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

Optimal. Leaf size=203 \[ \frac{(b c-a d) (a d+2 b c) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{9 c^{5/3} d^{7/3}}-\frac{2 (b c-a d) (a d+2 b c) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{7/3}}+\frac{2 (b c-a d) (a d+2 b c) \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt{3} \sqrt [3]{c}}\right )}{3 \sqrt{3} c^{5/3} d^{7/3}}+\frac{x (b c-a d)^2}{3 c d^2 \left (c+d x^3\right )}+\frac{b^2 x}{d^2} \]

[Out]

(b^2*x)/d^2 + ((b*c - a*d)^2*x)/(3*c*d^2*(c + d*x^3)) + (2*(b*c - a*d)*(2*b*c + a*d)*ArcTan[(c^(1/3) - 2*d^(1/
3)*x)/(Sqrt[3]*c^(1/3))])/(3*Sqrt[3]*c^(5/3)*d^(7/3)) - (2*(b*c - a*d)*(2*b*c + a*d)*Log[c^(1/3) + d^(1/3)*x])
/(9*c^(5/3)*d^(7/3)) + ((b*c - a*d)*(2*b*c + a*d)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/(9*c^(5/3)*d
^(7/3))

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Rubi [A]  time = 0.243648, antiderivative size = 203, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.421, Rules used = {390, 385, 200, 31, 634, 617, 204, 628} \[ \frac{(b c-a d) (a d+2 b c) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{9 c^{5/3} d^{7/3}}-\frac{2 (b c-a d) (a d+2 b c) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{7/3}}+\frac{2 (b c-a d) (a d+2 b c) \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt{3} \sqrt [3]{c}}\right )}{3 \sqrt{3} c^{5/3} d^{7/3}}+\frac{x (b c-a d)^2}{3 c d^2 \left (c+d x^3\right )}+\frac{b^2 x}{d^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^3)^2/(c + d*x^3)^2,x]

[Out]

(b^2*x)/d^2 + ((b*c - a*d)^2*x)/(3*c*d^2*(c + d*x^3)) + (2*(b*c - a*d)*(2*b*c + a*d)*ArcTan[(c^(1/3) - 2*d^(1/
3)*x)/(Sqrt[3]*c^(1/3))])/(3*Sqrt[3]*c^(5/3)*d^(7/3)) - (2*(b*c - a*d)*(2*b*c + a*d)*Log[c^(1/3) + d^(1/3)*x])
/(9*c^(5/3)*d^(7/3)) + ((b*c - a*d)*(2*b*c + a*d)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/(9*c^(5/3)*d
^(7/3))

Rule 390

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 385

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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^3\right )^2}{\left (c+d x^3\right )^2} \, dx &=\int \left (\frac{b^2}{d^2}-\frac{b^2 c^2-a^2 d^2+2 b d (b c-a d) x^3}{d^2 \left (c+d x^3\right )^2}\right ) \, dx\\ &=\frac{b^2 x}{d^2}-\frac{\int \frac{b^2 c^2-a^2 d^2+2 b d (b c-a d) x^3}{\left (c+d x^3\right )^2} \, dx}{d^2}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{3 c d^2 \left (c+d x^3\right )}-\frac{(2 (b c-a d) (2 b c+a d)) \int \frac{1}{c+d x^3} \, dx}{3 c d^2}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{3 c d^2 \left (c+d x^3\right )}-\frac{(2 (b c-a d) (2 b c+a d)) \int \frac{1}{\sqrt [3]{c}+\sqrt [3]{d} x} \, dx}{9 c^{5/3} d^2}-\frac{(2 (b c-a d) (2 b c+a d)) \int \frac{2 \sqrt [3]{c}-\sqrt [3]{d} x}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{9 c^{5/3} d^2}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{3 c d^2 \left (c+d x^3\right )}-\frac{2 (b c-a d) (2 b c+a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{7/3}}+\frac{((b c-a d) (2 b c+a d)) \int \frac{-\sqrt [3]{c} \sqrt [3]{d}+2 d^{2/3} x}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{9 c^{5/3} d^{7/3}}-\frac{((b c-a d) (2 b c+a d)) \int \frac{1}{c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2} \, dx}{3 c^{4/3} d^2}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{3 c d^2 \left (c+d x^3\right )}-\frac{2 (b c-a d) (2 b c+a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{7/3}}+\frac{(b c-a d) (2 b c+a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{9 c^{5/3} d^{7/3}}-\frac{(2 (b c-a d) (2 b c+a d)) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{d} x}{\sqrt [3]{c}}\right )}{3 c^{5/3} d^{7/3}}\\ &=\frac{b^2 x}{d^2}+\frac{(b c-a d)^2 x}{3 c d^2 \left (c+d x^3\right )}+\frac{2 (b c-a d) (2 b c+a d) \tan ^{-1}\left (\frac{\sqrt [3]{c}-2 \sqrt [3]{d} x}{\sqrt{3} \sqrt [3]{c}}\right )}{3 \sqrt{3} c^{5/3} d^{7/3}}-\frac{2 (b c-a d) (2 b c+a d) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{9 c^{5/3} d^{7/3}}+\frac{(b c-a d) (2 b c+a d) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{9 c^{5/3} d^{7/3}}\\ \end{align*}

