3.1.20 \(\int \frac {(a+b x^2) (c+d x^2)^3}{(e+f x^2)^2} \, dx\) [20]

Optimal. Leaf size=242 \[ -\frac {d \left (5 a f \left (15 d^2 e^2-22 c d e f+3 c^2 f^2\right )-b e \left (105 d^2 e^2-190 c d e f+81 c^2 f^2\right )\right ) x}{30 e f^4}-\frac {d (b e (35 d e-33 c f)-5 a f (5 d e-3 c f)) x \left (c+d x^2\right )}{30 e f^3}+\frac {d (7 b e-5 a f) x \left (c+d x^2\right )^2}{10 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{2 e f \left (e+f x^2\right )}-\frac {(d e-c f)^2 (b e (7 d e-c f)-a f (5 d e+c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{9/2}} \]

[Out]

-1/30*d*(5*a*f*(3*c^2*f^2-22*c*d*e*f+15*d^2*e^2)-b*e*(81*c^2*f^2-190*c*d*e*f+105*d^2*e^2))*x/e/f^4-1/30*d*(b*e
*(-33*c*f+35*d*e)-5*a*f*(-3*c*f+5*d*e))*x*(d*x^2+c)/e/f^3+1/10*d*(-5*a*f+7*b*e)*x*(d*x^2+c)^2/e/f^2-1/2*(-a*f+
b*e)*x*(d*x^2+c)^3/e/f/(f*x^2+e)-1/2*(-c*f+d*e)^2*(b*e*(-c*f+7*d*e)-a*f*(c*f+5*d*e))*arctan(x*f^(1/2)/e^(1/2))
/e^(3/2)/f^(9/2)

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Rubi [A]
time = 0.28, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {540, 542, 396, 211} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right ) (d e-c f)^2 (b e (7 d e-c f)-a f (c f+5 d e))}{2 e^{3/2} f^{9/2}}-\frac {d x \left (5 a f \left (3 c^2 f^2-22 c d e f+15 d^2 e^2\right )-b e \left (81 c^2 f^2-190 c d e f+105 d^2 e^2\right )\right )}{30 e f^4}-\frac {d x \left (c+d x^2\right ) (b e (35 d e-33 c f)-5 a f (5 d e-3 c f))}{30 e f^3}+\frac {d x \left (c+d x^2\right )^2 (7 b e-5 a f)}{10 e f^2}-\frac {x \left (c+d x^2\right )^3 (b e-a f)}{2 e f \left (e+f x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/30*(d*(5*a*f*(15*d^2*e^2 - 22*c*d*e*f + 3*c^2*f^2) - b*e*(105*d^2*e^2 - 190*c*d*e*f + 81*c^2*f^2))*x)/(e*f^
4) - (d*(b*e*(35*d*e - 33*c*f) - 5*a*f*(5*d*e - 3*c*f))*x*(c + d*x^2))/(30*e*f^3) + (d*(7*b*e - 5*a*f)*x*(c +
d*x^2)^2)/(10*e*f^2) - ((b*e - a*f)*x*(c + d*x^2)^3)/(2*e*f*(e + f*x^2)) - ((d*e - c*f)^2*(b*e*(7*d*e - c*f) -
a*f*(5*d*e + c*f))*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(2*e^(3/2)*f^(9/2))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 396

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 540

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]

