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

Optimal. Leaf size=207 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-c^2 f^2-6 c d e f+15 d^2 e^2\right )-a f \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{8 e^{5/2} f^{7/2}}+\frac {d x (b e (15 d e-c f)-3 a f (c f+d e))}{8 e^2 f^3}-\frac {x \left (c+d x^2\right ) (b e (5 d e-c f)-a f (3 c f+d e))}{8 e^2 f^2 \left (e+f x^2\right )}-\frac {x \left (c+d x^2\right )^2 (b e-a f)}{4 e f \left (e+f x^2\right )^2} \]

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Rubi [A]  time = 0.24, antiderivative size = 207, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-c^2 f^2-6 c d e f+15 d^2 e^2\right )-a f \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{8 e^{5/2} f^{7/2}}-\frac {x \left (c+d x^2\right ) (b e (5 d e-c f)-a f (3 c f+d e))}{8 e^2 f^2 \left (e+f x^2\right )}+\frac {d x (b e (15 d e-c f)-3 a f (c f+d e))}{8 e^2 f^3}-\frac {x \left (c+d x^2\right )^2 (b e-a f)}{4 e f \left (e+f x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(b*e*(15*d*e - c*f) - 3*a*f*(d*e + c*f))*x)/(8*e^2*f^3) - ((b*e - a*f)*x*(c + d*x^2)^2)/(4*e*f*(e + f*x^2)^
2) - ((b*e*(5*d*e - c*f) - a*f*(d*e + 3*c*f))*x*(c + d*x^2))/(8*e^2*f^2*(e + f*x^2)) - ((b*e*(15*d^2*e^2 - 6*c
*d*e*f - c^2*f^2) - a*f*(3*d^2*e^2 + 2*c*d*e*f + 3*c^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(8*e^(5/2)*f^(7/2))

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

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Mathematica [A]  time = 0.13, size = 183, normalized size = 0.88 \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) \left (b e \left (-c^2 f^2-6 c d e f+15 d^2 e^2\right )-a f \left (3 c^2 f^2+2 c d e f+3 d^2 e^2\right )\right )}{8 e^{5/2} f^{7/2}}+\frac {x (d e-c f) (b e (9 d e-c f)-a f (3 c f+5 d e))}{8 e^2 f^3 \left (e+f x^2\right )}-\frac {x (b e-a f) (d e-c f)^2}{4 e f^3 \left (e+f x^2\right )^2}+\frac {b d^2 x}{f^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d^2*x)/f^3 - ((b*e - a*f)*(d*e - c*f)^2*x)/(4*e*f^3*(e + f*x^2)^2) + ((d*e - c*f)*(b*e*(9*d*e - c*f) - a*f*
(5*d*e + 3*c*f))*x)/(8*e^2*f^3*(e + f*x^2)) - ((b*e*(15*d^2*e^2 - 6*c*d*e*f - c^2*f^2) - a*f*(3*d^2*e^2 + 2*c*
d*e*f + 3*c^2*f^2))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(8*e^(5/2)*f^(7/2))

