3.1573 \(\int \frac{1}{(a+b x)^{3/2} (c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=383 \[ \frac{2 \sqrt{2-\sqrt{3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),4 \sqrt{3}-7\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt{a+b x} (b c-a d) \sqrt{-\frac{\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{2 \sqrt [3]{c+d x}}{\sqrt{a+b x} (b c-a d)} \]

[Out]

(-2*(c + d*x)^(1/3))/((b*c - a*d)*Sqrt[a + b*x]) + (2*Sqrt[2 - Sqrt[3]]*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)
^(1/3))*Sqrt[((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 - S
qrt[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^(1/3)*(b*c - a*d)*Sqrt[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 = 0.293588, antiderivative size = 383, 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 = {51, 63, 219} \[ \frac{2 \sqrt{2-\sqrt{3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\sin ^{-1}\left (\frac{\left (1+\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right )|-7+4 \sqrt{3}\right )}{\sqrt [4]{3} \sqrt [3]{b} \sqrt{a+b x} (b c-a d) \sqrt{-\frac{\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt{3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{2 \sqrt [3]{c+d x}}{\sqrt{a+b x} (b c-a d)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(c + d*x)^(1/3))/((b*c - a*d)*Sqrt[a + b*x]) + (2*Sqrt[2 - Sqrt[3]]*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)
^(1/3))*Sqrt[((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 - S
qrt[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^(1/3)*(b*c - a*d)*Sqrt[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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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{1}{(a+b x)^{3/2} (c+d x)^{2/3}} \, dx &=-\frac{2 \sqrt [3]{c+d x}}{(b c-a d) \sqrt{a+b x}}-\frac{d \int \frac{1}{\sqrt{a+b x} (c+d x)^{2/3}} \, dx}{3 (b c-a d)}\\ &=-\frac{2 \sqrt [3]{c+d x}}{(b c-a d) \sqrt{a+b x}}-\frac{\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 c-a d}\\ &=-\frac{2 \sqrt [3]{c+d x}}{(b c-a d) \sqrt{a+b x}}+\frac{2 \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} \sqrt [3]{b} (b c-a d) \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 = 0.0215687, size = 71, normalized size = 0.19 \[ -\frac{2 \left (\frac{b (c+d x)}{b c-a d}\right )^{2/3} \, _2F_1\left (-\frac{1}{2},\frac{2}{3};\frac{1}{2};\frac{d (a+b x)}{a d-b c}\right )}{b \sqrt{a+b x} (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.04, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{-{\frac{3}{2}}} \left ( dx+c \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x + a)^(3/2)*(d*x + c)^(2/3)), 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} d x^{3} + a^{2} c +{\left (b^{2} c + 2 \, a b d\right )} x^{2} +{\left (2 \, a b c + a^{2} d\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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