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

Optimal. Leaf size=227 \[ -\frac {x \left (c+d x^2\right ) \left (7 a d f (5 d e-9 c f)-b \left (24 c^2 f^2-63 c d e f+35 d^2 e^2\right )\right )}{105 f^3}+\frac {x \left (7 a d f \left (33 c^2 f^2-40 c d e f+15 d^2 e^2\right )-b \left (-48 c^3 f^3+231 c^2 d e f^2-280 c d^2 e^2 f+105 d^3 e^3\right )\right )}{105 f^4}+\frac {(b e-a f) (d e-c f)^3 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{9/2}}-\frac {x \left (c+d x^2\right )^2 (-7 a d f-6 b c f+7 b d e)}{35 f^2}+\frac {b x \left (c+d x^2\right )^3}{7 f} \]

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Rubi [A]  time = 0.37, antiderivative size = 227, 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 = {528, 388, 205} \begin {gather*} -\frac {x \left (c+d x^2\right ) \left (7 a d f (5 d e-9 c f)-b \left (24 c^2 f^2-63 c d e f+35 d^2 e^2\right )\right )}{105 f^3}+\frac {x \left (7 a d f \left (33 c^2 f^2-40 c d e f+15 d^2 e^2\right )-b \left (231 c^2 d e f^2-48 c^3 f^3-280 c d^2 e^2 f+105 d^3 e^3\right )\right )}{105 f^4}-\frac {x \left (c+d x^2\right )^2 (-7 a d f-6 b c f+7 b d e)}{35 f^2}+\frac {(b e-a f) (d e-c f)^3 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{9/2}}+\frac {b x \left (c+d x^2\right )^3}{7 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*a*d*f*(15*d^2*e^2 - 40*c*d*e*f + 33*c^2*f^2) - b*(105*d^3*e^3 - 280*c*d^2*e^2*f + 231*c^2*d*e*f^2 - 48*c^3
*f^3))*x)/(105*f^4) - ((7*a*d*f*(5*d*e - 9*c*f) - b*(35*d^2*e^2 - 63*c*d*e*f + 24*c^2*f^2))*x*(c + d*x^2))/(10
5*f^3) - ((7*b*d*e - 6*b*c*f - 7*a*d*f)*x*(c + d*x^2)^2)/(35*f^2) + (b*x*(c + d*x^2)^3)/(7*f) + ((b*e - a*f)*(
d*e - c*f)^3*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(Sqrt[e]*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 528

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}{e+f x^2} \, dx &=\frac {b x \left (c+d x^2\right )^3}{7 f}+\frac {\int \frac {\left (c+d x^2\right )^2 \left (-c (b e-7 a f)+(-7 b d e+6 b c f+7 a d f) x^2\right )}{e+f x^2} \, dx}{7 f}\\ &=-\frac {(7 b d e-6 b c f-7 a d f) x \left (c+d x^2\right )^2}{35 f^2}+\frac {b x \left (c+d x^2\right )^3}{7 f}+\frac {\int \frac {\left (c+d x^2\right ) \left (c (b e (7 d e-11 c f)-7 a f (d e-5 c f))+\left (-7 a d f (5 d e-9 c f)+b \left (35 d^2 e^2-63 c d e f+24 c^2 f^2\right )\right ) x^2\right )}{e+f x^2} \, dx}{35 f^2}\\ &=-\frac {\left (7 a d f (5 d e-9 c f)-b \left (35 d^2 e^2-63 c d e f+24 c^2 f^2\right )\right ) x \left (c+d x^2\right )}{105 f^3}-\frac {(7 b d e-6 b c f-7 a d f) x \left (c+d x^2\right )^2}{35 f^2}+\frac {b x \left (c+d x^2\right )^3}{7 f}+\frac {\int \frac {c \left (7 a f \left (5 d^2 e^2-12 c d e f+15 c^2 f^2\right )-b e \left (35 d^2 e^2-84 c d e f+57 c^2 f^2\right )\right )+\left (7 a d f \left (15 d^2 e^2-40 c d e f+33 c^2 f^2\right )-b \left (105 d^3 e^3-280 c d^2 e^2 f+231 c^2 d e f^2-48 c^3 f^3\right )\right ) x^2}{e+f x^2} \, dx}{105 f^3}\\ &=\frac {\left (7 a d f \left (15 d^2 e^2-40 c d e f+33 c^2 f^2\right )-b \left (105 d^3 e^3-280 c d^2 e^2 f+231 c^2 d e f^2-48 c^3 f^3\right )\right ) x}{105 f^4}-\frac {\left (7 a d f (5 d e-9 c f)-b \left (35 d^2 e^2-63 c d e f+24 c^2 f^2\right )\right ) x \left (c+d x^2\right )}{105 f^3}-\frac {(7 b d e-6 b c f-7 a d f) x \left (c+d x^2\right )^2}{35 f^2}+\frac {b x \left (c+d x^2\right )^3}{7 f}+\frac {\left ((b e-a f) (d e-c f)^3\right ) \int \frac {1}{e+f x^2} \, dx}{f^4}\\ &=\frac {\left (7 a d f \left (15 d^2 e^2-40 c d e f+33 c^2 f^2\right )-b \left (105 d^3 e^3-280 c d^2 e^2 f+231 c^2 d e f^2-48 c^3 f^3\right )\right ) x}{105 f^4}-\frac {\left (7 a d f (5 d e-9 c f)-b \left (35 d^2 e^2-63 c d e f+24 c^2 f^2\right )\right ) x \left (c+d x^2\right )}{105 f^3}-\frac {(7 b d e-6 b c f-7 a d f) x \left (c+d x^2\right )^2}{35 f^2}+\frac {b x \left (c+d x^2\right )^3}{7 f}+\frac {(b e-a f) (d e-c f)^3 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 179, normalized size = 0.79 \begin {gather*} \frac {x \left (a d f \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )-b (d e-c f)^3\right )}{f^4}+\frac {d x^3 \left (a d f (3 c f-d e)+b \left (3 c^2 f^2-3 c d e f+d^2 e^2\right )\right )}{3 f^3}+\frac {d^2 x^5 (a d f+3 b c f-b d e)}{5 f^2}+\frac {(b e-a f) (d e-c f)^3 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{9/2}}+\frac {b d^3 x^7}{7 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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 )^3}{e+f x^2} \, 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),x]