Mathematica [A]  time = 0.194235, size = 210, normalized size = 1.03 \[ \frac{\frac{\left (-a^2 d^2-a b c d+2 b^2 c^2\right ) \log \left (c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2\right )}{c^{5/3}}-\frac{2 \left (-a^2 d^2-a b c d+2 b^2 c^2\right ) \log \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{c^{5/3}}+\frac{2 \sqrt{3} \left (-a^2 d^2-a b c d+2 b^2 c^2\right ) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{d} x}{\sqrt [3]{c}}}{\sqrt{3}}\right )}{c^{5/3}}+\frac{3 \sqrt [3]{d} x (b c-a d)^2}{c \left (c+d x^3\right )}+9 b^2 \sqrt [3]{d} x}{9 d^{7/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*b^2*d^(1/3)*x + (3*d^(1/3)*(b*c - a*d)^2*x)/(c*(c + d*x^3)) + (2*Sqrt[3]*(2*b^2*c^2 - a*b*c*d - a^2*d^2)*Ar
cTan[(1 - (2*d^(1/3)*x)/c^(1/3))/Sqrt[3]])/c^(5/3) - (2*(2*b^2*c^2 - a*b*c*d - a^2*d^2)*Log[c^(1/3) + d^(1/3)*
x])/c^(5/3) + ((2*b^2*c^2 - a*b*c*d - a^2*d^2)*Log[c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2])/c^(5/3))/(9*d^(
7/3))