Rule 542

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
f*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(b*(n*(p + q + 1) + 1))), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 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 )^2} \, dx &=-\frac {(b e-a f) x \left (c+d x^2\right )^3}{2 e f \left (e+f x^2\right )}-\frac {\int \frac {\left (c+d x^2\right )^2 \left (-c (b e+a f)-d (7 b e-5 a f) x^2\right )}{e+f x^2} \, dx}{2 e f}\\ &=\frac {d (7 b e-5 a f) x \left (c+d x^2\right )^2}{10 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{2 e f \left (e+f x^2\right )}-\frac {\int \frac {\left (c+d x^2\right ) \left (c (b e (7 d e-5 c f)-5 a f (d e+c f))+d (b e (35 d e-33 c f)-5 a f (5 d e-3 c f)) x^2\right )}{e+f x^2} \, dx}{10 e f^2}\\ &=-\frac {d (b e (35 d e-33 c f)-5 a f (5 d e-3 c f)) x \left (c+d x^2\right )}{30 e f^3}+\frac {d (7 b e-5 a f) x \left (c+d x^2\right )^2}{10 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{2 e f \left (e+f x^2\right )}-\frac {\int \frac {c \left (5 a f \left (5 d^2 e^2-6 c d e f-3 c^2 f^2\right )-b e \left (35 d^2 e^2-54 c d e f+15 c^2 f^2\right )\right )+d \left (5 a f \left (15 d^2 e^2-22 c d e f+3 c^2 f^2\right )-b e \left (105 d^2 e^2-190 c d e f+81 c^2 f^2\right )\right ) x^2}{e+f x^2} \, dx}{30 e f^3}\\ &=-\frac {d \left (5 a f \left (15 d^2 e^2-22 c d e f+3 c^2 f^2\right )-b e \left (105 d^2 e^2-190 c d e f+81 c^2 f^2\right )\right ) x}{30 e f^4}-\frac {d (b e (35 d e-33 c f)-5 a f (5 d e-3 c f)) x \left (c+d x^2\right )}{30 e f^3}+\frac {d (7 b e-5 a f) x \left (c+d x^2\right )^2}{10 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{2 e f \left (e+f x^2\right )}-\frac {\left ((d e-c f)^2 (b e (7 d e-c f)-a f (5 d e+c f))\right ) \int \frac {1}{e+f x^2} \, dx}{2 e f^4}\\ &=-\frac {d \left (5 a f \left (15 d^2 e^2-22 c d e f+3 c^2 f^2\right )-b e \left (105 d^2 e^2-190 c d e f+81 c^2 f^2\right )\right ) x}{30 e f^4}-\frac {d (b e (35 d e-33 c f)-5 a f (5 d e-3 c f)) x \left (c+d x^2\right )}{30 e f^3}+\frac {d (7 b e-5 a f) x \left (c+d x^2\right )^2}{10 e f^2}-\frac {(b e-a f) x \left (c+d x^2\right )^3}{2 e f \left (e+f x^2\right )}-\frac {(d e-c f)^2 (b e (7 d e-c f)-a f (5 d e+c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{9/2}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 176, normalized size = 0.73 \begin {gather*} \frac {d \left (3 b (d e-c f)^2+a d f (-2 d e+3 c f)\right ) x}{f^4}+\frac {d^2 (-2 b d e+3 b c f+a d f) x^3}{3 f^3}+\frac {b d^3 x^5}{5 f^2}+\frac {(b e-a f) (d e-c f)^3 x}{2 e f^4 \left (e+f x^2\right )}-\frac {(d e-c f)^2 (b e (7 d e-c f)-a f (5 d e+c f)) \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{2 e^{3/2} f^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.17, size = 308, normalized size = 1.27