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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 )^2}{\left (e+f x^2\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.43, size = 777, normalized size = 3.75 \begin {gather*} \left [\frac {16 \, b d^{2} e^{3} f^{3} x^{5} + 2 \, {\left (25 \, b d^{2} e^{4} f^{2} + 3 \, a c^{2} e f^{5} - 5 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f^{3} + {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{4}\right )} x^{3} + {\left (15 \, b d^{2} e^{5} - 3 \, a c^{2} e^{2} f^{3} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{4} f - {\left (b c^{2} + 2 \, a c d\right )} e^{3} f^{2} + {\left (15 \, b d^{2} e^{3} f^{2} - 3 \, a c^{2} f^{5} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{3} - {\left (b c^{2} + 2 \, a c d\right )} e f^{4}\right )} x^{4} + 2 \, {\left (15 \, b d^{2} e^{4} f - 3 \, a c^{2} e f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f^{2} - {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{3}\right )} x^{2}\right )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) + 2 \, {\left (15 \, b d^{2} e^{5} f + 5 \, a c^{2} e^{2} f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{4} f^{2} - {\left (b c^{2} + 2 \, a c d\right )} e^{3} f^{3}\right )} x}{16 \, {\left (e^{3} f^{6} x^{4} + 2 \, e^{4} f^{5} x^{2} + e^{5} f^{4}\right )}}, \frac {8 \, b d^{2} e^{3} f^{3} x^{5} + {\left (25 \, b d^{2} e^{4} f^{2} + 3 \, a c^{2} e f^{5} - 5 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f^{3} + {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{4}\right )} x^{3} - {\left (15 \, b d^{2} e^{5} - 3 \, a c^{2} e^{2} f^{3} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{4} f - {\left (b c^{2} + 2 \, a c d\right )} e^{3} f^{2} + {\left (15 \, b d^{2} e^{3} f^{2} - 3 \, a c^{2} f^{5} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{3} - {\left (b c^{2} + 2 \, a c d\right )} e f^{4}\right )} x^{4} + 2 \, {\left (15 \, b d^{2} e^{4} f - 3 \, a c^{2} e f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f^{2} - {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{3}\right )} x^{2}\right )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) + {\left (15 \, b d^{2} e^{5} f + 5 \, a c^{2} e^{2} f^{4} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{4} f^{2} - {\left (b c^{2} + 2 \, a c d\right )} e^{3} f^{3}\right )} x}{8 \, {\left (e^{3} f^{6} x^{4} + 2 \, e^{4} f^{5} x^{2} + e^{5} f^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(16*b*d^2*e^3*f^3*x^5 + 2*(25*b*d^2*e^4*f^2 + 3*a*c^2*e*f^5 - 5*(2*b*c*d + a*d^2)*e^3*f^3 + (b*c^2 + 2*a
*c*d)*e^2*f^4)*x^3 + (15*b*d^2*e^5 - 3*a*c^2*e^2*f^3 - 3*(2*b*c*d + a*d^2)*e^4*f - (b*c^2 + 2*a*c*d)*e^3*f^2 +
(15*b*d^2*e^3*f^2 - 3*a*c^2*f^5 - 3*(2*b*c*d + a*d^2)*e^2*f^3 - (b*c^2 + 2*a*c*d)*e*f^4)*x^4 + 2*(15*b*d^2*e^
4*f - 3*a*c^2*e*f^4 - 3*(2*b*c*d + a*d^2)*e^3*f^2 - (b*c^2 + 2*a*c*d)*e^2*f^3)*x^2)*sqrt(-e*f)*log((f*x^2 - 2*
sqrt(-e*f)*x - e)/(f*x^2 + e)) + 2*(15*b*d^2*e^5*f + 5*a*c^2*e^2*f^4 - 3*(2*b*c*d + a*d^2)*e^4*f^2 - (b*c^2 +
2*a*c*d)*e^3*f^3)*x)/(e^3*f^6*x^4 + 2*e^4*f^5*x^2 + e^5*f^4), 1/8*(8*b*d^2*e^3*f^3*x^5 + (25*b*d^2*e^4*f^2 + 3
*a*c^2*e*f^5 - 5*(2*b*c*d + a*d^2)*e^3*f^3 + (b*c^2 + 2*a*c*d)*e^2*f^4)*x^3 - (15*b*d^2*e^5 - 3*a*c^2*e^2*f^3
- 3*(2*b*c*d + a*d^2)*e^4*f - (b*c^2 + 2*a*c*d)*e^3*f^2 + (15*b*d^2*e^3*f^2 - 3*a*c^2*f^5 - 3*(2*b*c*d + a*d^2
)*e^2*f^3 - (b*c^2 + 2*a*c*d)*e*f^4)*x^4 + 2*(15*b*d^2*e^4*f - 3*a*c^2*e*f^4 - 3*(2*b*c*d + a*d^2)*e^3*f^2 - (
b*c^2 + 2*a*c*d)*e^2*f^3)*x^2)*sqrt(e*f)*arctan(sqrt(e*f)*x/e) + (15*b*d^2*e^5*f + 5*a*c^2*e^2*f^4 - 3*(2*b*c*
d + a*d^2)*e^4*f^2 - (b*c^2 + 2*a*c*d)*e^3*f^3)*x)/(e^3*f^6*x^4 + 2*e^4*f^5*x^2 + e^5*f^4)]