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Maple [B]  time = 0.01, size = 367, normalized size = 1.8 \begin{align*}{\frac{{b}^{2}x}{{d}^{2}}}+{\frac{{a}^{2}x}{3\,c \left ( d{x}^{3}+c \right ) }}-{\frac{2\,xab}{3\,d \left ( d{x}^{3}+c \right ) }}+{\frac{cx{b}^{2}}{3\,{d}^{2} \left ( d{x}^{3}+c \right ) }}+{\frac{2\,{a}^{2}}{9\,cd}\ln \left ( x+\sqrt [3]{{\frac{c}{d}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,ab}{9\,{d}^{2}}\ln \left ( x+\sqrt [3]{{\frac{c}{d}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}-{\frac{4\,{b}^{2}c}{9\,{d}^{3}}\ln \left ( x+\sqrt [3]{{\frac{c}{d}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}-{\frac{{a}^{2}}{9\,cd}\ln \left ({x}^{2}-\sqrt [3]{{\frac{c}{d}}}x+ \left ({\frac{c}{d}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}-{\frac{ab}{9\,{d}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{c}{d}}}x+ \left ({\frac{c}{d}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,{b}^{2}c}{9\,{d}^{3}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{c}{d}}}x+ \left ({\frac{c}{d}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,\sqrt{3}{a}^{2}}{9\,cd}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{c}{d}}}}}}-1 \right ) } \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,\sqrt{3}ab}{9\,{d}^{2}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{c}{d}}}}}}-1 \right ) } \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}}-{\frac{4\,c\sqrt{3}{b}^{2}}{9\,{d}^{3}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{c}{d}}}}}}-1 \right ) } \right ) \left ({\frac{c}{d}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x/d^2+1/3/c*x/(d*x^3+c)*a^2-2/3/d*x/(d*x^3+c)*a*b+1/3/d^2*c*x/(d*x^3+c)*b^2+2/9/d/c/(c/d)^(2/3)*ln(x+(c/d)
^(1/3))*a^2+2/9/d^2/(c/d)^(2/3)*ln(x+(c/d)^(1/3))*a*b-4/9/d^3*c/(c/d)^(2/3)*ln(x+(c/d)^(1/3))*b^2-1/9/d/c/(c/d
)^(2/3)*ln(x^2-(c/d)^(1/3)*x+(c/d)^(2/3))*a^2-1/9/d^2/(c/d)^(2/3)*ln(x^2-(c/d)^(1/3)*x+(c/d)^(2/3))*a*b+2/9/d^
3*c/(c/d)^(2/3)*ln(x^2-(c/d)^(1/3)*x+(c/d)^(2/3))*b^2+2/9/d/c/(c/d)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(c/d)^
(1/3)*x-1))*a^2+2/9/d^2/(c/d)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(c/d)^(1/3)*x-1))*a*b-4/9/d^3*c/(c/d)^(2/3)*
3^(1/2)*arctan(1/3*3^(1/2)*(2/(c/d)^(1/3)*x-1))*b^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.72141, size = 1635, normalized size = 8.05 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/9*(9*b^2*c^3*d^2*x^4 - 3*sqrt(1/3)*(2*b^2*c^4*d - a*b*c^3*d^2 - a^2*c^2*d^3 + (2*b^2*c^3*d^2 - a*b*c^2*d^3
- a^2*c*d^4)*x^3)*sqrt(-(c^2*d)^(1/3)/d)*log((2*c*d*x^3 - 3*(c^2*d)^(1/3)*c*x - c^2 + 3*sqrt(1/3)*(2*c*d*x^2 +
(c^2*d)^(2/3)*x - (c^2*d)^(1/3)*c)*sqrt(-(c^2*d)^(1/3)/d))/(d*x^3 + c)) + (2*b^2*c^3 - a*b*c^2*d - a^2*c*d^2
+ (2*b^2*c^2*d - a*b*c*d^2 - a^2*d^3)*x^3)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x + (c^2*d)^(1/3)*c) - 2*
(2*b^2*c^3 - a*b*c^2*d - a^2*c*d^2 + (2*b^2*c^2*d - a*b*c*d^2 - a^2*d^3)*x^3)*(c^2*d)^(2/3)*log(c*d*x + (c^2*d
)^(2/3)) + 3*(4*b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*x)/(c^3*d^4*x^3 + c^4*d^3), 1/9*(9*b^2*c^3*d^2*x^4 -
6*sqrt(1/3)*(2*b^2*c^4*d - a*b*c^3*d^2 - a^2*c^2*d^3 + (2*b^2*c^3*d^2 - a*b*c^2*d^3 - a^2*c*d^4)*x^3)*sqrt((c^
2*d)^(1/3)/d)*arctan(sqrt(1/3)*(2*(c^2*d)^(2/3)*x - (c^2*d)^(1/3)*c)*sqrt((c^2*d)^(1/3)/d)/c^2) + (2*b^2*c^3 -
a*b*c^2*d - a^2*c*d^2 + (2*b^2*c^2*d - a*b*c*d^2 - a^2*d^3)*x^3)*(c^2*d)^(2/3)*log(c*d*x^2 - (c^2*d)^(2/3)*x
+ (c^2*d)^(1/3)*c) - 2*(2*b^2*c^3 - a*b*c^2*d - a^2*c*d^2 + (2*b^2*c^2*d - a*b*c*d^2 - a^2*d^3)*x^3)*(c^2*d)^(
2/3)*log(c*d*x + (c^2*d)^(2/3)) + 3*(4*b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3)*x)/(c^3*d^4*x^3 + c^4*d^3)]

