Integrand size = 19, antiderivative size = 449 \[ \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx=-\frac {27 (b c-a d)^2 \sqrt {a+b x} \sqrt [6]{c+d x}}{320 b d^2}+\frac {3 (b c-a d) (a+b x)^{3/2} \sqrt [6]{c+d x}}{80 b d}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 b}+\frac {27\ 3^{3/4} (b c-a d)^{8/3} \sqrt [6]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{640 b d^3 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \]
3/80*(-a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/6)/b/d+3/8*(b*x+a)^(5/2)*(d*x+c)^ (1/6)/b-27/320*(-a*d+b*c)^2*(d*x+c)^(1/6)*(b*x+a)^(1/2)/b/d^2+27/640*3^(3/ 4)*(-a*d+b*c)^(8/3)*(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^3/(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)
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 = 0.16 \[ \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx=\frac {2 (a+b x)^{5/2} \sqrt [6]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {5}{2},\frac {7}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b \sqrt [6]{\frac {b (c+d x)}{b c-a d}}} \]
(2*(a + b*x)^(5/2)*(c + d*x)^(1/6)*Hypergeometric2F1[-1/6, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*((b*(c + d*x))/(b*c - a*d))^(1/6))
Time = 0.37 (sec) , antiderivative size = 468, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {60, 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.
\(\displaystyle \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{(c+d x)^{5/6}}dx}{16 b}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}-\frac {9 (b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{5/6}}dx}{10 d}\right )}{16 b}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}-\frac {9 (b c-a d) \left (\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{5/6}}dx}{4 d}\right )}{10 d}\right )}{16 b}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(b c-a d) \left (\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}-\frac {9 (b c-a d) \left (\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 d}-\frac {9 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [6]{c+d x}}{2 d^2}\right )}{10 d}\right )}{16 b}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 b}\) |
\(\Big \downarrow \) 766 |
\(\displaystyle \frac {(b c-a d) \left (\frac {3 (a+b x)^{3/2} \sqrt [6]{c+d x}}{5 d}-\frac {9 (b c-a d) \left (\frac {3 \sqrt {a+b x} \sqrt [6]{c+d x}}{2 d}-\frac {3\ 3^{3/4} \sqrt [6]{c+d x} (b c-a d)^{2/3} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arccos \left (\frac {\sqrt [3]{b c-a d}-\left (1-\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}}\right ),\frac {1}{4} \left (2+\sqrt {3}\right )\right )}{4 d^2 \sqrt {-\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\sqrt [3]{b c-a d}-\left (1+\sqrt {3}\right ) \sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{10 d}\right )}{16 b}+\frac {3 (a+b x)^{5/2} \sqrt [6]{c+d x}}{8 b}\) |
(3*(a + b*x)^(5/2)*(c + d*x)^(1/6))/(8*b) + ((b*c - a*d)*((3*(a + b*x)^(3/ 2)*(c + d*x)^(1/6))/(5*d) - (9*(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*d)))/(16*b)
3.18.37.3.1 Defintions of rubi rules used
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]
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]
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 ]
\[\int \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {1}{6}}d x\]
\[ \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{6}} \,d x } \]
\[ \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx=\int \left (a + b x\right )^{\frac {3}{2}} \sqrt [6]{c + d x}\, dx \]
\[ \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{6}} \,d x } \]
\[ \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{6}} \,d x } \]
Timed out. \[ \int (a+b x)^{3/2} \sqrt [6]{c+d x} \, dx=\int {\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{1/6} \,d x \]