3.3016 \(\int \frac{(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=369 \[ -\frac{\log (c+d x) \left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right )}{18 b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \log \left (\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{6 b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt{3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{7/3} d^{8/3}}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (-4 a d f-5 b c f+9 b d e)}{6 b^2 d^2}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d} \]

[Out]

(f*(9*b*d*e - 5*b*c*f - 4*a*d*f)*(a + b*x)^(2/3)*(c + d*x)^(1/3))/(6*b^2*d^2) + (f*(a + b*x)^(2/3)*(c + d*x)^(
1/3)*(e + f*x))/(2*b*d) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2)
)*ArcTan[1/Sqrt[3] + (2*d^(1/3)*(a + b*x)^(1/3))/(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))])/(3*Sqrt[3]*b^(7/3)*d^(8/3
)) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[c + d*x])/(18*b
^(7/3)*d^(8/3)) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[-1
+ (d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x)^(1/3))])/(6*b^(7/3)*d^(8/3))

________________________________________________________________________________________

Rubi [A]  time = 0.296912, antiderivative size = 369, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {90, 80, 59} \[ -\frac{\log (c+d x) \left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right )}{18 b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \log \left (\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{6 b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \tan ^{-1}\left (\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt{3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{7/3} d^{8/3}}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (-4 a d f-5 b c f+9 b d e)}{6 b^2 d^2}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(9*b*d*e - 5*b*c*f - 4*a*d*f)*(a + b*x)^(2/3)*(c + d*x)^(1/3))/(6*b^2*d^2) + (f*(a + b*x)^(2/3)*(c + d*x)^(
1/3)*(e + f*x))/(2*b*d) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2)
)*ArcTan[1/Sqrt[3] + (2*d^(1/3)*(a + b*x)^(1/3))/(Sqrt[3]*b^(1/3)*(c + d*x)^(1/3))])/(3*Sqrt[3]*b^(7/3)*d^(8/3
)) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[c + d*x])/(18*b
^(7/3)*d^(8/3)) - ((2*a^2*d^2*f^2 - 2*a*b*d*f*(3*d*e - c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*Log[-1
+ (d^(1/3)*(a + b*x)^(1/3))/(b^(1/3)*(c + d*x)^(1/3))])/(6*b^(7/3)*d^(8/3))

Rule 90

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a + b*
x)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 3)), x] + Dist[1/(d*f*(n + p + 3)), Int[(c + d*x)^n*(e +
f*x)^p*Simp[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(n + p + 4) - b*(d*e*(
n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, -Simp[(Sqrt
[3]*q*ArcTan[(2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3)) + 1/Sqrt[3]])/d, x] + (-Simp[(3*q*Log[(q*(a + b*x
)^(1/3))/(c + d*x)^(1/3) - 1])/(2*d), x] - Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[
b*c - a*d, 0] && PosQ[d/b]

Rubi steps

\begin{align*} \int \frac{(e+f x)^2}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx &=\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}+\frac{\int \frac{\frac{1}{3} \left (6 b d e^2-f (2 b c e+a d e+3 a c f)\right )+\frac{1}{3} f (9 b d e-5 b c f-4 a d f) x}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx}{2 b d}\\ &=\frac{f (9 b d e-5 b c f-4 a d f) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 b^2 d^2}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}+\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \int \frac{1}{\sqrt [3]{a+b x} (c+d x)^{2/3}} \, dx}{9 b^2 d^2}\\ &=\frac{f (9 b d e-5 b c f-4 a d f) (a+b x)^{2/3} \sqrt [3]{c+d x}}{6 b^2 d^2}+\frac{f (a+b x)^{2/3} \sqrt [3]{c+d x} (e+f x)}{2 b d}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \tan ^{-1}\left (\frac{1}{\sqrt{3}}+\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt{3} \sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{3 \sqrt{3} b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log (c+d x)}{18 b^{7/3} d^{8/3}}-\frac{\left (2 a^2 d^2 f^2-2 a b d f (3 d e-c f)+b^2 \left (9 d^2 e^2-12 c d e f+5 c^2 f^2\right )\right ) \log \left (-1+\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{6 b^{7/3} d^{8/3}}\\ \end{align*}

