Integrand size = 19, antiderivative size = 366 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx=-\frac {2 \sqrt [3]{c+d x}}{b \sqrt {a+b x}}-\frac {4 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} b^{4/3} \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \]
-2*(d*x+c)^(1/3)/b/(b*x+a)^(1/2)-4/3*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/ 3))*EllipticF((-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)+(-a*d+b*c)^(1/3)*(1-3^(1/2))),2*I-I*3^(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))/(-b^ (1/3)*(d*x+c)^(1/3)+(-a*d+b*c)^(1/3)*(1-3^(1/2)))^2)^(1/2)*(1/2*6^(1/2)-1/ 2*2^(1/2))*3^(3/4)/b^(4/3)/(b*x+a)^(1/2)/(-(-a*d+b*c)^(1/3)*((-a*d+b*c)^(1 /3)-b^(1/3)*(d*x+c)^(1/3))/(-b^(1/3)*(d*x+c)^(1/3)+(-a*d+b*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 = 71, normalized size of antiderivative = 0.19 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx=-\frac {2 \sqrt [3]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{3},\frac {1}{2},\frac {d (a+b x)}{-b c+a d}\right )}{b \sqrt {a+b x} \sqrt [3]{\frac {b (c+d x)}{b c-a d}}} \]
(-2*(c + d*x)^(1/3)*Hypergeometric2F1[-1/2, -1/3, 1/2, (d*(a + b*x))/(-(b* c) + a*d)])/(b*Sqrt[a + b*x]*((b*(c + d*x))/(b*c - a*d))^(1/3))
Time = 0.30 (sec) , antiderivative size = 380, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {57, 73, 760}
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.
\(\displaystyle \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {2 d \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}}dx}{3 b}-\frac {2 \sqrt [3]{c+d x}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [3]{c+d x}}{b}-\frac {2 \sqrt [3]{c+d x}}{b \sqrt {a+b x}}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle -\frac {4 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} b^{4/3} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [3]{c+d x}}{b \sqrt {a+b x}}\) |
(-2*(c + d*x)^(1/3))/(b*Sqrt[a + b*x]) - (4*Sqrt[2 - Sqrt[3]]*((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))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3] )*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))/((1 - Sqrt[3])*(b*c - a*d)^ (1/3) - b^(1/3)*(c + d*x)^(1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*b^(4/3)*Sqrt[ -(((b*c - a*d)^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)]*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])
3.16.62.3.1 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 + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[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_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
\[\int \frac {\left (d x +c \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx=\int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{3/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{1/3}}{{\left (a+b\,x\right )}^{3/2}} \,d x \]