3.1.19 \(\int \frac {(a+b x^2)^3}{(c+d x^2)^3} \, dx\) [19]

Optimal. Leaf size=130 \[ \frac {b^3 x}{d^3}-\frac {(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac {3 (b c-a d)^2 (3 b c+a d) x}{8 c^2 d^3 \left (c+d x^2\right )}-\frac {3 (b c-a d) \left (4 b^2 c^2+(b c+a d)^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{7/2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {398, 1171, 393, 211} \begin {gather*} -\frac {3 (b c-a d) \left ((a d+b c)^2+4 b^2 c^2\right ) \text {ArcTan}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{7/2}}+\frac {3 x (b c-a d)^2 (a d+3 b c)}{8 c^2 d^3 \left (c+d x^2\right )}-\frac {x (b c-a d)^3}{4 c d^3 \left (c+d x^2\right )^2}+\frac {b^3 x}{d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^3/(c + d*x^2)^3,x]

[Out]

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

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 393

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

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+b x^2\right )^3}{\left (c+d x^2\right )^3} \, dx &=\int \left (\frac {b^3}{d^3}-\frac {b^3 c^3-a^3 d^3+3 b d (b c-a d) (b c+a d) x^2+3 b^2 d^2 (b c-a d) x^4}{d^3 \left (c+d x^2\right )^3}\right ) \, dx\\ &=\frac {b^3 x}{d^3}-\frac {\int \frac {b^3 c^3-a^3 d^3+3 b d (b c-a d) (b c+a d) x^2+3 b^2 d^2 (b c-a d) x^4}{\left (c+d x^2\right )^3} \, dx}{d^3}\\ &=\frac {b^3 x}{d^3}-\frac {(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac {\int \frac {-3 (b c-a d) (b c+a d)^2-12 b^2 c d (b c-a d) x^2}{\left (c+d x^2\right )^2} \, dx}{4 c d^3}\\ &=\frac {b^3 x}{d^3}-\frac {(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac {3 (b c-a d)^2 (3 b c+a d) x}{8 c^2 d^3 \left (c+d x^2\right )}-\frac {\left (3 (b c-a d) \left (4 b^2 c^2+(b c+a d)^2\right )\right ) \int \frac {1}{c+d x^2} \, dx}{8 c^2 d^3}\\ &=\frac {b^3 x}{d^3}-\frac {(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac {3 (b c-a d)^2 (3 b c+a d) x}{8 c^2 d^3 \left (c+d x^2\right )}-\frac {3 (b c-a d) \left (4 b^2 c^2+(b c+a d)^2\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 141, normalized size = 1.08 \begin {gather*} \frac {b^3 x}{d^3}-\frac {(b c-a d)^3 x}{4 c d^3 \left (c+d x^2\right )^2}+\frac {3 (b c-a d)^2 (3 b c+a d) x}{8 c^2 d^3 \left (c+d x^2\right )}-\frac {3 \left (5 b^3 c^3-3 a b^2 c^2 d-a^2 b c d^2-a^3 d^3\right ) \tan ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c}}\right )}{8 c^{5/2} d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.09, size = 167, normalized size = 1.28

method	result	size
default	\(\frac {b^{3} x}{d^{3}}+\frac {\frac {\frac {3 d \left (a^{3} d^{3}+a^{2} b c \,d^{2}-5 a \,b^{2} c^{2} d +3 b^{3} c^{3}\right ) x^{3}}{8 c^{2}}+\frac {\left (5 a^{3} d^{3}-3 a^{2} b c \,d^{2}-9 a \,b^{2} c^{2} d +7 b^{3} c^{3}\right ) x}{8 c}}{\left (d \,x^{2}+c \right )^{2}}+\frac {3 \left (a^{3} d^{3}+a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 c^{2} \sqrt {c d}}}{d^{3}}\)	\(167\)
risch	\(\frac {b^{3} x}{d^{3}}+\frac {\frac {3 d \left (a^{3} d^{3}+a^{2} b c \,d^{2}-5 a \,b^{2} c^{2} d +3 b^{3} c^{3}\right ) x^{3}}{8 c^{2}}+\frac {\left (5 a^{3} d^{3}-3 a^{2} b c \,d^{2}-9 a \,b^{2} c^{2} d +7 b^{3} c^{3}\right ) x}{8 c}}{d^{3} \left (d \,x^{2}+c \right )^{2}}-\frac {3 \ln \left (d x +\sqrt {-c d}\right ) a^{3}}{16 \sqrt {-c d}\, c^{2}}-\frac {3 \ln \left (d x +\sqrt {-c d}\right ) a^{2} b}{16 d \sqrt {-c d}\, c}-\frac {9 \ln \left (d x +\sqrt {-c d}\right ) a \,b^{2}}{16 d^{2} \sqrt {-c d}}+\frac {15 c \ln \left (d x +\sqrt {-c d}\right ) b^{3}}{16 d^{3} \sqrt {-c d}}+\frac {3 \ln \left (-d x +\sqrt {-c d}\right ) a^{3}}{16 \sqrt {-c d}\, c^{2}}+\frac {3 \ln \left (-d x +\sqrt {-c d}\right ) a^{2} b}{16 d \sqrt {-c d}\, c}+\frac {9 \ln \left (-d x +\sqrt {-c d}\right ) a \,b^{2}}{16 d^{2} \sqrt {-c d}}-\frac {15 c \ln \left (-d x +\sqrt {-c d}\right ) b^{3}}{16 d^{3} \sqrt {-c d}}\)	\(327\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^3/(d*x^2+c)^3,x,method=_RETURNVERBOSE)

