\(\int \sqrt {a+b x} \sqrt [6]{c+d x} \, dx\) [489]

Optimal result

Integrand size = 19, antiderivative size = 411 \[ \int \sqrt {a+b x} \sqrt [6]{c+d x} \, dx=\frac {3 (b c-a d) \sqrt {a+b x} \sqrt [6]{c+d x}}{20 b d}+\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 b}-\frac {3\ 3^{3/4} (b c-a d)^{5/3} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{40 b d^2 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \] Output:

3/20*(-a*d+b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/6)/b/d+3/5*(b*x+a)^(3/2)*(d*x+c)^ 
(1/6)/b-3/40*3^(3/4)*(-a*d+b*c)^(5/3)*(d*x+c)^(1/6)*((-a*d+b*c)^(1/3)-b^(1 
/3)*(d*x+c)^(1/3))*(((-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)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1 
/3))^2)^(1/2)*InverseJacobiAM(arccos(((-a*d+b*c)^(1/3)-(1-3^(1/2))*b^(1/3) 
*(d*x+c)^(1/3))/((-a*d+b*c)^(1/3)-(1+3^(1/2))*b^(1/3)*(d*x+c)^(1/3))),1/4* 
6^(1/2)+1/4*2^(1/2))/b/d^2/(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)-(1+3^(1/2))*b^(1/3)*(d*x 
+c)^(1/3))^2)^(1/2)
 

Mathematica [C] (verified)

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

Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.18 \[ \int \sqrt {a+b x} \sqrt [6]{c+d x} \, dx=\frac {2 (a+b x)^{3/2} \sqrt [6]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {3}{2},\frac {5}{2},\frac {d (a+b x)}{-b c+a d}\right )}{3 b \sqrt [6]{\frac {b (c+d x)}{b c-a d}}} \] Input:

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

Output:

(2*(a + b*x)^(3/2)*(c + d*x)^(1/6)*Hypergeometric2F1[-1/6, 3/2, 5/2, (d*(a 
 + b*x))/(-(b*c) + a*d)])/(3*b*((b*(c + d*x))/(b*c - a*d))^(1/6))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 427, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {60, 60, 73, 766}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{5/6}}dx}{10 b}+\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 b}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(b c-a d) \left (\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}}dx}{4 d}\right )}{10 b}+\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {(b c-a d) \left (\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 d}-\frac {9 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{2 d^2}\right )}{10 b}+\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 b}\)

\(\Big \downarrow \) 766

\(\displaystyle \frac {(b c-a d) \left (\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 d}-\frac {3\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{2/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 d^2 \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{10 b}+\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 b}\)

Input:

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

Output:

(3*(a + b*x)^(3/2)*(c + d*x)^(1/6))/(5*b) + ((b*c - a*d)*((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[ 
-((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)]*Sqrt[a - ( 
b*c)/d + (b*(c + d*x))/d])))/(10*b)
 

Defintions of rubi rules used

rule 60

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] + Simp[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] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 766

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 
]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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