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Sympy [A]  time = 1.75829, size = 189, normalized size = 0.93 \begin{align*} \frac{b^{2} x}{d^{2}} + \frac{x \left (a^{2} d^{2} - 2 a b c d + b^{2} c^{2}\right )}{3 c^{2} d^{2} + 3 c d^{3} x^{3}} + \operatorname{RootSum}{\left (729 t^{3} c^{5} d^{7} - 8 a^{6} d^{6} - 24 a^{5} b c d^{5} + 24 a^{4} b^{2} c^{2} d^{4} + 88 a^{3} b^{3} c^{3} d^{3} - 48 a^{2} b^{4} c^{4} d^{2} - 96 a b^{5} c^{5} d + 64 b^{6} c^{6}, \left ( t \mapsto t \log{\left (\frac{9 t c^{2} d^{2}}{2 a^{2} d^{2} + 2 a b c d - 4 b^{2} c^{2}} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**2*x/d**2 + x*(a**2*d**2 - 2*a*b*c*d + b**2*c**2)/(3*c**2*d**2 + 3*c*d**3*x**3) + RootSum(729*_t**3*c**5*d**
7 - 8*a**6*d**6 - 24*a**5*b*c*d**5 + 24*a**4*b**2*c**2*d**4 + 88*a**3*b**3*c**3*d**3 - 48*a**2*b**4*c**4*d**2
- 96*a*b**5*c**5*d + 64*b**6*c**6, Lambda(_t, _t*log(9*_t*c**2*d**2/(2*a**2*d**2 + 2*a*b*c*d - 4*b**2*c**2) +
x)))

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Giac [A]  time = 1.11978, size = 358, normalized size = 1.76 \begin{align*} \frac{b^{2} x}{d^{2}} + \frac{2 \,{\left (2 \, b^{2} c^{2} - a b c d - a^{2} d^{2}\right )} \left (-\frac{c}{d}\right )^{\frac{1}{3}} \log \left ({\left | x - \left (-\frac{c}{d}\right )^{\frac{1}{3}} \right |}\right )}{9 \, c^{2} d^{2}} - \frac{2 \, \sqrt{3}{\left (2 \, \left (-c d^{2}\right )^{\frac{1}{3}} b^{2} c^{2} - \left (-c d^{2}\right )^{\frac{1}{3}} a b c d - \left (-c d^{2}\right )^{\frac{1}{3}} a^{2} d^{2}\right )} \arctan \left (\frac{\sqrt{3}{\left (2 \, x + \left (-\frac{c}{d}\right )^{\frac{1}{3}}\right )}}{3 \, \left (-\frac{c}{d}\right )^{\frac{1}{3}}}\right )}{9 \, c^{2} d^{3}} + \frac{b^{2} c^{2} x - 2 \, a b c d x + a^{2} d^{2} x}{3 \,{\left (d x^{3} + c\right )} c d^{2}} - \frac{{\left (2 \, \left (-c d^{2}\right )^{\frac{1}{3}} b^{2} c^{2} - \left (-c d^{2}\right )^{\frac{1}{3}} a b c d - \left (-c d^{2}\right )^{\frac{1}{3}} a^{2} d^{2}\right )} \log \left (x^{2} + x \left (-\frac{c}{d}\right )^{\frac{1}{3}} + \left (-\frac{c}{d}\right )^{\frac{2}{3}}\right )}{9 \, c^{2} d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*x/d^2 + 2/9*(2*b^2*c^2 - a*b*c*d - a^2*d^2)*(-c/d)^(1/3)*log(abs(x - (-c/d)^(1/3)))/(c^2*d^2) - 2/9*sqrt(3
)*(2*(-c*d^2)^(1/3)*b^2*c^2 - (-c*d^2)^(1/3)*a*b*c*d - (-c*d^2)^(1/3)*a^2*d^2)*arctan(1/3*sqrt(3)*(2*x + (-c/d
)^(1/3))/(-c/d)^(1/3))/(c^2*d^3) + 1/3*(b^2*c^2*x - 2*a*b*c*d*x + a^2*d^2*x)/((d*x^3 + c)*c*d^2) - 1/9*(2*(-c*
d^2)^(1/3)*b^2*c^2 - (-c*d^2)^(1/3)*a*b*c*d - (-c*d^2)^(1/3)*a^2*d^2)*log(x^2 + x*(-c/d)^(1/3) + (-c/d)^(2/3))
/(c^2*d^3)