method	result	size
default	\(\frac {d \left (\frac {1}{5} b \,d^{2} x^{5} f^{2}+\frac {1}{3} a \,d^{2} f^{2} x^{3}+b c d \,f^{2} x^{3}-\frac {2}{3} b \,d^{2} e f \,x^{3}+3 a c d \,f^{2} x -2 a \,d^{2} e f x +3 b \,c^{2} f^{2} x -6 b c d e f x +3 b \,d^{2} e^{2} x \right )}{f^{4}}+\frac {\frac {\left (a \,c^{3} f^{4}-3 a \,c^{2} d e \,f^{3}+3 a c \,d^{2} e^{2} f^{2}-a \,d^{3} e^{3} f -b \,c^{3} e \,f^{3}+3 b \,c^{2} d \,e^{2} f^{2}-3 b c \,d^{2} e^{3} f +b \,d^{3} e^{4}\right ) x}{2 e \left (f \,x^{2}+e \right )}+\frac {\left (a \,c^{3} f^{4}+3 a \,c^{2} d e \,f^{3}-9 a c \,d^{2} e^{2} f^{2}+5 a \,d^{3} e^{3} f +b \,c^{3} e \,f^{3}-9 b \,c^{2} d \,e^{2} f^{2}+15 b c \,d^{2} e^{3} f -7 b \,d^{3} e^{4}\right ) \arctan \left (\frac {f x}{\sqrt {f e}}\right )}{2 e \sqrt {f e}}}{f^{4}}\)	\(308\)
risch	\(-\frac {3 \ln \left (f x +\sqrt {-f e}\right ) a \,c^{2} d}{4 f \sqrt {-f e}}-\frac {5 e^{2} \ln \left (f x +\sqrt {-f e}\right ) a \,d^{3}}{4 f^{3} \sqrt {-f e}}+\frac {7 e^{3} \ln \left (f x +\sqrt {-f e}\right ) b \,d^{3}}{4 f^{4} \sqrt {-f e}}+\frac {d^{3} b \,x^{5}}{5 f^{2}}+\frac {d^{3} a \,x^{3}}{3 f^{2}}+\frac {3 d^{2} a c x}{f^{2}}-\frac {2 d^{3} a e x}{f^{3}}+\frac {3 d b \,c^{2} x}{f^{2}}+\frac {3 d^{3} b \,e^{2} x}{f^{4}}-\frac {\ln \left (f x +\sqrt {-f e}\right ) a \,c^{3}}{4 \sqrt {-f e}\, e}-\frac {\ln \left (f x +\sqrt {-f e}\right ) b \,c^{3}}{4 f \sqrt {-f e}}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) a \,c^{3}}{4 \sqrt {-f e}\, e}+\frac {\ln \left (-f x +\sqrt {-f e}\right ) b \,c^{3}}{4 f \sqrt {-f e}}+\frac {9 e \ln \left (f x +\sqrt {-f e}\right ) a c \,d^{2}}{4 f^{2} \sqrt {-f e}}+\frac {9 e \ln \left (f x +\sqrt {-f e}\right ) b \,c^{2} d}{4 f^{2} \sqrt {-f e}}-\frac {15 e^{2} \ln \left (f x +\sqrt {-f e}\right ) b c \,d^{2}}{4 f^{3} \sqrt {-f e}}-\frac {9 e \ln \left (-f x +\sqrt {-f e}\right ) a c \,d^{2}}{4 f^{2} \sqrt {-f e}}-\frac {9 e \ln \left (-f x +\sqrt {-f e}\right ) b \,c^{2} d}{4 f^{2} \sqrt {-f e}}+\frac {15 e^{2} \ln \left (-f x +\sqrt {-f e}\right ) b c \,d^{2}}{4 f^{3} \sqrt {-f e}}+\frac {\left (a \,c^{3} f^{4}-3 a \,c^{2} d e \,f^{3}+3 a c \,d^{2} e^{2} f^{2}-a \,d^{3} e^{3} f -b \,c^{3} e \,f^{3}+3 b \,c^{2} d \,e^{2} f^{2}-3 b c \,d^{2} e^{3} f +b \,d^{3} e^{4}\right ) x}{2 e \,f^{4} \left (f \,x^{2}+e \right )}+\frac {3 \ln \left (-f x +\sqrt {-f e}\right ) a \,c^{2} d}{4 f \sqrt {-f e}}+\frac {5 e^{2} \ln \left (-f x +\sqrt {-f e}\right ) a \,d^{3}}{4 f^{3} \sqrt {-f e}}-\frac {7 e^{3} \ln \left (-f x +\sqrt {-f e}\right ) b \,d^{3}}{4 f^{4} \sqrt {-f e}}+\frac {d^{2} b c \,x^{3}}{f^{2}}-\frac {2 d^{3} b e \,x^{3}}{3 f^{3}}-\frac {6 d^{2} b c e x}{f^{3}}\)	\(666\)

