3.1585 \(\int \frac{\sqrt [3]{c+d x}}{(a+b x)^{5/3}} \, dx\)

Optimal. Leaf size=568 \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} d^{2/3} ((a+b x) (c+d x))^{2/3} \sqrt{(a d+b c+2 b d x)^2} \left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{2/3}\right ) \sqrt{\frac{2 \sqrt [3]{2} b^{2/3} d^{2/3} ((a+b x) (c+d x))^{2/3}-2^{2/3} \sqrt [3]{b} \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{4/3}}{\left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1+\sqrt{3}\right ) (b c-a d)^{2/3}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1-\sqrt{3}\right ) (b c-a d)^{2/3}}{2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1+\sqrt{3}\right ) (b c-a d)^{2/3}}\right ),-7-4 \sqrt{3}\right )}{2^{2/3} b^{4/3} (a+b x)^{2/3} (c+d x)^{2/3} (a d+b c+2 b d x) \sqrt{\frac{(b c-a d)^{2/3} \left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{2/3}\right )}{\left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1+\sqrt{3}\right ) (b c-a d)^{2/3}\right )^2}} \sqrt{(a d+b (c+2 d x))^2}}-\frac{3 \sqrt [3]{c+d x}}{2 b (a+b x)^{2/3}} \]

[Out]

(-3*(c + d*x)^(1/3))/(2*b*(a + b*x)^(2/3)) + (3^(3/4)*Sqrt[2 + Sqrt[3]]*d^(2/3)*((a + b*x)*(c + d*x))^(2/3)*Sq
rt[(b*c + a*d + 2*b*d*x)^2]*((b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3))*Sqrt[((b
*c - a*d)^(4/3) - 2^(2/3)*b^(1/3)*d^(1/3)*(b*c - a*d)^(2/3)*((a + b*x)*(c + d*x))^(1/3) + 2*2^(1/3)*b^(2/3)*d^
(2/3)*((a + b*x)*(c + d*x))^(2/3))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c +
d*x))^(1/3))^2]*EllipticF[ArcSin[((1 - Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*
x))^(1/3))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3))], -7 - 4*Sq
rt[3]])/(2^(2/3)*b^(4/3)*(a + b*x)^(2/3)*(c + d*x)^(2/3)*(b*c + a*d + 2*b*d*x)*Sqrt[((b*c - a*d)^(2/3)*((b*c -
a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3)))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)
*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3))^2]*Sqrt[(a*d + b*(c + 2*d*x))^2])

________________________________________________________________________________________

Rubi [A]  time = 0.594428, antiderivative size = 568, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {47, 62, 623, 218} \[ \frac{3^{3/4} \sqrt{2+\sqrt{3}} d^{2/3} ((a+b x) (c+d x))^{2/3} \sqrt{(a d+b c+2 b d x)^2} \left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{2/3}\right ) \sqrt{\frac{2 \sqrt [3]{2} b^{2/3} d^{2/3} ((a+b x) (c+d x))^{2/3}-2^{2/3} \sqrt [3]{b} \sqrt [3]{d} (b c-a d)^{2/3} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{4/3}}{\left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1+\sqrt{3}\right ) (b c-a d)^{2/3}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}{\left (1+\sqrt{3}\right ) (b c-a d)^{2/3}+2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}}\right )|-7-4 \sqrt{3}\right )}{2^{2/3} b^{4/3} (a+b x)^{2/3} (c+d x)^{2/3} (a d+b c+2 b d x) \sqrt{\frac{(b c-a d)^{2/3} \left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+(b c-a d)^{2/3}\right )}{\left (2^{2/3} \sqrt [3]{b} \sqrt [3]{d} \sqrt [3]{(a+b x) (c+d x)}+\left (1+\sqrt{3}\right ) (b c-a d)^{2/3}\right )^2}} \sqrt{(a d+b (c+2 d x))^2}}-\frac{3 \sqrt [3]{c+d x}}{2 b (a+b x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(1/3)/(a + b*x)^(5/3),x]

