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

Optimal. Leaf size=142 \[ -\frac {x \left (5 a d f (3 d e-5 c f)-b \left (8 c^2 f^2-25 c d e f+15 d^2 e^2\right )\right )}{15 f^3}-\frac {(b e-a f) (d e-c f)^2 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{7/2}}-\frac {x \left (c+d x^2\right ) (-5 a d f-4 b c f+5 b d e)}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f} \]

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Rubi [A]  time = 0.21, antiderivative size = 142, 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 = {528, 388, 205} \begin {gather*} -\frac {x \left (5 a d f (3 d e-5 c f)-b \left (8 c^2 f^2-25 c d e f+15 d^2 e^2\right )\right )}{15 f^3}-\frac {x \left (c+d x^2\right ) (-5 a d f-4 b c f+5 b d e)}{15 f^2}-\frac {(b e-a f) (d e-c f)^2 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{7/2}}+\frac {b x \left (c+d x^2\right )^2}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((5*a*d*f*(3*d*e - 5*c*f) - b*(15*d^2*e^2 - 25*c*d*e*f + 8*c^2*f^2))*x)/(15*f^3) - ((5*b*d*e - 4*b*c*f - 5*a*
d*f)*x*(c + d*x^2))/(15*f^2) + (b*x*(c + d*x^2)^2)/(5*f) - ((b*e - a*f)*(d*e - c*f)^2*ArcTan[(Sqrt[f]*x)/Sqrt[
e]])/(Sqrt[e]*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 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 )^2}{e+f x^2} \, dx &=\frac {b x \left (c+d x^2\right )^2}{5 f}+\frac {\int \frac {\left (c+d x^2\right ) \left (-c (b e-5 a f)+(-5 b d e+4 b c f+5 a d f) x^2\right )}{e+f x^2} \, dx}{5 f}\\ &=-\frac {(5 b d e-4 b c f-5 a d f) x \left (c+d x^2\right )}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f}+\frac {\int \frac {c (b e (5 d e-7 c f)-5 a f (d e-3 c f))-\left (5 a d f (3 d e-5 c f)-b \left (15 d^2 e^2-25 c d e f+8 c^2 f^2\right )\right ) x^2}{e+f x^2} \, dx}{15 f^2}\\ &=-\frac {\left (5 a d f (3 d e-5 c f)-b \left (15 d^2 e^2-25 c d e f+8 c^2 f^2\right )\right ) x}{15 f^3}-\frac {(5 b d e-4 b c f-5 a d f) x \left (c+d x^2\right )}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f}-\frac {\left ((b e-a f) (d e-c f)^2\right ) \int \frac {1}{e+f x^2} \, dx}{f^3}\\ &=-\frac {\left (5 a d f (3 d e-5 c f)-b \left (15 d^2 e^2-25 c d e f+8 c^2 f^2\right )\right ) x}{15 f^3}-\frac {(5 b d e-4 b c f-5 a d f) x \left (c+d x^2\right )}{15 f^2}+\frac {b x \left (c+d x^2\right )^2}{5 f}-\frac {(b e-a f) (d e-c f)^2 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 115, normalized size = 0.81 \begin {gather*} -\frac {(b e-a f) (d e-c f)^2 \tan ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e}}\right )}{\sqrt {e} f^{7/2}}+\frac {x \left (a d f (2 c f-d e)+b (d e-c f)^2\right )}{f^3}+\frac {d x^3 (a d f+2 b c f-b d e)}{3 f^2}+\frac {b d^2 x^5}{5 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((b*(d*e - c*f)^2 + a*d*f*(-(d*e) + 2*c*f))*x)/f^3 + (d*(-(b*d*e) + 2*b*c*f + a*d*f)*x^3)/(3*f^2) + (b*d^2*x^5
)/(5*f) - ((b*e - a*f)*(d*e - c*f)^2*ArcTan[(Sqrt[f]*x)/Sqrt[e]])/(Sqrt[e]*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}{e+f x^2} \, 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),x]

[Out]

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

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fricas [A]  time = 1.19, size = 366, normalized size = 2.58 \begin {gather*} \left [\frac {6 \, b d^{2} e f^{3} x^{5} - 10 \, {\left (b d^{2} e^{2} f^{2} - {\left (2 \, b c d + a d^{2}\right )} e f^{3}\right )} x^{3} + 15 \, {\left (b d^{2} e^{3} - a c^{2} f^{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 )} \sqrt {-e f} \log \left (\frac {f x^{2} - 2 \, \sqrt {-e f} x - e}{f x^{2} + e}\right ) + 30 \, {\left (b d^{2} e^{3} f - {\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}{30 \, e f^{4}}, \frac {3 \, b d^{2} e f^{3} x^{5} - 5 \, {\left (b d^{2} e^{2} f^{2} - {\left (2 \, b c d + a d^{2}\right )} e f^{3}\right )} x^{3} - 15 \, {\left (b d^{2} e^{3} - a c^{2} f^{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 )} \sqrt {e f} \arctan \left (\frac {\sqrt {e f} x}{e}\right ) + 15 \, {\left (b d^{2} e^{3} f - {\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}{15 \, e f^{4}}\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),x, algorithm="fricas")

[Out]