Mathematica [C]  time = 0.141879, size = 174, normalized size = 0.47 \[ \frac{(a+b x)^{2/3} \left (2 \left (\frac{b (c+d x)}{b c-a d}\right )^{2/3} \left (2 a^2 d^2 f^2+2 a b d f (c f-3 d e)+b^2 \left (5 c^2 f^2-12 c d e f+9 d^2 e^2\right )\right ) \, _2F_1\left (\frac{2}{3},\frac{2}{3};\frac{5}{3};\frac{d (a+b x)}{a d-b c}\right )-2 b f (c+d x) (4 a d f+5 b c f-9 b d e)+6 b^2 d f (c+d x) (e+f x)\right )}{12 b^3 d^2 (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(2/3)*(-2*b*f*(-9*b*d*e + 5*b*c*f + 4*a*d*f)*(c + d*x) + 6*b^2*d*f*(c + d*x)*(e + f*x) + 2*(2*a^2*d
^2*f^2 + 2*a*b*d*f*(-3*d*e + c*f) + b^2*(9*d^2*e^2 - 12*c*d*e*f + 5*c^2*f^2))*((b*(c + d*x))/(b*c - a*d))^(2/3
)*Hypergeometric2F1[2/3, 2/3, 5/3, (d*(a + b*x))/(-(b*c) + a*d)]))/(12*b^3*d^2*(c + d*x)^(2/3))

________________________________________________________________________________________

Maple [F]  time = 0.025, size = 0, normalized size = 0. \begin{align*} \int{ \left ( fx+e \right ) ^{2}{\frac{1}{\sqrt [3]{bx+a}}} \left ( dx+c \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2}}{{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.53401, size = 2367, normalized size = 6.41 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*(9*b^3*d^3*e^2 - 6*(2*b^3*c*d^2 + a*b^2*d^3)*e*f + (5*b^3*c^2*d + 2*a*b^2*c*d^2 + 2*a^2*b*d
^3)*f^2)*sqrt((-b*d^2)^(1/3)/b)*log(-3*b*d^2*x - 2*b*c*d - a*d^2 - 3*(-b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^
(1/3)*d - 3*sqrt(1/3)*(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3)
+ (-b*d^2)^(1/3)*(b*d*x + a*d))*sqrt((-b*d^2)^(1/3)/b)) - 2*(9*b^2*d^2*e^2 - 6*(2*b^2*c*d + a*b*d^2)*e*f + (5*
b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*(-b*d^2)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (-b*d^2)^(2/3)
*(b*x + a))/(b*x + a)) + (9*b^2*d^2*e^2 - 6*(2*b^2*c*d + a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^
2)*(-b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) -
(-b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(3*b^2*d^3*f^2*x + 12*b^2*d^3*e*f - (5*b^2*c*d^2 + 4*a*b*d^3)*f^2
)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^3*d^4), 1/18*(6*sqrt(1/3)*(9*b^3*d^3*e^2 - 6*(2*b^3*c*d^2 + a*b^2*d^3)*e
*f + (5*b^3*c^2*d + 2*a*b^2*c*d^2 + 2*a^2*b*d^3)*f^2)*sqrt(-(-b*d^2)^(1/3)/b)*arctan(sqrt(1/3)*(2*(-b*d^2)^(2/
3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d))*sqrt(-(-b*d^2)^(1/3)/b)/(b*d^2*x + a*d^2))
- 2*(9*b^2*d^2*e^2 - 6*(2*b^2*c*d + a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*(-b*d^2)^(2/3)*log
(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (-b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (9*b^2*d^2*e^2 - 6*(2*b^2*c*d +
a*b*d^2)*e*f + (5*b^2*c^2 + 2*a*b*c*d + 2*a^2*d^2)*f^2)*(-b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b
*d + (-b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (-b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(3*b^2*d^3*
f^2*x + 12*b^2*d^3*e*f - (5*b^2*c*d^2 + 4*a*b*d^3)*f^2)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^3*d^4)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2}}{{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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