[Out]

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

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fricas [A]  time = 1.00, size = 586, normalized size = 2.58 \begin {gather*} \left [\frac {30 \, b d^{3} e f^{4} x^{7} - 42 \, {\left (b d^{3} e^{2} f^{3} - {\left (3 \, b c d^{2} + a d^{3}\right )} e f^{4}\right )} x^{5} + 70 \, {\left (b d^{3} e^{3} f^{2} - {\left (3 \, b c d^{2} + a d^{3}\right )} e^{2} f^{3} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} e f^{4}\right )} x^{3} - 105 \, {\left (b d^{3} e^{4} + a c^{3} f^{4} - {\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 )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) - 210 \, {\left (b d^{3} e^{4} f - {\left (3 \, b c d^{2} + a d^{3}\right )} e^{3} f^{2} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} e^{2} f^{3} - {\left (b c^{3} + 3 \, a c^{2} d\right )} e f^{4}\right )} x}{210 \, e f^{5}}, \frac {15 \, b d^{3} e f^{4} x^{7} - 21 \, {\left (b d^{3} e^{2} f^{3} - {\left (3 \, b c d^{2} + a d^{3}\right )} e f^{4}\right )} x^{5} + 35 \, {\left (b d^{3} e^{3} f^{2} - {\left (3 \, b c d^{2} + a d^{3}\right )} e^{2} f^{3} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} e f^{4}\right )} x^{3} + 105 \, {\left (b d^{3} e^{4} + a c^{3} f^{4} - {\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 )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) - 105 \, {\left (b d^{3} e^{4} f - {\left (3 \, b c d^{2} + a d^{3}\right )} e^{3} f^{2} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} e^{2} f^{3} - {\left (b c^{3} + 3 \, a c^{2} d\right )} e f^{4}\right )} x}{105 \, e f^{5}}\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),x, algorithm="fricas")

[Out]

[1/210*(30*b*d^3*e*f^4*x^7 - 42*(b*d^3*e^2*f^3 - (3*b*c*d^2 + a*d^3)*e*f^4)*x^5 + 70*(b*d^3*e^3*f^2 - (3*b*c*d
^2 + a*d^3)*e^2*f^3 + 3*(b*c^2*d + a*c*d^2)*e*f^4)*x^3 - 105*(b*d^3*e^4 + a*c^3*f^4 - (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)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x - e)/(f*
x^2 + e)) - 210*(b*d^3*e^4*f - (3*b*c*d^2 + a*d^3)*e^3*f^2 + 3*(b*c^2*d + a*c*d^2)*e^2*f^3 - (b*c^3 + 3*a*c^2*
d)*e*f^4)*x)/(e*f^5), 1/105*(15*b*d^3*e*f^4*x^7 - 21*(b*d^3*e^2*f^3 - (3*b*c*d^2 + a*d^3)*e*f^4)*x^5 + 35*(b*d
^3*e^3*f^2 - (3*b*c*d^2 + a*d^3)*e^2*f^3 + 3*(b*c^2*d + a*c*d^2)*e*f^4)*x^3 + 105*(b*d^3*e^4 + a*c^3*f^4 - (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)*sqrt(e*f)*arctan(sqrt(e*f)
*x/e) - 105*(b*d^3*e^4*f - (3*b*c*d^2 + a*d^3)*e^3*f^2 + 3*(b*c^2*d + a*c*d^2)*e^2*f^3 - (b*c^3 + 3*a*c^2*d)*e
*f^4)*x)/(e*f^5)]

