\(\int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx\) [1739]

Optimal result

Integrand size = 19, antiderivative size = 375 \[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 b}+\frac {3^{3/4} (b c-a d)^{2/3} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 b d \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \]

[Out]

3/2*(d*x+c)^(1/6)*(b*x+a)^(1/2)/b+1/4*3^(3/4)*(-a*d+b*c)^(2/3)*(d*x+c)^(1/6)*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)
^(1/3))*(((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1-3^(1/2)))^2/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1+3^(
1/2)))^2)^(1/2)/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1-3^(1/2)))*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(
1+3^(1/2)))*EllipticF((1-((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1-3^(1/2)))^2/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x
+c)^(1/3)*(1+3^(1/2)))^2)^(1/2),1/4*6^(1/2)+1/4*2^(1/2))*(((-a*d+b*c)^(2/3)+b^(1/3)*(-a*d+b*c)^(1/3)*(d*x+c)^(
1/3)+b^(2/3)*(d*x+c)^(2/3))/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1+3^(1/2)))^2)^(1/2)/b/d/(b*x+a)^(1/2)/(-
b^(1/3)*(d*x+c)^(1/3)*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))/((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3)*(1+3^(1
/2)))^2)^(1/2)

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {52, 65, 231} \[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\frac {3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{2/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 b d \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}}+\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 b} \]

[In]

Int[(c + d*x)^(1/6)/Sqrt[a + b*x],x]

[Out]

(3*Sqrt[a + b*x]*(c + d*x)^(1/6))/(2*b) + (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*b*d*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 52

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 65

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 231

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]/(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)]))*EllipticF[ArcCos[(s + (1 - Sqrt[3])*r*x^2)/(s + (1 + Sqrt[3])*r*x^
2)], (2 + Sqrt[3])/4], x]] /; FreeQ[{a, b}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 b}+\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}} \, dx}{4 b} \\ & = \frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 b}+\frac {(3 (b c-a d)) \text {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 b d} \\ & = \frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 b}+\frac {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 b d \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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\frac {2 \sqrt {a+b x} \sqrt [6]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b \sqrt [6]{\frac {b (c+d x)}{b c-a d}}} \]

[In]

Integrate[(c + d*x)^(1/6)/Sqrt[a + b*x],x]

[Out]

(2*Sqrt[a + b*x]*(c + d*x)^(1/6)*Hypergeometric2F1[-1/6, 1/2, 3/2, (d*(a + b*x))/(-(b*c) + a*d)])/(b*((b*(c +
d*x))/(b*c - a*d))^(1/6))

Maple [F]

\[\int \frac {\left (d x +c \right )^{\frac {1}{6}}}{\sqrt {b x +a}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{6}}}{\sqrt {b x + a}} \,d x } \]

[In]

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

[Out]

integral((d*x + c)^(1/6)/sqrt(b*x + a), x)

Sympy [F]

\[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\int \frac {\sqrt [6]{c + d x}}{\sqrt {a + b x}}\, dx \]

[In]

integrate((d*x+c)**(1/6)/(b*x+a)**(1/2),x)

[Out]

Integral((c + d*x)**(1/6)/sqrt(a + b*x), x)

Maxima [F]

\[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{6}}}{\sqrt {b x + a}} \,d x } \]

[In]

integrate((d*x+c)^(1/6)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)^(1/6)/sqrt(b*x + a), x)

Giac [F]

\[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{6}}}{\sqrt {b x + a}} \,d x } \]

[In]

integrate((d*x+c)^(1/6)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate((d*x + c)^(1/6)/sqrt(b*x + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt [6]{c+d x}}{\sqrt {a+b x}} \, dx=\int \frac {{\left (c+d\,x\right )}^{1/6}}{\sqrt {a+b\,x}} \,d x \]

[In]

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

[Out]

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