \frac {b d^{3} x}{f^{4}} + \frac {3 \, {\left (29 \, b d^{3} e^{4} f^{2} + 5 \, a c^{3} f^{6} - 11 \, {\left (3 \, b c d^{2} + a d^{3}\right )} e^{3} f^{3} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} e^{2} f^{4} + {\left (b c^{3} + 3 \, a c^{2} d\right )} e f^{5}\right )} x^{5} + 8 \, {\left (17 \, b d^{3} e^{5} f + 5 \, a c^{3} e f^{5} - 5 \, {\left (3 \, b c d^{2} + a d^{3}\right )} e^{4} f^{2} - 3 \, {\left (b c^{2} d + a c d^{2}\right )} e^{3} f^{3} + {\left (b c^{3} + 3 \, a c^{2} d\right )} e^{2} f^{4}\right )} x^{3} + 3 \, {\left (19 \, b d^{3} e^{6} + 11 \, a c^{3} e^{2} f^{4} - 5 \, {\left (3 \, b c d^{2} + a d^{3}\right )} e^{5} f - 3 \, {\left (b c^{2} d + a c d^{2}\right )} e^{4} f^{2} - {\left (b c^{3} + 3 \, a c^{2} d\right )} e^{3} f^{3}\right )} x}{48 \, {\left (e^{3} f^{7} x^{6} + 3 \, e^{4} f^{6} x^{4} + 3 \, e^{5} f^{5} x^{2} + e^{6} f^{4}\right )}} - \frac {{\left (35 \, b d^{3} e^{4} - 5 \, a c^{3} f^{4} - 5 \, {\left (3 \, b c d^{2} + a d^{3}\right )} e^{3} f - 3 \, {\left (b c^{2} d + a c d^{2}\right )} e^{2} f^{2} - {\left (b c^{3} + 3 \, a c^{2} d\right )} e f^{3}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{16 \, \sqrt {e f} e^{3} f^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d^3*x/f^4 + 1/48*(3*(29*b*d^3*e^4*f^2 + 5*a*c^3*f^6 - 11*(3*b*c*d^2 + a*d^3)*e^3*f^3 + 3*(b*c^2*d + a*c*d^2)
*e^2*f^4 + (b*c^3 + 3*a*c^2*d)*e*f^5)*x^5 + 8*(17*b*d^3*e^5*f + 5*a*c^3*e*f^5 - 5*(3*b*c*d^2 + a*d^3)*e^4*f^2
- 3*(b*c^2*d + a*c*d^2)*e^3*f^3 + (b*c^3 + 3*a*c^2*d)*e^2*f^4)*x^3 + 3*(19*b*d^3*e^6 + 11*a*c^3*e^2*f^4 - 5*(3
*b*c*d^2 + a*d^3)*e^5*f - 3*(b*c^2*d + a*c*d^2)*e^4*f^2 - (b*c^3 + 3*a*c^2*d)*e^3*f^3)*x)/(e^3*f^7*x^6 + 3*e^4
*f^6*x^4 + 3*e^5*f^5*x^2 + e^6*f^4) - 1/16*(35*b*d^3*e^4 - 5*a*c^3*f^4 - 5*(3*b*c*d^2 + a*d^3)*e^3*f - 3*(b*c^
2*d + a*c*d^2)*e^2*f^2 - (b*c^3 + 3*a*c^2*d)*e*f^3)*arctan(f*x/sqrt(e*f))/(sqrt(e*f)*e^3*f^4)

________________________________________________________________________________________

mupad [B]  time = 1.19, size = 444, normalized size = 1.28 \begin {gather*} \frac {\frac {x^3\,\left (b\,c^3\,e\,f^4+5\,a\,c^3\,f^5-3\,b\,c^2\,d\,e^2\,f^3+3\,a\,c^2\,d\,e\,f^4-15\,b\,c\,d^2\,e^3\,f^2-3\,a\,c\,d^2\,e^2\,f^3+17\,b\,d^3\,e^4\,f-5\,a\,d^3\,e^3\,f^2\right )}{6\,e^2}+\frac {x^5\,\left (b\,c^3\,e\,f^5+5\,a\,c^3\,f^6+3\,b\,c^2\,d\,e^2\,f^4+3\,a\,c^2\,d\,e\,f^5-33\,b\,c\,d^2\,e^3\,f^3+3\,a\,c\,d^2\,e^2\,f^4+29\,b\,d^3\,e^4\,f^2-11\,a\,d^3\,e^3\,f^3\right )}{16\,e^3}-\frac {x\,\left (b\,c^3\,e\,f^3-11\,a\,c^3\,f^4+3\,b\,c^2\,d\,e^2\,f^2+3\,a\,c^2\,d\,e\,f^3+15\,b\,c\,d^2\,e^3\,f+3\,a\,c\,d^2\,e^2\,f^2-19\,b\,d^3\,e^4+5\,a\,d^3\,e^3\,f\right )}{16\,e}}{e^3\,f^4+3\,e^2\,f^5\,x^2+3\,e\,f^6\,x^4+f^7\,x^6}+\frac {b\,d^3\,x}{f^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (b\,c^3\,e\,f^3+5\,a\,c^3\,f^4+3\,b\,c^2\,d\,e^2\,f^2+3\,a\,c^2\,d\,e\,f^3+15\,b\,c\,d^2\,e^3\,f+3\,a\,c\,d^2\,e^2\,f^2-35\,b\,d^3\,e^4+5\,a\,d^3\,e^3\,f\right )}{16\,e^{7/2}\,f^{9/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x^2)*(c + d*x^2)^3)/(e + f*x^2)^4,x)

[Out]

((x^3*(5*a*c^3*f^5 - 5*a*d^3*e^3*f^2 + b*c^3*e*f^4 + 17*b*d^3*e^4*f + 3*a*c^2*d*e*f^4 - 3*a*c*d^2*e^2*f^3 - 15
*b*c*d^2*e^3*f^2 - 3*b*c^2*d*e^2*f^3))/(6*e^2) + (x^5*(5*a*c^3*f^6 - 11*a*d^3*e^3*f^3 + 29*b*d^3*e^4*f^2 + b*c
^3*e*f^5 + 3*a*c^2*d*e*f^5 + 3*a*c*d^2*e^2*f^4 - 33*b*c*d^2*e^3*f^3 + 3*b*c^2*d*e^2*f^4))/(16*e^3) - (x*(5*a*d
^3*e^3*f - 19*b*d^3*e^4 - 11*a*c^3*f^4 + b*c^3*e*f^3 + 3*a*c^2*d*e*f^3 + 15*b*c*d^2*e^3*f + 3*a*c*d^2*e^2*f^2
+ 3*b*c^2*d*e^2*f^2))/(16*e))/(e^3*f^4 + f^7*x^6 + 3*e*f^6*x^4 + 3*e^2*f^5*x^2) + (b*d^3*x)/f^4 + (atan((f^(1/
2)*x)/e^(1/2))*(5*a*c^3*f^4 - 35*b*d^3*e^4 + 5*a*d^3*e^3*f + b*c^3*e*f^3 + 3*a*c^2*d*e*f^3 + 15*b*c*d^2*e^3*f
+ 3*a*c*d^2*e^2*f^2 + 3*b*c^2*d*e^2*f^2))/(16*e^(7/2)*f^(9/2))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)*(d*x**2+c)**3/(f*x**2+e)**4,x)

[Out]

Timed out

________________________________________________________________________________________