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)^2,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 298, normalized size = 1.23 \begin {gather*} \frac {{\left (a c^{3} f^{4} + b d^{3} e^{4} - {\left (b c^{3} e + 3 \, a c^{2} d e\right )} f^{3} + 3 \, {\left (b c^{2} d e^{2} + a c d^{2} e^{2}\right )} f^{2} - {\left (3 \, b c d^{2} e^{3} + a d^{3} e^{3}\right )} f\right )} x}{2 \, {\left (f^{5} x^{2} e + f^{4} e^{2}\right )}} + \frac {{\left (a c^{3} f^{4} - 7 \, b d^{3} e^{4} + {\left (b c^{3} e + 3 \, a c^{2} d e\right )} f^{3} - 9 \, {\left (b c^{2} d e^{2} + a c d^{2} e^{2}\right )} f^{2} + 5 \, {\left (3 \, b c d^{2} e^{3} + a d^{3} e^{3}\right )} f\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {3}{2}\right )}}{2 \, f^{\frac {9}{2}}} + \frac {3 \, b d^{3} f^{2} x^{5} - 5 \, {\left (2 \, b d^{3} f e - {\left (3 \, b c d^{2} + a d^{3}\right )} f^{2}\right )} x^{3} + 15 \, {\left (3 \, b d^{3} e^{2} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} f^{2} - 2 \, {\left (3 \, b c d^{2} e + a d^{3} e\right )} f\right )} x}{15 \, 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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.18, size = 824, normalized size = 3.40 \begin {gather*} \left [\frac {30 \, a c^{3} f^{5} x e + 210 \, b d^{3} f x e^{5} + 15 \, {\left (a c^{3} f^{5} x^{2} - 7 \, b d^{3} e^{5} - {\left (7 \, b d^{3} f x^{2} - 5 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f\right )} e^{4} + {\left (5 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{2} x^{2} - 9 \, {\left (b c^{2} d + a c d^{2}\right )} f^{2}\right )} e^{3} - {\left (9 \, {\left (b c^{2} d + a c d^{2}\right )} f^{3} x^{2} - {\left (b c^{3} + 3 \, a c^{2} d\right )} f^{3}\right )} e^{2} + {\left (a c^{3} f^{4} + {\left (b c^{3} + 3 \, a c^{2} d\right )} f^{4} x^{2}\right )} e\right )} \sqrt {-f e} \log \left (\frac {f x^{2} + 2 \, \sqrt {-f e} x - e}{f x^{2} + e}\right ) + 10 \, {\left (14 \, b d^{3} f^{2} x^{3} - 15 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{2} x\right )} e^{4} - 2 \, {\left (14 \, b d^{3} f^{3} x^{5} + 50 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{3} x^{3} - 135 \, {\left (b c^{2} d + a c d^{2}\right )} f^{3} x\right )} e^{3} + 2 \, {\left (6 \, b d^{3} f^{4} x^{7} + 10 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{4} x^{5} + 90 \, {\left (b c^{2} d + a c d^{2}\right )} f^{4} x^{3} - 15 \, {\left (b c^{3} + 3 \, a c^{2} d\right )} f^{4} x\right )} e^{2}}{60 \, {\left (f^{6} x^{2} e^{2} + f^{5} e^{3}\right )}}, \frac {15 \, a c^{3} f^{5} x e + 105 \, b d^{3} f x e^{5} + 15 \, {\left (a c^{3} f^{5} x^{2} - 7 \, b d^{3} e^{5} - {\left (7 \, b d^{3} f x^{2} - 5 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f\right )} e^{4} + {\left (5 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{2} x^{2} - 9 \, {\left (b c^{2} d + a c d^{2}\right )} f^{2}\right )} e^{3} - {\left (9 \, {\left (b c^{2} d + a c d^{2}\right )} f^{3} x^{2} - {\left (b c^{3} + 3 \, a c^{2} d\right )} f^{3}\right )} e^{2} + {\left (a c^{3} f^{4} + {\left (b c^{3} + 3 \, a c^{2} d\right )} f^{4} x^{2}\right )} e\right )} \sqrt {f} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\frac {1}{2}} + 5 \, {\left (14 \, b d^{3} f^{2} x^{3} - 15 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{2} x\right )} e^{4} - {\left (14 \, b d^{3} f^{3} x^{5} + 50 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{3} x^{3} - 135 \, {\left (b c^{2} d + a c d^{2}\right )} f^{3} x\right )} e^{3} + {\left (6 \, b d^{3} f^{4} x^{7} + 10 \, {\left (3 \, b c d^{2} + a d^{3}\right )} f^{4} x^{5} + 90 \, {\left (b c^{2} d + a c d^{2}\right )} f^{4} x^{3} - 15 \, {\left (b c^{3} + 3 \, a c^{2} d\right )} f^{4} x\right )} e^{2}}{30 \, {\left (f^{6} x^{2} e^{2} + f^{5} e^{3}\right )}}\right ] \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)^2,x, algorithm="fricas")

