3.1562 \(\int \frac{\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx\)

Optimal. Leaf size=366 \[ -\frac{4 \sqrt{2-\sqrt{3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),4 \sqrt{3}-7\right )}{\sqrt [4]{3} b^{4/3} \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{2 \sqrt [3]{c+d x}}{b \sqrt{a+b x}} \]

[Out]

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

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Rubi [A]  time = 0.261813, antiderivative size = 366, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {47, 63, 219} \[ -\frac{4 \sqrt{2-\sqrt{3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt{3}\right )}{\sqrt [4]{3} b^{4/3} \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{2 \sqrt [3]{c+d x}}{b \sqrt{a+b x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c + d*x)^(1/3))/(b*Sqrt[a + b*x]) - (4*Sqrt[2 - Sqrt[3]]*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))*Sq
rt[((b*c - a*d)^(2/3) + b^(1/3)*(b*c - a*d)^(1/3)*(c + d*x)^(1/3) + b^(2/3)*(c + d*x)^(2/3))/((1 - Sqrt[3])*(b
*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c +
d*x)^(1/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*b^(4/3)*S
qrt[a + b*x]*Sqrt[-(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/((1 - Sqrt[3])*(b*c - a*
d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 219

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 - S
qrt[3])*s + r*x)^2)]), x]] /; FreeQ[{a, b}, x] && NegQ[a]

Rubi steps

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

Mathematica [C]  time = 0.0227167, size = 71, normalized size = 0.19 \[ -\frac{2 \sqrt [3]{c+d x} \, _2F_1\left (-\frac{1}{2},-\frac{1}{3};\frac{1}{2};\frac{d (a+b x)}{a d-b c}\right )}{b \sqrt{a+b x} \sqrt [3]{\frac{b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [3]{dx+c} \left ( bx+a \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x + a}{\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)^(3/2),x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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