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giac [A]  time = 0.45, size = 238, normalized size = 1.15 \begin {gather*} \frac {b d^{2} x}{f^{3}} + \frac {{\left (3 \, a c^{2} f^{3} + b c^{2} f^{2} e + 2 \, a c d f^{2} e + 6 \, b c d f e^{2} + 3 \, a d^{2} f e^{2} - 15 \, b d^{2} e^{3}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {5}{2}\right )}}{8 \, f^{\frac {7}{2}}} + \frac {{\left (3 \, a c^{2} f^{4} x^{3} + b c^{2} f^{3} x^{3} e + 2 \, a c d f^{3} x^{3} e - 10 \, b c d f^{2} x^{3} e^{2} - 5 \, a d^{2} f^{2} x^{3} e^{2} + 9 \, b d^{2} f x^{3} e^{3} + 5 \, a c^{2} f^{3} x e - b c^{2} f^{2} x e^{2} - 2 \, a c d f^{2} x e^{2} - 6 \, b c d f x e^{3} - 3 \, a d^{2} f x e^{3} + 7 \, b d^{2} x e^{4}\right )} e^{\left (-2\right )}}{8 \, {\left (f x^{2} + e\right )}^{2} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d^2*x/f^3 + 1/8*(3*a*c^2*f^3 + b*c^2*f^2*e + 2*a*c*d*f^2*e + 6*b*c*d*f*e^2 + 3*a*d^2*f*e^2 - 15*b*d^2*e^3)*a
rctan(sqrt(f)*x*e^(-1/2))*e^(-5/2)/f^(7/2) + 1/8*(3*a*c^2*f^4*x^3 + b*c^2*f^3*x^3*e + 2*a*c*d*f^3*x^3*e - 10*b
*c*d*f^2*x^3*e^2 - 5*a*d^2*f^2*x^3*e^2 + 9*b*d^2*f*x^3*e^3 + 5*a*c^2*f^3*x*e - b*c^2*f^2*x*e^2 - 2*a*c*d*f^2*x
*e^2 - 6*b*c*d*f*x*e^3 - 3*a*d^2*f*x*e^3 + 7*b*d^2*x*e^4)*e^(-2)/((f*x^2 + e)^2*f^3)

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maple [B]  time = 0.01, size = 397, normalized size = 1.92 \begin {gather*} \frac {3 a \,c^{2} f \,x^{3}}{8 \left (f \,x^{2}+e \right )^{2} e^{2}}+\frac {a c d \,x^{3}}{4 \left (f \,x^{2}+e \right )^{2} e}-\frac {5 a \,d^{2} x^{3}}{8 \left (f \,x^{2}+e \right )^{2} f}+\frac {b \,c^{2} x^{3}}{8 \left (f \,x^{2}+e \right )^{2} e}-\frac {5 b c d \,x^{3}}{4 \left (f \,x^{2}+e \right )^{2} f}+\frac {9 b \,d^{2} e \,x^{3}}{8 \left (f \,x^{2}+e \right )^{2} f^{2}}+\frac {5 a \,c^{2} x}{8 \left (f \,x^{2}+e \right )^{2} e}-\frac {a c d x}{4 \left (f \,x^{2}+e \right )^{2} f}-\frac {3 a \,d^{2} e x}{8 \left (f \,x^{2}+e \right )^{2} f^{2}}-\frac {b \,c^{2} x}{8 \left (f \,x^{2}+e \right )^{2} f}-\frac {3 b c d e x}{4 \left (f \,x^{2}+e \right )^{2} f^{2}}+\frac {7 b \,d^{2} e^{2} x}{8 \left (f \,x^{2}+e \right )^{2} f^{3}}+\frac {3 a \,c^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 \sqrt {e f}\, e^{2}}+\frac {a c d \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{4 \sqrt {e f}\, e f}+\frac {3 a \,d^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 \sqrt {e f}\, f^{2}}+\frac {b \,c^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 \sqrt {e f}\, e f}+\frac {3 b c d \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{4 \sqrt {e f}\, f^{2}}-\frac {15 b \,d^{2} e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 \sqrt {e f}\, f^{3}}+\frac {b \,d^{2} x}{f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d^2/f^3*x+3/8*f/(f*x^2+e)^2/e^2*x^3*a*c^2+1/4/(f*x^2+e)^2/e*x^3*a*c*d-5/8/f/(f*x^2+e)^2*x^3*a*d^2+1/8/(f*x^2
+e)^2/e*x^3*b*c^2-5/4/f/(f*x^2+e)^2*x^3*b*c*d+9/8/f^2/(f*x^2+e)^2*x^3*b*d^2*e+5/8/(f*x^2+e)^2/e*x*a*c^2-1/4/f/
(f*x^2+e)^2*a*c*d*x-3/8/f^2/(f*x^2+e)^2*a*d^2*e*x-1/8/f/(f*x^2+e)^2*b*c^2*x-3/4/f^2/(f*x^2+e)^2*b*c*d*e*x+7/8/
f^3/(f*x^2+e)^2*b*d^2*e^2*x+3/8/e^2/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*a*c^2+1/4/f/e/(e*f)^(1/2)*arctan(1/(
e*f)^(1/2)*f*x)*a*c*d+3/8/f^2/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*a*d^2+1/8/f/e/(e*f)^(1/2)*arctan(1/(e*f)^(
1/2)*f*x)*b*c^2+3/4/f^2/(e*f)^(1/2)*arctan(1/(e*f)^(1/2)*f*x)*b*c*d-15/8/f^3*e/(e*f)^(1/2)*arctan(1/(e*f)^(1/2
)*f*x)*b*d^2