[Out]

b^3*x/d^3+1/d^3*((3/8*d*(a^3*d^3+a^2*b*c*d^2-5*a*b^2*c^2*d+3*b^3*c^3)/c^2*x^3+1/8*(5*a^3*d^3-3*a^2*b*c*d^2-9*a
*b^2*c^2*d+7*b^3*c^3)/c*x)/(d*x^2+c)^2+3/8*(a^3*d^3+a^2*b*c*d^2+3*a*b^2*c^2*d-5*b^3*c^3)/c^2/(c*d)^(1/2)*arcta
n(d*x/(c*d)^(1/2)))

________________________________________________________________________________________

Maxima [A]
time = 0.52, size = 187, normalized size = 1.44 \begin {gather*} \frac {b^{3} x}{d^{3}} + \frac {3 \, {\left (3 \, b^{3} c^{3} d - 5 \, a b^{2} c^{2} d^{2} + a^{2} b c d^{3} + a^{3} d^{4}\right )} x^{3} + {\left (7 \, b^{3} c^{4} - 9 \, a b^{2} c^{3} d - 3 \, a^{2} b c^{2} d^{2} + 5 \, a^{3} c d^{3}\right )} x}{8 \, {\left (c^{2} d^{5} x^{4} + 2 \, c^{3} d^{4} x^{2} + c^{4} d^{3}\right )}} - \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 298 vs. \(2 (116) = 232\).
time = 0.47, size = 618, normalized size = 4.75 \begin {gather*} \left [\frac {16 \, b^{3} c^{3} d^{3} x^{5} + 2 \, {\left (25 \, b^{3} c^{4} d^{2} - 15 \, a b^{2} c^{3} d^{3} + 3 \, a^{2} b c^{2} d^{4} + 3 \, a^{3} c d^{5}\right )} x^{3} + 3 \, {\left (5 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3} + {\left (5 \, b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} - a^{2} b c d^{4} - a^{3} d^{5}\right )} x^{4} + 2 \, {\left (5 \, b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} - a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} x^{2}\right )} \sqrt {-c d} \log \left (\frac {d x^{2} - 2 \, \sqrt {-c d} x - c}{d x^{2} + c}\right ) + 2 \, {\left (15 \, b^{3} c^{5} d - 9 \, a b^{2} c^{4} d^{2} - 3 \, a^{2} b c^{3} d^{3} + 5 \, a^{3} c^{2} d^{4}\right )} x}{16 \, {\left (c^{3} d^{6} x^{4} + 2 \, c^{4} d^{5} x^{2} + c^{5} d^{4}\right )}}, \frac {8 \, b^{3} c^{3} d^{3} x^{5} + {\left (25 \, b^{3} c^{4} d^{2} - 15 \, a b^{2} c^{3} d^{3} + 3 \, a^{2} b c^{2} d^{4} + 3 \, a^{3} c d^{5}\right )} x^{3} - 3 \, {\left (5 \, b^{3} c^{5} - 3 \, a b^{2} c^{4} d - a^{2} b c^{3} d^{2} - a^{3} c^{2} d^{3} + {\left (5 \, b^{3} c^{3} d^{2} - 3 \, a b^{2} c^{2} d^{3} - a^{2} b c d^{4} - a^{3} d^{5}\right )} x^{4} + 2 \, {\left (5 \, b^{3} c^{4} d - 3 \, a b^{2} c^{3} d^{2} - a^{2} b c^{2} d^{3} - a^{3} c d^{4}\right )} x^{2}\right )} \sqrt {c d} \arctan \left (\frac {\sqrt {c d} x}{c}\right ) + {\left (15 \, b^{3} c^{5} d - 9 \, a b^{2} c^{4} d^{2} - 3 \, a^{2} b c^{3} d^{3} + 5 \, a^{3} c^{2} d^{4}\right )} x}{8 \, {\left (c^{3} d^{6} x^{4} + 2 \, c^{4} d^{5} x^{2} + c^{5} d^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(16*b^3*c^3*d^3*x^5 + 2*(25*b^3*c^4*d^2 - 15*a*b^2*c^3*d^3 + 3*a^2*b*c^2*d^4 + 3*a^3*c*d^5)*x^3 + 3*(5*b
^3*c^5 - 3*a*b^2*c^4*d - a^2*b*c^3*d^2 - a^3*c^2*d^3 + (5*b^3*c^3*d^2 - 3*a*b^2*c^2*d^3 - a^2*b*c*d^4 - a^3*d^
5)*x^4 + 2*(5*b^3*c^4*d - 3*a*b^2*c^3*d^2 - a^2*b*c^2*d^3 - a^3*c*d^4)*x^2)*sqrt(-c*d)*log((d*x^2 - 2*sqrt(-c*
d)*x - c)/(d*x^2 + c)) + 2*(15*b^3*c^5*d - 9*a*b^2*c^4*d^2 - 3*a^2*b*c^3*d^3 + 5*a^3*c^2*d^4)*x)/(c^3*d^6*x^4
+ 2*c^4*d^5*x^2 + c^5*d^4), 1/8*(8*b^3*c^3*d^3*x^5 + (25*b^3*c^4*d^2 - 15*a*b^2*c^3*d^3 + 3*a^2*b*c^2*d^4 + 3*
a^3*c*d^5)*x^3 - 3*(5*b^3*c^5 - 3*a*b^2*c^4*d - a^2*b*c^3*d^2 - a^3*c^2*d^3 + (5*b^3*c^3*d^2 - 3*a*b^2*c^2*d^3
- a^2*b*c*d^4 - a^3*d^5)*x^4 + 2*(5*b^3*c^4*d - 3*a*b^2*c^3*d^2 - a^2*b*c^2*d^3 - a^3*c*d^4)*x^2)*sqrt(c*d)*a
rctan(sqrt(c*d)*x/c) + (15*b^3*c^5*d - 9*a*b^2*c^4*d^2 - 3*a^2*b*c^3*d^3 + 5*a^3*c^2*d^4)*x)/(c^3*d^6*x^4 + 2*
c^4*d^5*x^2 + c^5*d^4)]