[Out]

[1/60*(30*a*c^3*f^5*x*e + 210*b*d^3*f*x*e^5 + 15*(a*c^3*f^5*x^2 - 7*b*d^3*e^5 - (7*b*d^3*f*x^2 - 5*(3*b*c*d^2
+ a*d^3)*f)*e^4 + (5*(3*b*c*d^2 + a*d^3)*f^2*x^2 - 9*(b*c^2*d + a*c*d^2)*f^2)*e^3 - (9*(b*c^2*d + a*c*d^2)*f^3
*x^2 - (b*c^3 + 3*a*c^2*d)*f^3)*e^2 + (a*c^3*f^4 + (b*c^3 + 3*a*c^2*d)*f^4*x^2)*e)*sqrt(-f*e)*log((f*x^2 + 2*s
qrt(-f*e)*x - e)/(f*x^2 + e)) + 10*(14*b*d^3*f^2*x^3 - 15*(3*b*c*d^2 + a*d^3)*f^2*x)*e^4 - 2*(14*b*d^3*f^3*x^5
+ 50*(3*b*c*d^2 + a*d^3)*f^3*x^3 - 135*(b*c^2*d + a*c*d^2)*f^3*x)*e^3 + 2*(6*b*d^3*f^4*x^7 + 10*(3*b*c*d^2 +
a*d^3)*f^4*x^5 + 90*(b*c^2*d + a*c*d^2)*f^4*x^3 - 15*(b*c^3 + 3*a*c^2*d)*f^4*x)*e^2)/(f^6*x^2*e^2 + f^5*e^3),
1/30*(15*a*c^3*f^5*x*e + 105*b*d^3*f*x*e^5 + 15*(a*c^3*f^5*x^2 - 7*b*d^3*e^5 - (7*b*d^3*f*x^2 - 5*(3*b*c*d^2 +
a*d^3)*f)*e^4 + (5*(3*b*c*d^2 + a*d^3)*f^2*x^2 - 9*(b*c^2*d + a*c*d^2)*f^2)*e^3 - (9*(b*c^2*d + a*c*d^2)*f^3*
x^2 - (b*c^3 + 3*a*c^2*d)*f^3)*e^2 + (a*c^3*f^4 + (b*c^3 + 3*a*c^2*d)*f^4*x^2)*e)*sqrt(f)*arctan(sqrt(f)*x*e^(
-1/2))*e^(1/2) + 5*(14*b*d^3*f^2*x^3 - 15*(3*b*c*d^2 + a*d^3)*f^2*x)*e^4 - (14*b*d^3*f^3*x^5 + 50*(3*b*c*d^2 +
a*d^3)*f^3*x^3 - 135*(b*c^2*d + a*c*d^2)*f^3*x)*e^3 + (6*b*d^3*f^4*x^7 + 10*(3*b*c*d^2 + a*d^3)*f^4*x^5 + 90*
(b*c^2*d + a*c*d^2)*f^4*x^3 - 15*(b*c^3 + 3*a*c^2*d)*f^4*x)*e^2)/(f^6*x^2*e^2 + f^5*e^3)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 661 vs. \(2 (231) = 462\).
time = 2.65, size = 661, normalized size = 2.73 \begin {gather*} \frac {b d^{3} x^{5}}{5 f^{2}} + x^{3} \left (\frac {a d^{3}}{3 f^{2}} + \frac {b c d^{2}}{f^{2}} - \frac {2 b d^{3} e}{3 f^{3}}\right ) + x \left (\frac {3 a c d^{2}}{f^{2}} - \frac {2 a d^{3} e}{f^{3}} + \frac {3 b c^{2} d}{f^{2}} - \frac {6 b c d^{2} e}{f^{3}} + \frac {3 b d^{3} e^{2}}{f^{4}}\right ) + \frac {x \left (a c^{3} f^{4} - 3 a c^{2} d e f^{3} + 3 a c d^{2} e^{2} f^{2} - a d^{3} e^{3} f - b c^{3} e f^{3} + 3 b c^{2} d e^{2} f^{2} - 3 b c d^{2} e^{3} f + b d^{3} e^{4}\right )}{2 e^{2} f^{4} + 2 e f^{5} x^{2}} - \frac {\sqrt {- \frac {1}{e^{3} f^{9}}} \left (c f - d e\right )^{2} \left (a c f^{2} + 5 a d e f + b c e f - 7 b d e^{2}\right ) \log {\left (- \frac {e^{2} f^{4} \sqrt {- \frac {1}{e^{3} f^{9}}} \left (c f - d e\right )^{2} \left (a c f^{2} + 5 a d e f + b c e f - 7 b d e^{2}\right )}{a c^{3} f^{4} + 3 a c^{2} d e f^{3} - 9 a c d^{2} e^{2} f^{2} + 5 a d^{3} e^{3} f + b c^{3} e f^{3} - 9 b c^{2} d e^{2} f^{2} + 15 b c d^{2} e^{3} f - 7 b d^{3} e^{4}} + x \right )}}{4} + \frac {\sqrt {- \frac {1}{e^{3} f^{9}}} \left (c f - d e\right )^{2} \left (a c f^{2} + 5 a d e f + b c e f - 7 b d e^{2}\right ) \log {\left (\frac {e^{2} f^{4} \sqrt {- \frac {1}{e^{3} f^{9}}} \left (c f - d e\right )^{2} \left (a c f^{2} + 5 a d e f + b c e f - 7 b d e^{2}\right )}{a c^{3} f^{4} + 3 a c^{2} d e f^{3} - 9 a c d^{2} e^{2} f^{2} + 5 a d^{3} e^{3} f + b c^{3} e f^{3} - 9 b c^{2} d e^{2} f^{2} + 15 b c d^{2} e^{3} f - 7 b d^{3} e^{4}} + x \right )}}{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)**2,x)

[Out]

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

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Giac [A]
time = 1.25, size = 321, normalized size = 1.33 \begin {gather*} \frac {{\left (a c^{3} f^{4} + b c^{3} f^{3} e + 3 \, a c^{2} d f^{3} e - 9 \, b c^{2} d f^{2} e^{2} - 9 \, a c d^{2} f^{2} e^{2} + 15 \, b c d^{2} f e^{3} + 5 \, a d^{3} f e^{3} - 7 \, b d^{3} e^{4}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {3}{2}\right )}}{2 \, f^{\frac {9}{2}}} + \frac {{\left (a c^{3} f^{4} x - b c^{3} f^{3} x e - 3 \, a c^{2} d f^{3} x e + 3 \, b c^{2} d f^{2} x e^{2} + 3 \, a c d^{2} f^{2} x e^{2} - 3 \, b c d^{2} f x e^{3} - a d^{3} f x e^{3} + b d^{3} x e^{4}\right )} e^{\left (-1\right )}}{2 \, {\left (f x^{2} + e\right )} f^{4}} + \frac {3 \, b d^{3} f^{8} x^{5} + 15 \, b c d^{2} f^{8} x^{3} + 5 \, a d^{3} f^{8} x^{3} - 10 \, b d^{3} f^{7} x^{3} e + 45 \, b c^{2} d f^{8} x + 45 \, a c d^{2} f^{8} x - 90 \, b c d^{2} f^{7} x e - 30 \, a d^{3} f^{7} x e + 45 \, b d^{3} f^{6} x e^{2}}{15 \, f^{10}} \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)^2,x, algorithm="giac")