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maxima [A]  time = 2.01, size = 236, normalized size = 1.14 \begin {gather*} \frac {b d^{2} x}{f^{3}} + \frac {{\left (9 \, b d^{2} e^{3} f + 3 \, a c^{2} f^{4} - 5 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f^{2} + {\left (b c^{2} + 2 \, a c d\right )} e f^{3}\right )} x^{3} + {\left (7 \, b d^{2} e^{4} + 5 \, a c^{2} e f^{3} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{3} f - {\left (b c^{2} + 2 \, a c d\right )} e^{2} f^{2}\right )} x}{8 \, {\left (e^{2} f^{5} x^{4} + 2 \, e^{3} f^{4} x^{2} + e^{4} f^{3}\right )}} - \frac {{\left (15 \, b d^{2} e^{3} - 3 \, a c^{2} f^{3} - 3 \, {\left (2 \, b c d + a d^{2}\right )} e^{2} f - {\left (b c^{2} + 2 \, a c d\right )} e f^{2}\right )} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{8 \, \sqrt {e f} e^{2} f^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d^2*x/f^3 + 1/8*((9*b*d^2*e^3*f + 3*a*c^2*f^4 - 5*(2*b*c*d + a*d^2)*e^2*f^2 + (b*c^2 + 2*a*c*d)*e*f^3)*x^3 +
(7*b*d^2*e^4 + 5*a*c^2*e*f^3 - 3*(2*b*c*d + a*d^2)*e^3*f - (b*c^2 + 2*a*c*d)*e^2*f^2)*x)/(e^2*f^5*x^4 + 2*e^3
*f^4*x^2 + e^4*f^3) - 1/8*(15*b*d^2*e^3 - 3*a*c^2*f^3 - 3*(2*b*c*d + a*d^2)*e^2*f - (b*c^2 + 2*a*c*d)*e*f^2)*a
rctan(f*x/sqrt(e*f))/(sqrt(e*f)*e^2*f^3)