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giac [A]  time = 0.49, size = 307, normalized size = 1.35 \begin {gather*} \frac {{\left (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} - 3 \, b c d^{2} f e^{3} - a d^{3} f e^{3} + b d^{3} e^{4}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {1}{2}\right )}}{f^{\frac {9}{2}}} + \frac {15 \, b d^{3} f^{6} x^{7} + 63 \, b c d^{2} f^{6} x^{5} + 21 \, a d^{3} f^{6} x^{5} - 21 \, b d^{3} f^{5} x^{5} e + 105 \, b c^{2} d f^{6} x^{3} + 105 \, a c d^{2} f^{6} x^{3} - 105 \, b c d^{2} f^{5} x^{3} e - 35 \, a d^{3} f^{5} x^{3} e + 35 \, b d^{3} f^{4} x^{3} e^{2} + 105 \, b c^{3} f^{6} x + 315 \, a c^{2} d f^{6} x - 315 \, b c^{2} d f^{5} x e - 315 \, a c d^{2} f^{5} x e + 315 \, b c d^{2} f^{4} x e^{2} + 105 \, a d^{3} f^{4} x e^{2} - 105 \, b d^{3} f^{3} x e^{3}}{105 \, f^{7}} \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),x, algorithm="giac")

[Out]

(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 - 3*b*c*d^2*f*e^3 - a*d^3*f
*e^3 + b*d^3*e^4)*arctan(sqrt(f)*x*e^(-1/2))*e^(-1/2)/f^(9/2) + 1/105*(15*b*d^3*f^6*x^7 + 63*b*c*d^2*f^6*x^5 +
21*a*d^3*f^6*x^5 - 21*b*d^3*f^5*x^5*e + 105*b*c^2*d*f^6*x^3 + 105*a*c*d^2*f^6*x^3 - 105*b*c*d^2*f^5*x^3*e - 3
5*a*d^3*f^5*x^3*e + 35*b*d^3*f^4*x^3*e^2 + 105*b*c^3*f^6*x + 315*a*c^2*d*f^6*x - 315*b*c^2*d*f^5*x*e - 315*a*c
*d^2*f^5*x*e + 315*b*c*d^2*f^4*x*e^2 + 105*a*d^3*f^4*x*e^2 - 105*b*d^3*f^3*x*e^3)/f^7

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maple [A]  time = 0.01, size = 401, normalized size = 1.77 \begin {gather*} \frac {b \,d^{3} x^{7}}{7 f}+\frac {a \,d^{3} x^{5}}{5 f}+\frac {3 b c \,d^{2} x^{5}}{5 f}-\frac {b \,d^{3} e \,x^{5}}{5 f^{2}}+\frac {a c \,d^{2} x^{3}}{f}-\frac {a \,d^{3} e \,x^{3}}{3 f^{2}}+\frac {b \,c^{2} d \,x^{3}}{f}-\frac {b c \,d^{2} e \,x^{3}}{f^{2}}+\frac {b \,d^{3} e^{2} x^{3}}{3 f^{3}}+\frac {a \,c^{3} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}}-\frac {3 a \,c^{2} d e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f}+\frac {3 a c \,d^{2} e^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{2}}-\frac {a \,d^{3} e^{3} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{3}}-\frac {b \,c^{3} e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f}+\frac {3 b \,c^{2} d \,e^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{2}}-\frac {3 b c \,d^{2} e^{3} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{3}}+\frac {b \,d^{3} e^{4} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{4}}+\frac {3 a \,c^{2} d x}{f}-\frac {3 a c \,d^{2} e x}{f^{2}}+\frac {a \,d^{3} e^{2} x}{f^{3}}+\frac {b \,c^{3} x}{f}-\frac {3 b \,c^{2} d e x}{f^{2}}+\frac {3 b c \,d^{2} e^{2} x}{f^{3}}-\frac {b \,d^{3} e^{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),x)

[Out]

