3.1756 \(\int \frac{\sqrt{a+b x}}{(c+d x)^{5/6}} \, dx\)

Optimal. Leaf size=372 \[ \frac{3 \sqrt{a+b x} \sqrt [6]{c+d x}}{2 d}-\frac{3\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{2/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \text{EllipticF}\left (\cos ^{-1}\left (\frac{\sqrt [3]{b c-a d}-\left (1-\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{4 d^2 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \]

[Out]

(3*Sqrt[a + b*x]*(c + d*x)^(1/6))/(2*d) - (3*3^(3/4)*(b*c - a*d)^(2/3)*(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])/(4*d^2*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 = 0.235514, antiderivative size = 372, 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 = {50, 63, 225} \[ \frac{3 \sqrt{a+b x} \sqrt [6]{c+d x}}{2 d}-\frac{3\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{2/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt{\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} F\left (\cos ^{-1}\left (\frac{\sqrt [3]{b c-a d}-\left (1-\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right )|\frac{1}{4} \left (2+\sqrt{3}\right )\right )}{4 d^2 \sqrt{a+b x} \sqrt{-\frac{\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt{3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]/(c + d*x)^(5/6),x]

[Out]

(3*Sqrt[a + b*x]*(c + d*x)^(1/6))/(2*d) - (3*3^(3/4)*(b*c - a*d)^(2/3)*(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])/(4*d^2*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{\sqrt{a+b x}}{(c+d x)^{5/6}} \, dx &=\frac{3 \sqrt{a+b x} \sqrt [6]{c+d x}}{2 d}-\frac{(3 (b c-a d)) \int \frac{1}{\sqrt{a+b x} (c+d x)^{5/6}} \, dx}{4 d}\\ &=\frac{3 \sqrt{a+b x} \sqrt [6]{c+d x}}{2 d}-\frac{(9 (b c-a d)) \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 )}{2 d^2}\\ &=\frac{3 \sqrt{a+b x} \sqrt [6]{c+d x}}{2 d}-\frac{3\ 3^{3/4} (b c-a d)^{2/3} \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 )}{4 d^2 \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 = 0.0246816, size = 73, normalized size = 0.2 \[ \frac{2 (a+b x)^{3/2} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/6} \, _2F_1\left (\frac{5}{6},\frac{3}{2};\frac{5}{2};\frac{d (a+b x)}{a d-b c}\right )}{3 b (c+d x)^{5/6}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.016, size = 0, normalized size = 0. \begin{align*} \int{\sqrt{bx+a} \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)^(1/2)/(d*x+c)^(5/6),x)

[Out]

int((b*x+a)^(1/2)/(d*x+c)^(5/6),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x + a}}{{\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)^(1/2)/(d*x+c)^(5/6),x, algorithm="maxima")

[Out]

integrate(sqrt(b*x + a)/(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{\sqrt{b x + a}}{{\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)^(1/2)/(d*x+c)^(5/6),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{a + b x}}{\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)**(1/2)/(d*x+c)**(5/6),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{b x + a}}{{\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)^(1/2)/(d*x+c)^(5/6),x, algorithm="giac")

[Out]

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