[Out]

(-3*(c + d*x)^(1/3))/(2*b*(a + b*x)^(2/3)) + (3^(3/4)*Sqrt[2 + Sqrt[3]]*d^(2/3)*((a + b*x)*(c + d*x))^(2/3)*Sq
rt[(b*c + a*d + 2*b*d*x)^2]*((b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3))*Sqrt[((b
*c - a*d)^(4/3) - 2^(2/3)*b^(1/3)*d^(1/3)*(b*c - a*d)^(2/3)*((a + b*x)*(c + d*x))^(1/3) + 2*2^(1/3)*b^(2/3)*d^
(2/3)*((a + b*x)*(c + d*x))^(2/3))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c +
d*x))^(1/3))^2]*EllipticF[ArcSin[((1 - Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*
x))^(1/3))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3))], -7 - 4*Sq
rt[3]])/(2^(2/3)*b^(4/3)*(a + b*x)^(2/3)*(c + d*x)^(2/3)*(b*c + a*d + 2*b*d*x)*Sqrt[((b*c - a*d)^(2/3)*((b*c -
a*d)^(2/3) + 2^(2/3)*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3)))/((1 + Sqrt[3])*(b*c - a*d)^(2/3) + 2^(2/3)
*b^(1/3)*d^(1/3)*((a + b*x)*(c + d*x))^(1/3))^2]*Sqrt[(a*d + b*(c + 2*d*x))^2])

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 62

Int[((a_.) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Dist[((a + b*x)^m*(c + d*x)^m)/((a + b*x)
*(c + d*x))^m, Int[(a*c + (b*c + a*d)*x + b*d*x^2)^m, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] &&
LtQ[-1, m, 0] && LeQ[3, Denominator[m], 4]

Rule 623

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[(d*Sqrt[(b + 2*c*x)
^2])/(b + 2*c*x), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 218

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(2*Sqr
t[2 + Sqrt[3]]*(s + r*x)*Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3
])*s + r*x)/((1 + Sqrt[3])*s + r*x)], -7 - 4*Sqrt[3]])/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(s*(s + r*x))/((1 + Sqr
t[3])*s + r*x)^2]), x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rubi steps

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

Mathematica [C]  time = 0.0236049, size = 73, normalized size = 0.13 \[ -\frac{3 \sqrt [3]{c+d x} \, _2F_1\left (-\frac{2}{3},-\frac{1}{3};\frac{1}{3};\frac{d (a+b x)}{a d-b c}\right )}{2 b (a+b x)^{2/3} \sqrt [3]{\frac{b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(1/3)/(a + b*x)^(5/3),x]

[Out]

(-3*(c + d*x)^(1/3)*Hypergeometric2F1[-2/3, -1/3, 1/3, (d*(a + b*x))/(-(b*c) + a*d)])/(2*b*(a + b*x)^(2/3)*((b
*(c + d*x))/(b*c - a*d))^(1/3))

________________________________________________________________________________________

Maple [F]  time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [3]{dx+c} \left ( bx+a \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(1/3)/(b*x+a)^(5/3),x)

[Out]

int((d*x+c)^(1/3)/(b*x+a)^(5/3),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{1}{3}}}{{\left (b x + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(5/3),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(5/3), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(5/3),x, algorithm="fricas")

[Out]

integral((b*x + a)^(1/3)*(d*x + c)^(1/3)/(b^2*x^2 + 2*a*b*x + a^2), x)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(1/3)/(b*x+a)**(5/3),x)

[Out]

Integral((c + d*x)**(1/3)/(a + b*x)**(5/3), x)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{\frac{1}{3}}}{{\left (b x + a\right )}^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(1/3)/(b*x+a)^(5/3),x, algorithm="giac")

[Out]

integrate((d*x + c)^(1/3)/(b*x + a)^(5/3), x)