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

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maxima [A]  time = 2.03, size = 266, normalized size = 1.17 \begin {gather*} \frac {{\left (b d^{3} e^{4} + a c^{3} f^{4} - {\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 )}{\sqrt {e f} f^{4}} + \frac {15 \, b d^{3} f^{3} x^{7} - 21 \, {\left (b d^{3} e f^{2} - {\left (3 \, b c d^{2} + a d^{3}\right )} f^{3}\right )} x^{5} + 35 \, {\left (b d^{3} e^{2} f - {\left (3 \, b c d^{2} + a d^{3}\right )} e f^{2} + 3 \, {\left (b c^{2} d + a c d^{2}\right )} f^{3}\right )} x^{3} - 105 \, {\left (b d^{3} e^{3} - {\left (3 \, b c d^{2} + a d^{3}\right )} e^{2} f + 3 \, {\left (b c^{2} d + a c d^{2}\right )} e f^{2} - {\left (b c^{3} + 3 \, a c^{2} d\right )} f^{3}\right )} x}{105 \, 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),x, algorithm="maxima")

[Out]

(b*d^3*e^4 + a*c^3*f^4 - (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)*f^4) + 1/105*(15*b*d^3*f^3*x^7 - 21*(b*d^3*e*f^2 - (3*b*c*d^2 + a*d^3)*f^3)
*x^5 + 35*(b*d^3*e^2*f - (3*b*c*d^2 + a*d^3)*e*f^2 + 3*(b*c^2*d + a*c*d^2)*f^3)*x^3 - 105*(b*d^3*e^3 - (3*b*c*
d^2 + a*d^3)*e^2*f + 3*(b*c^2*d + a*c*d^2)*e*f^2 - (b*c^3 + 3*a*c^2*d)*f^3)*x)/f^4

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mupad [B]  time = 0.85, size = 312, normalized size = 1.37 \begin {gather*} x\,\left (\frac {b\,c^3+3\,a\,d\,c^2}{f}+\frac {e\,\left (\frac {e\,\left (\frac {a\,d^3+3\,b\,c\,d^2}{f}-\frac {b\,d^3\,e}{f^2}\right )}{f}-\frac {3\,c\,d\,\left (a\,d+b\,c\right )}{f}\right )}{f}\right )+x^5\,\left (\frac {a\,d^3+3\,b\,c\,d^2}{5\,f}-\frac {b\,d^3\,e}{5\,f^2}\right )-x^3\,\left (\frac {e\,\left (\frac {a\,d^3+3\,b\,c\,d^2}{f}-\frac {b\,d^3\,e}{f^2}\right )}{3\,f}-\frac {c\,d\,\left (a\,d+b\,c\right )}{f}\right )+\frac {b\,d^3\,x^7}{7\,f}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x\,\left (a\,f-b\,e\right )\,{\left (c\,f-d\,e\right )}^3}{\sqrt {e}\,\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 )}\right )\,\left (a\,f-b\,e\right )\,{\left (c\,f-d\,e\right )}^3}{\sqrt {e}\,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),x)

[Out]

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

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sympy [B]  time = 1.49, size = 508, normalized size = 2.24 \begin {gather*} \frac {b d^{3} x^{7}}{7 f} + x^{5} \left (\frac {a d^{3}}{5 f} + \frac {3 b c d^{2}}{5 f} - \frac {b d^{3} e}{5 f^{2}}\right ) + x^{3} \left (\frac {a c d^{2}}{f} - \frac {a d^{3} e}{3 f^{2}} + \frac {b c^{2} d}{f} - \frac {b c d^{2} e}{f^{2}} + \frac {b d^{3} e^{2}}{3 f^{3}}\right ) + x \left (\frac {3 a c^{2} d}{f} - \frac {3 a c d^{2} e}{f^{2}} + \frac {a d^{3} e^{2}}{f^{3}} + \frac {b c^{3}}{f} - \frac {3 b c^{2} d e}{f^{2}} + \frac {3 b c d^{2} e^{2}}{f^{3}} - \frac {b d^{3} e^{3}}{f^{4}}\right ) - \frac {\sqrt {- \frac {1}{e f^{9}}} \left (a f - b e\right ) \left (c f - d e\right )^{3} \log {\left (- \frac {e f^{4} \sqrt {- \frac {1}{e f^{9}}} \left (a f - b e\right ) \left (c f - d e\right )^{3}}{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}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{e f^{9}}} \left (a f - b e\right ) \left (c f - d e\right )^{3} \log {\left (\frac {e f^{4} \sqrt {- \frac {1}{e f^{9}}} \left (a f - b e\right ) \left (c f - d e\right )^{3}}{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}} + x \right )}}{2} \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),x)

[Out]

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

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