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 422 vs. \(2 (122) = 244\).
time = 1.00, size = 422, normalized size = 3.25 \begin {gather*} \frac {b^{3} x}{d^{3}} - \frac {3 \sqrt {- \frac {1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right ) \log {\left (- \frac {3 c^{3} d^{3} \sqrt {- \frac {1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right )}{3 a^{3} d^{3} + 3 a^{2} b c d^{2} + 9 a b^{2} c^{2} d - 15 b^{3} c^{3}} + x \right )}}{16} + \frac {3 \sqrt {- \frac {1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right ) \log {\left (\frac {3 c^{3} d^{3} \sqrt {- \frac {1}{c^{5} d^{7}}} \left (a d - b c\right ) \left (a^{2} d^{2} + 2 a b c d + 5 b^{2} c^{2}\right )}{3 a^{3} d^{3} + 3 a^{2} b c d^{2} + 9 a b^{2} c^{2} d - 15 b^{3} c^{3}} + x \right )}}{16} + \frac {x^{3} \cdot \left (3 a^{3} d^{4} + 3 a^{2} b c d^{3} - 15 a b^{2} c^{2} d^{2} + 9 b^{3} c^{3} d\right ) + x \left (5 a^{3} c d^{3} - 3 a^{2} b c^{2} d^{2} - 9 a b^{2} c^{3} d + 7 b^{3} c^{4}\right )}{8 c^{4} d^{3} + 16 c^{3} d^{4} x^{2} + 8 c^{2} d^{5} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.23, size = 180, normalized size = 1.38 \begin {gather*} \frac {b^{3} x}{d^{3}} - \frac {3 \, {\left (5 \, b^{3} c^{3} - 3 \, a b^{2} c^{2} d - a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {d x}{\sqrt {c d}}\right )}{8 \, \sqrt {c d} c^{2} d^{3}} + \frac {9 \, b^{3} c^{3} d x^{3} - 15 \, a b^{2} c^{2} d^{2} x^{3} + 3 \, a^{2} b c d^{3} x^{3} + 3 \, a^{3} d^{4} x^{3} + 7 \, b^{3} c^{4} x - 9 \, a b^{2} c^{3} d x - 3 \, a^{2} b c^{2} d^{2} x + 5 \, a^{3} c d^{3} x}{8 \, {\left (d x^{2} + c\right )}^{2} c^{2} d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [B]
time = 4.96, size = 240, normalized size = 1.85 \begin {gather*} \frac {\frac {x\,\left (5\,a^3\,d^3-3\,a^2\,b\,c\,d^2-9\,a\,b^2\,c^2\,d+7\,b^3\,c^3\right )}{8\,c}+\frac {3\,x^3\,\left (a^3\,d^4+a^2\,b\,c\,d^3-5\,a\,b^2\,c^2\,d^2+3\,b^3\,c^3\,d\right )}{8\,c^2}}{c^2\,d^3+2\,c\,d^4\,x^2+d^5\,x^4}+\frac {b^3\,x}{d^3}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {d}\,x\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2+2\,a\,b\,c\,d+5\,b^2\,c^2\right )}{\sqrt {c}\,\left (a^3\,d^3+a^2\,b\,c\,d^2+3\,a\,b^2\,c^2\,d-5\,b^3\,c^3\right )}\right )\,\left (a\,d-b\,c\right )\,\left (a^2\,d^2+2\,a\,b\,c\,d+5\,b^2\,c^2\right )}{8\,c^{5/2}\,d^{7/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^3/(c + d*x^2)^3,x)

[Out]

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

________________________________________________________________________________________