[1/30*(6*b*d^2*e*f^3*x^5 - 10*(b*d^2*e^2*f^2 - (2*b*c*d + a*d^2)*e*f^3)*x^3 + 15*(b*d^2*e^3 - a*c^2*f^3 - (2*b
*c*d + a*d^2)*e^2*f + (b*c^2 + 2*a*c*d)*e*f^2)*sqrt(-e*f)*log((f*x^2 - 2*sqrt(-e*f)*x - e)/(f*x^2 + e)) + 30*(
b*d^2*e^3*f - (2*b*c*d + a*d^2)*e^2*f^2 + (b*c^2 + 2*a*c*d)*e*f^3)*x)/(e*f^4), 1/15*(3*b*d^2*e*f^3*x^5 - 5*(b*
d^2*e^2*f^2 - (2*b*c*d + a*d^2)*e*f^3)*x^3 - 15*(b*d^2*e^3 - a*c^2*f^3 - (2*b*c*d + a*d^2)*e^2*f + (b*c^2 + 2*
a*c*d)*e*f^2)*sqrt(e*f)*arctan(sqrt(e*f)*x/e) + 15*(b*d^2*e^3*f - (2*b*c*d + a*d^2)*e^2*f^2 + (b*c^2 + 2*a*c*d
)*e*f^3)*x)/(e*f^4)]

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giac [A]  time = 0.29, size = 178, normalized size = 1.25 \begin {gather*} \frac {{\left (a c^{2} f^{3} - b c^{2} f^{2} e - 2 \, a c d f^{2} e + 2 \, b c d f e^{2} + a d^{2} f e^{2} - b d^{2} e^{3}\right )} \arctan \left (\sqrt {f} x e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {1}{2}\right )}}{f^{\frac {7}{2}}} + \frac {3 \, b d^{2} f^{4} x^{5} + 10 \, b c d f^{4} x^{3} + 5 \, a d^{2} f^{4} x^{3} - 5 \, b d^{2} f^{3} x^{3} e + 15 \, b c^{2} f^{4} x + 30 \, a c d f^{4} x - 30 \, b c d f^{3} x e - 15 \, a d^{2} f^{3} x e + 15 \, b d^{2} f^{2} x e^{2}}{15 \, f^{5}} \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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 243, normalized size = 1.71 \begin {gather*} \frac {b \,d^{2} x^{5}}{5 f}+\frac {a \,d^{2} x^{3}}{3 f}+\frac {2 b c d \,x^{3}}{3 f}-\frac {b \,d^{2} e \,x^{3}}{3 f^{2}}+\frac {a \,c^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}}-\frac {2 a c d e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f}+\frac {a \,d^{2} e^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{2}}-\frac {b \,c^{2} e \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f}+\frac {2 b c d \,e^{2} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{2}}-\frac {b \,d^{2} e^{3} \arctan \left (\frac {f x}{\sqrt {e f}}\right )}{\sqrt {e f}\, f^{3}}+\frac {2 a c d x}{f}-\frac {a \,d^{2} e x}{f^{2}}+\frac {b \,c^{2} x}{f}-\frac {2 b c d e x}{f^{2}}+\frac {b \,d^{2} e^{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),x)

[Out]

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

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maxima [A]  time = 1.35, size = 160, normalized size = 1.13 \begin {gather*} -\frac {{\left (b d^{2} e^{3} - a c^{2} f^{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 )}{\sqrt {e f} f^{3}} + \frac {3 \, b d^{2} f^{2} x^{5} - 5 \, {\left (b d^{2} e f - {\left (2 \, b c d + a d^{2}\right )} f^{2}\right )} x^{3} + 15 \, {\left (b d^{2} e^{2} - {\left (2 \, b c d + a d^{2}\right )} e f + {\left (b c^{2} + 2 \, a c d\right )} f^{2}\right )} x}{15 \, 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),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.86, size = 203, normalized size = 1.43 \begin {gather*} x^3\,\left (\frac {a\,d^2+2\,b\,c\,d}{3\,f}-\frac {b\,d^2\,e}{3\,f^2}\right )+x\,\left (\frac {b\,c^2+2\,a\,d\,c}{f}-\frac {e\,\left (\frac {a\,d^2+2\,b\,c\,d}{f}-\frac {b\,d^2\,e}{f^2}\right )}{f}\right )+\frac {b\,d^2\,x^5}{5\,f}+\frac {\mathrm {atan}\left (\frac {\sqrt {f}\,x\,\left (a\,f-b\,e\right )\,{\left (c\,f-d\,e\right )}^2}{\sqrt {e}\,\left (-b\,c^2\,e\,f^2+a\,c^2\,f^3+2\,b\,c\,d\,e^2\,f-2\,a\,c\,d\,e\,f^2-b\,d^2\,e^3+a\,d^2\,e^2\,f\right )}\right )\,\left (a\,f-b\,e\right )\,{\left (c\,f-d\,e\right )}^2}{\sqrt {e}\,f^{7/2}} \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),x)

[Out]

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

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sympy [B]  time = 1.04, size = 347, normalized size = 2.44 \begin {gather*} \frac {b d^{2} x^{5}}{5 f} + x^{3} \left (\frac {a d^{2}}{3 f} + \frac {2 b c d}{3 f} - \frac {b d^{2} e}{3 f^{2}}\right ) + x \left (\frac {2 a c d}{f} - \frac {a d^{2} e}{f^{2}} + \frac {b c^{2}}{f} - \frac {2 b c d e}{f^{2}} + \frac {b d^{2} e^{2}}{f^{3}}\right ) - \frac {\sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2} \log {\left (- \frac {e f^{3} \sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2}}{a c^{2} f^{3} - 2 a c d e f^{2} + a d^{2} e^{2} f - b c^{2} e f^{2} + 2 b c d e^{2} f - b d^{2} e^{3}} + x \right )}}{2} + \frac {\sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2} \log {\left (\frac {e f^{3} \sqrt {- \frac {1}{e f^{7}}} \left (a f - b e\right ) \left (c f - d e\right )^{2}}{a c^{2} f^{3} - 2 a c d e f^{2} + a d^{2} e^{2} f - b c^{2} e f^{2} + 2 b c d e^{2} f - b d^{2} e^{3}} + 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)**2/(f*x**2+e),x)

[Out]

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

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