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mupad [B]  time = 0.98, size = 243, normalized size = 1.17 \begin {gather*} \frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x}{\sqrt {e}}\right )\,\left (b\,c^2\,e\,f^2+3\,a\,c^2\,f^3+6\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-15\,b\,d^2\,e^3+3\,a\,d^2\,e^2\,f\right )}{8\,e^{5/2}\,f^{7/2}}-\frac {\frac {x\,\left (b\,c^2\,e\,f^2-5\,a\,c^2\,f^3+6\,b\,c\,d\,e^2\,f+2\,a\,c\,d\,e\,f^2-7\,b\,d^2\,e^3+3\,a\,d^2\,e^2\,f\right )}{8\,e}-\frac {x^3\,\left (b\,c^2\,e\,f^3+3\,a\,c^2\,f^4-10\,b\,c\,d\,e^2\,f^2+2\,a\,c\,d\,e\,f^3+9\,b\,d^2\,e^3\,f-5\,a\,d^2\,e^2\,f^2\right )}{8\,e^2}}{e^2\,f^3+2\,e\,f^4\,x^2+f^5\,x^4}+\frac {b\,d^2\,x}{f^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((f^(1/2)*x)/e^(1/2))*(3*a*c^2*f^3 - 15*b*d^2*e^3 + 3*a*d^2*e^2*f + b*c^2*e*f^2 + 2*a*c*d*e*f^2 + 6*b*c*d
*e^2*f))/(8*e^(5/2)*f^(7/2)) - ((x*(3*a*d^2*e^2*f - 7*b*d^2*e^3 - 5*a*c^2*f^3 + b*c^2*e*f^2 + 2*a*c*d*e*f^2 +
6*b*c*d*e^2*f))/(8*e) - (x^3*(3*a*c^2*f^4 - 5*a*d^2*e^2*f^2 + b*c^2*e*f^3 + 9*b*d^2*e^3*f - 10*b*c*d*e^2*f^2 +
2*a*c*d*e*f^3))/(8*e^2))/(e^2*f^3 + f^5*x^4 + 2*e*f^4*x^2) + (b*d^2*x)/f^3

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sympy [B]  time = 11.98, size = 400, normalized size = 1.93 \begin {gather*} \frac {b d^{2} x}{f^{3}} - \frac {\sqrt {- \frac {1}{e^{5} f^{7}}} \left (3 a c^{2} f^{3} + 2 a c d e f^{2} + 3 a d^{2} e^{2} f + b c^{2} e f^{2} + 6 b c d e^{2} f - 15 b d^{2} e^{3}\right ) \log {\left (- e^{3} f^{3} \sqrt {- \frac {1}{e^{5} f^{7}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{e^{5} f^{7}}} \left (3 a c^{2} f^{3} + 2 a c d e f^{2} + 3 a d^{2} e^{2} f + b c^{2} e f^{2} + 6 b c d e^{2} f - 15 b d^{2} e^{3}\right ) \log {\left (e^{3} f^{3} \sqrt {- \frac {1}{e^{5} f^{7}}} + x \right )}}{16} + \frac {x^{3} \left (3 a c^{2} f^{4} + 2 a c d e f^{3} - 5 a d^{2} e^{2} f^{2} + b c^{2} e f^{3} - 10 b c d e^{2} f^{2} + 9 b d^{2} e^{3} f\right ) + x \left (5 a c^{2} e f^{3} - 2 a c d e^{2} f^{2} - 3 a d^{2} e^{3} f - b c^{2} e^{2} f^{2} - 6 b c d e^{3} f + 7 b d^{2} e^{4}\right )}{8 e^{4} f^{3} + 16 e^{3} f^{4} x^{2} + 8 e^{2} f^{5} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*d**2*x/f**3 - sqrt(-1/(e**5*f**7))*(3*a*c**2*f**3 + 2*a*c*d*e*f**2 + 3*a*d**2*e**2*f + b*c**2*e*f**2 + 6*b*c
*d*e**2*f - 15*b*d**2*e**3)*log(-e**3*f**3*sqrt(-1/(e**5*f**7)) + x)/16 + sqrt(-1/(e**5*f**7))*(3*a*c**2*f**3
+ 2*a*c*d*e*f**2 + 3*a*d**2*e**2*f + b*c**2*e*f**2 + 6*b*c*d*e**2*f - 15*b*d**2*e**3)*log(e**3*f**3*sqrt(-1/(e
**5*f**7)) + x)/16 + (x**3*(3*a*c**2*f**4 + 2*a*c*d*e*f**3 - 5*a*d**2*e**2*f**2 + b*c**2*e*f**3 - 10*b*c*d*e**
2*f**2 + 9*b*d**2*e**3*f) + x*(5*a*c**2*e*f**3 - 2*a*c*d*e**2*f**2 - 3*a*d**2*e**3*f - b*c**2*e**2*f**2 - 6*b*
c*d*e**3*f + 7*b*d**2*e**4))/(8*e**4*f**3 + 16*e**3*f**4*x**2 + 8*e**2*f**5*x**4)

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