[Out]

1/2*(a*c^3*f^4 + b*c^3*f^3*e + 3*a*c^2*d*f^3*e - 9*b*c^2*d*f^2*e^2 - 9*a*c*d^2*f^2*e^2 + 15*b*c*d^2*f*e^3 + 5*
a*d^3*f*e^3 - 7*b*d^3*e^4)*arctan(sqrt(f)*x*e^(-1/2))*e^(-3/2)/f^(9/2) + 1/2*(a*c^3*f^4*x - b*c^3*f^3*x*e - 3*
a*c^2*d*f^3*x*e + 3*b*c^2*d*f^2*x*e^2 + 3*a*c*d^2*f^2*x*e^2 - 3*b*c*d^2*f*x*e^3 - a*d^3*f*x*e^3 + b*d^3*x*e^4)
*e^(-1)/((f*x^2 + e)*f^4) + 1/15*(3*b*d^3*f^8*x^5 + 15*b*c*d^2*f^8*x^3 + 5*a*d^3*f^8*x^3 - 10*b*d^3*f^7*x^3*e
+ 45*b*c^2*d*f^8*x + 45*a*c*d^2*f^8*x - 90*b*c*d^2*f^7*x*e - 30*a*d^3*f^7*x*e + 45*b*d^3*f^6*x*e^2)/f^10

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Mupad [B]
time = 0.98, size = 389, normalized size = 1.61 \begin {gather*} x^3\,\left (\frac {a\,d^3+3\,b\,c\,d^2}{3\,f^2}-\frac {2\,b\,d^3\,e}{3\,f^3}\right )-x\,\left (\frac {2\,e\,\left (\frac {a\,d^3+3\,b\,c\,d^2}{f^2}-\frac {2\,b\,d^3\,e}{f^3}\right )}{f}+\frac {b\,d^3\,e^2}{f^4}-\frac {3\,c\,d\,\left (a\,d+b\,c\right )}{f^2}\right )+\frac {b\,d^3\,x^5}{5\,f^2}+\frac {x\,\left (-b\,c^3\,e\,f^3+a\,c^3\,f^4+3\,b\,c^2\,d\,e^2\,f^2-3\,a\,c^2\,d\,e\,f^3-3\,b\,c\,d^2\,e^3\,f+3\,a\,c\,d^2\,e^2\,f^2+b\,d^3\,e^4-a\,d^3\,e^3\,f\right )}{2\,e\,\left (f^5\,x^2+e\,f^4\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x\,{\left (c\,f-d\,e\right )}^2\,\left (a\,c\,f^2-7\,b\,d\,e^2+5\,a\,d\,e\,f+b\,c\,e\,f\right )}{\sqrt {e}\,\left (b\,c^3\,e\,f^3+a\,c^3\,f^4-9\,b\,c^2\,d\,e^2\,f^2+3\,a\,c^2\,d\,e\,f^3+15\,b\,c\,d^2\,e^3\,f-9\,a\,c\,d^2\,e^2\,f^2-7\,b\,d^3\,e^4+5\,a\,d^3\,e^3\,f\right )}\right )\,{\left (c\,f-d\,e\right )}^2\,\left (a\,c\,f^2-7\,b\,d\,e^2+5\,a\,d\,e\,f+b\,c\,e\,f\right )}{2\,e^{3/2}\,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)^2,x)

[Out]

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

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