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

Optimal. Leaf size=304 \[ \frac {b \left (8 a^2 f^2-8 a b e f+3 b^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{8 d f^{5/2}}-\frac {b^2 x \sqrt {e+f x^2} (b c-a d)}{2 d^2 f}-\frac {3 b^2 x \sqrt {e+f x^2} (b e-2 a f)}{8 d f^2}+\frac {b^2 x \left (a+b x^2\right ) \sqrt {e+f x^2}}{4 d f}-\frac {(b c-a d)^3 \tan ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {c} \sqrt {e+f x^2}}\right )}{\sqrt {c} d^3 \sqrt {d e-c f}}+\frac {b (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{d^3 \sqrt {f}}+\frac {b (b c-a d) (b e-2 a f) \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{2 d^2 f^{3/2}} \]

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Rubi [A]  time = 0.30, antiderivative size = 304, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {545, 416, 388, 217, 206, 523, 377, 205} \begin {gather*} \frac {b \left (8 a^2 f^2-8 a b e f+3 b^2 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{8 d f^{5/2}}-\frac {b^2 x \sqrt {e+f x^2} (b c-a d)}{2 d^2 f}-\frac {3 b^2 x \sqrt {e+f x^2} (b e-2 a f)}{8 d f^2}+\frac {b^2 x \left (a+b x^2\right ) \sqrt {e+f x^2}}{4 d f}+\frac {b (b c-a d) (b e-2 a f) \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{2 d^2 f^{3/2}}-\frac {(b c-a d)^3 \tan ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {c} \sqrt {e+f x^2}}\right )}{\sqrt {c} d^3 \sqrt {d e-c f}}+\frac {b (b c-a d)^2 \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{d^3 \sqrt {f}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(b^2*(b*c - a*d)*x*Sqrt[e + f*x^2])/(2*d^2*f) - (3*b^2*(b*e - 2*a*f)*x*Sqrt[e + f*x^2])/(8*d*f^2) + (b^2*x*(a
+ b*x^2)*Sqrt[e + f*x^2])/(4*d*f) - ((b*c - a*d)^3*ArcTan[(Sqrt[d*e - c*f]*x)/(Sqrt[c]*Sqrt[e + f*x^2])])/(Sq
rt[c]*d^3*Sqrt[d*e - c*f]) + (b*(b*c - a*d)^2*ArcTanh[(Sqrt[f]*x)/Sqrt[e + f*x^2]])/(d^3*Sqrt[f]) + (b*(b*c -
a*d)*(b*e - 2*a*f)*ArcTanh[(Sqrt[f]*x)/Sqrt[e + f*x^2]])/(2*d^2*f^(3/2)) + (b*(3*b^2*e^2 - 8*a*b*e*f + 8*a^2*f
^2)*ArcTanh[(Sqrt[f]*x)/Sqrt[e + f*x^2]])/(8*d*f^(5/2))

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

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

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 416

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^(q - 1))/(b*(n*(p + q) + 1)), x] + Dist[1/(b*(n*(p + q) + 1)), Int[(a + b*x^n)^p*(c + d*x^n)^(q - 2)
*Simp[c*(b*c*(n*(p + q) + 1) - a*d) + d*(b*c*(n*(p + 2*q - 1) + 1) - a*d*(n*(q - 1) + 1))*x^n, x], x], x] /; F
reeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && GtQ[q, 1] && NeQ[n*(p + q) + 1, 0] &&  !IGtQ[p, 1] && IntB
inomialQ[a, b, c, d, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 545

Int[(((c_) + (d_.)*(x_)^2)^(q_)*((e_) + (f_.)*(x_)^2)^(r_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Dist[d/b, Int[
(c + d*x^2)^(q - 1)*(e + f*x^2)^r, x], x] + Dist[(b*c - a*d)/b, Int[((c + d*x^2)^(q - 1)*(e + f*x^2)^r)/(a + b
*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, r}, x] && GtQ[q, 1]

Rubi steps

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

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Mathematica [A]  time = 0.37, size = 265, normalized size = 0.87 \begin {gather*} \frac {\frac {8 b \left (3 a^2 d^2-3 a b c d+b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{\sqrt {f}}+\frac {4 b^2 d e (b c-3 a d) \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )}{f^{3/2}}-\frac {4 b^2 d x \sqrt {e+f x^2} (b c-3 a d)}{f}+\frac {8 (a d-b c)^3 \tan ^{-1}\left (\frac {x \sqrt {d e-c f}}{\sqrt {c} \sqrt {e+f x^2}}\right )}{\sqrt {c} \sqrt {d e-c f}}+\frac {3 b^3 d^2 e \left (e \tanh ^{-1}\left (\frac {\sqrt {f} x}{\sqrt {e+f x^2}}\right )-\sqrt {f} x \sqrt {e+f x^2}\right )}{f^{5/2}}+\frac {2 b^3 d^2 x^3 \sqrt {e+f x^2}}{f}}{8 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.67, size = 266, normalized size = 0.88 \begin {gather*} \frac {\log \left (\sqrt {e+f x^2}-\sqrt {f} x\right ) \left (-24 a^2 b d^2 f^2+24 a b^2 c d f^2+12 a b^2 d^2 e f-8 b^3 c^2 f^2-4 b^3 c d e f-3 b^3 d^2 e^2\right )}{8 d^3 f^{5/2}}+\frac {\left (-a^3 d^3+3 a^2 b c d^2-3 a b^2 c^2 d+b^3 c^3\right ) \tan ^{-1}\left (\frac {c \sqrt {f}-d x \sqrt {e+f x^2}+d \sqrt {f} x^2}{\sqrt {c} \sqrt {d e-c f}}\right )}{\sqrt {c} d^3 \sqrt {d e-c f}}+\frac {\sqrt {e+f x^2} \left (12 a b^2 d f x-4 b^3 c f x-3 b^3 d e x+2 b^3 d f x^3\right )}{8 d^2 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[e + f*x^2]*(-3*b^3*d*e*x - 4*b^3*c*f*x + 12*a*b^2*d*f*x + 2*b^3*d*f*x^3))/(8*d^2*f^2) + ((b^3*c^3 - 3*a*
b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*ArcTan[(c*Sqrt[f] + d*Sqrt[f]*x^2 - d*x*Sqrt[e + f*x^2])/(Sqrt[c]*Sqrt[d*
e - c*f])])/(Sqrt[c]*d^3*Sqrt[d*e - c*f]) + ((-3*b^3*d^2*e^2 - 4*b^3*c*d*e*f + 12*a*b^2*d^2*e*f - 8*b^3*c^2*f^
2 + 24*a*b^2*c*d*f^2 - 24*a^2*b*d^2*f^2)*Log[-(Sqrt[f]*x) + Sqrt[e + f*x^2]])/(8*d^3*f^(5/2))

