3.1755 \(\int \frac{(a+b x)^{3/2}}{(c+d x)^{5/6}} \, dx\)

Optimal. Leaf size=405 \[ \frac{27\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{5/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{b c-a d}-\left (1-\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 d^3 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{27 \sqrt{a+b x} \sqrt [6]{c+d x} (b c-a d)}{20 d^2}+\frac{3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d} \]

[Out]

(-27*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(1/6))/(20*d^2) + (3*(a + b*x)^(3/2)*(c + d*x)^(1/6))/(5*d) + (27*3^(
3/4)*(b*c - a*d)^(5/3)*(c + d*x)^(1/6)*((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))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/
3)*(c + d*x)^(1/3))^2]*EllipticF[ArcCos[((b*c - a*d)^(1/3) - (1 - Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))/((b*c - a*
d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))], (2 + Sqrt[3])/4])/(40*d^3*Sqrt[a + b*x]*Sqrt[-((b^(1/3)*(c
+ d*x)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d
*x)^(1/3))^2)])

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Rubi [A]  time = 0.279166, antiderivative size = 405, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {50, 63, 225} \[ \frac{27\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{5/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{b c-a d}-\left (1-\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 d^3 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}-\frac{27 \sqrt{a+b x} \sqrt [6]{c+d x} (b c-a d)}{20 d^2}+\frac{3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/(c + d*x)^(5/6),x]

[Out]

(-27*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(1/6))/(20*d^2) + (3*(a + b*x)^(3/2)*(c + d*x)^(1/6))/(5*d) + (27*3^(
3/4)*(b*c - a*d)^(5/3)*(c + d*x)^(1/6)*((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))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/
3)*(c + d*x)^(1/3))^2]*EllipticF[ArcCos[((b*c - a*d)^(1/3) - (1 - Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))/((b*c - a*
d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d*x)^(1/3))], (2 + Sqrt[3])/4])/(40*d^3*Sqrt[a + b*x]*Sqrt[-((b^(1/3)*(c
+ d*x)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/((b*c - a*d)^(1/3) - (1 + Sqrt[3])*b^(1/3)*(c + d
*x)^(1/3))^2)])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 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 225

Int[1/Sqrt[(a_) + (b_.)*(x_)^6], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[(x*(s
+ r*x^2)*Sqrt[(s^2 - r*s*x^2 + r^2*x^4)/(s + (1 + Sqrt[3])*r*x^2)^2]*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2
)/(s + (1 + Sqrt[3])*r*x^2)], (2 + Sqrt[3])/4])/(2*3^(1/4)*s*Sqrt[a + b*x^6]*Sqrt[(r*x^2*(s + r*x^2))/(s + (1
+ Sqrt[3])*r*x^2)^2]), x]] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{(a+b x)^{3/2}}{(c+d x)^{5/6}} \, dx &=\frac{3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}-\frac{(9 (b c-a d)) \int \frac{\sqrt{a+b x}}{(c+d x)^{5/6}} \, dx}{10 d}\\ &=-\frac{27 (b c-a d) \sqrt{a+b x} \sqrt [6]{c+d x}}{20 d^2}+\frac{3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}+\frac{\left (27 (b c-a d)^2\right ) \int \frac{1}{\sqrt{a+b x} (c+d x)^{5/6}} \, dx}{40 d^2}\\ &=-\frac{27 (b c-a d) \sqrt{a+b x} \sqrt [6]{c+d x}}{20 d^2}+\frac{3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}+\frac{\left (81 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-\frac{b c}{d}+\frac{b x^6}{d}}} \, dx,x,\sqrt [6]{c+d x}\right )}{20 d^3}\\ &=-\frac{27 (b c-a d) \sqrt{a+b x} \sqrt [6]{c+d x}}{20 d^2}+\frac{3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}+\frac{27\ 3^{3/4} (b c-a d)^{5/3} \sqrt [6]{c+d x} \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 (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{b c-a d}-\left (1-\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{40 d^3 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}\\ \end{align*}

Mathematica [C]  time = 0.0310273, size = 73, normalized size = 0.18 \[ \frac{2 (a+b x)^{5/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/6} \, _2F_1\left (\frac{5}{6},\frac{5}{2};\frac{7}{2};\frac{d (a+b x)}{a d-b c}\right )}{5 b (c+d x)^{5/6}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/(c + d*x)^(5/6),x]

[Out]

(2*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(5/6)*Hypergeometric2F1[5/6, 5/2, 7/2, (d*(a + b*x))/(-(b*c) +
a*d)])/(5*b*(c + d*x)^(5/6))

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Maple [F]  time = 0.019, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{3}{2}}} \left ( dx+c \right ) ^{-{\frac{5}{6}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/(d*x+c)^(5/6),x)

[Out]

int((b*x+a)^(3/2)/(d*x+c)^(5/6),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{3}{2}}}{{\left (d x + c\right )}^{\frac{5}{6}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(5/6),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(3/2)/(d*x + c)^(5/6), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{\frac{3}{2}}}{{\left (d x + c\right )}^{\frac{5}{6}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(5/6),x, algorithm="fricas")

[Out]

integral((b*x + a)^(3/2)/(d*x + c)^(5/6), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{3}{2}}}{\left (c + d x\right )^{\frac{5}{6}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/(d*x+c)**(5/6),x)

[Out]

Integral((a + b*x)**(3/2)/(c + d*x)**(5/6), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{3}{2}}}{{\left (d x + c\right )}^{\frac{5}{6}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/(d*x+c)^(5/6),x, algorithm="giac")

[Out]

integrate((b*x + a)^(3/2)/(d*x + c)^(5/6), x)