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fricas [A]  time = 20.97, size = 1718, normalized size = 5.65

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/(d*x^2+c)/(f*x^2+e)^(1/2),x, algorithm="fricas")

[Out]

[1/16*(4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-c*d*e + c^2*f)*f^3*log(((d^2*e^2 - 8*c*d*e*
f + 8*c^2*f^2)*x^4 + c^2*e^2 - 2*(3*c*d*e^2 - 4*c^2*e*f)*x^2 - 4*((d*e - 2*c*f)*x^3 - c*e*x)*sqrt(-c*d*e + c^2
*f)*sqrt(f*x^2 + e))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + (3*b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f +
4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3)*e*f^2 - 8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqr
t(f)*log(-2*f*x^2 - 2*sqrt(f*x^2 + e)*sqrt(f)*x - e) + 2*(2*(b^3*c*d^3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c
*d^3*e^2*f + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(
c*d^4*e*f^3 - c^2*d^3*f^4), -1/16*(8*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(c*d*e - c^2*f)*f
^3*arctan(1/2*sqrt(c*d*e - c^2*f)*((d*e - 2*c*f)*x^2 - c*e)*sqrt(f*x^2 + e)/((c*d*e*f - c^2*f^2)*x^3 + (c*d*e^
2 - c^2*e*f)*x)) - (3*b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f + 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 +
6*a^2*b*c*d^3)*e*f^2 - 8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqrt(f)*log(-2*f*x^2 - 2*sqrt(f*x^2
+ e)*sqrt(f)*x - e) - 2*(2*(b^3*c*d^3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c*d^3*e^2*f + (b^3*c^2*d^2 - 12*a*
b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(c*d^4*e*f^3 - c^2*d^3*f^4), 1/8*(
2*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(-c*d*e + c^2*f)*f^3*log(((d^2*e^2 - 8*c*d*e*f + 8*c
^2*f^2)*x^4 + c^2*e^2 - 2*(3*c*d*e^2 - 4*c^2*e*f)*x^2 - 4*((d*e - 2*c*f)*x^3 - c*e*x)*sqrt(-c*d*e + c^2*f)*sqr
t(f*x^2 + e))/(d^2*x^4 + 2*c*d*x^2 + c^2)) - (3*b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f + 4*(b^3*
c^3*d - 3*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3)*e*f^2 - 8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqrt(-f)*a
rctan(sqrt(-f)*x/sqrt(f*x^2 + e)) + (2*(b^3*c*d^3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c*d^3*e^2*f + (b^3*c^2
*d^2 - 12*a*b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(c*d^4*e*f^3 - c^2*d^3
*f^4), -1/8*(4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*sqrt(c*d*e - c^2*f)*f^3*arctan(1/2*sqrt(c*d
*e - c^2*f)*((d*e - 2*c*f)*x^2 - c*e)*sqrt(f*x^2 + e)/((c*d*e*f - c^2*f^2)*x^3 + (c*d*e^2 - c^2*e*f)*x)) + (3*
b^3*c*d^3*e^3 + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e^2*f + 4*(b^3*c^3*d - 3*a*b^2*c^2*d^2 + 6*a^2*b*c*d^3)*e*f^2 -
8*(b^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2)*f^3)*sqrt(-f)*arctan(sqrt(-f)*x/sqrt(f*x^2 + e)) - (2*(b^3*c*d^
3*e*f^2 - b^3*c^2*d^2*f^3)*x^3 - (3*b^3*c*d^3*e^2*f + (b^3*c^2*d^2 - 12*a*b^2*c*d^3)*e*f^2 - 4*(b^3*c^3*d - 3*
a*b^2*c^2*d^2)*f^3)*x)*sqrt(f*x^2 + e))/(c*d^4*e*f^3 - c^2*d^3*f^4)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/(d*x^2+c)/(f*x^2+e)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:

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maple [B]  time = 0.04, size = 1541, normalized size = 5.07

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^3/(d*x^2+c)/(f*x^2+e)^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^3/(d*x^2+c)/(f*x^2+e)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,x^2+a\right )}^3}{\left (d\,x^2+c\right )\,\sqrt {f\,x^2+e}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^3/((c + d*x^2)*(e + f*x^2)^(1/2)),x)

[Out]

int((a + b*x^2)^3/((c + d*x^2)*(e + f*x^2)^(1/2)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**3/(d*x**2+c)/(f*x**2+e)**(1/2),x)

[Out]

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

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