Integrand size = 16, antiderivative size = 152 \[ \int (c+d x)^{2/3} \cos (a+b x) \, dx=\frac {d e^{i \left (a-\frac {b c}{d}\right )} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},-\frac {i b (c+d x)}{d}\right )}{3 b^2 \sqrt [3]{c+d x}}+\frac {d e^{-i \left (a-\frac {b c}{d}\right )} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},\frac {i b (c+d x)}{d}\right )}{3 b^2 \sqrt [3]{c+d x}}+\frac {(c+d x)^{2/3} \sin (a+b x)}{b} \]
1/3*d*exp(I*(a-b*c/d))*(-I*b*(d*x+c)/d)^(1/3)*GAMMA(2/3,-I*b*(d*x+c)/d)/b^ 2/(d*x+c)^(1/3)+1/3*d*(I*b*(d*x+c)/d)^(1/3)*GAMMA(2/3,I*b*(d*x+c)/d)/b^2/e xp(I*(a-b*c/d))/(d*x+c)^(1/3)+(d*x+c)^(2/3)*sin(b*x+a)/b
Time = 0.05 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.80 \[ \int (c+d x)^{2/3} \cos (a+b x) \, dx=\frac {d e^{-\frac {i (b c+a d)}{d}} \left (e^{2 i a} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {5}{3},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (\frac {5}{3},\frac {i b (c+d x)}{d}\right )\right )}{2 b^2 \sqrt [3]{c+d x}} \]
(d*(E^((2*I)*a)*(((-I)*b*(c + d*x))/d)^(1/3)*Gamma[5/3, ((-I)*b*(c + d*x)) /d] + E^(((2*I)*b*c)/d)*((I*b*(c + d*x))/d)^(1/3)*Gamma[5/3, (I*b*(c + d*x ))/d]))/(2*b^2*E^((I*(b*c + a*d))/d)*(c + d*x)^(1/3))
Time = 0.43 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {3042, 3777, 25, 3042, 3789, 2612}
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.
\(\displaystyle \int (c+d x)^{2/3} \cos (a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (c+d x)^{2/3} \sin \left (a+b x+\frac {\pi }{2}\right )dx\) |
\(\Big \downarrow \) 3777 |
\(\displaystyle \frac {2 d \int -\frac {\sin (a+b x)}{\sqrt [3]{c+d x}}dx}{3 b}+\frac {(c+d x)^{2/3} \sin (a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {(c+d x)^{2/3} \sin (a+b x)}{b}-\frac {2 d \int \frac {\sin (a+b x)}{\sqrt [3]{c+d x}}dx}{3 b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(c+d x)^{2/3} \sin (a+b x)}{b}-\frac {2 d \int \frac {\sin (a+b x)}{\sqrt [3]{c+d x}}dx}{3 b}\) |
\(\Big \downarrow \) 3789 |
\(\displaystyle \frac {(c+d x)^{2/3} \sin (a+b x)}{b}-\frac {2 d \left (\frac {1}{2} i \int \frac {e^{-i (a+b x)}}{\sqrt [3]{c+d x}}dx-\frac {1}{2} i \int \frac {e^{i (a+b x)}}{\sqrt [3]{c+d x}}dx\right )}{3 b}\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle \frac {(c+d x)^{2/3} \sin (a+b x)}{b}-\frac {2 d \left (-\frac {e^{i \left (a-\frac {b c}{d}\right )} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},-\frac {i b (c+d x)}{d}\right )}{2 b \sqrt [3]{c+d x}}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},\frac {i b (c+d x)}{d}\right )}{2 b \sqrt [3]{c+d x}}\right )}{3 b}\) |
(-2*d*(-1/2*(E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(1/3)*Gamma[2/3, ( (-I)*b*(c + d*x))/d])/(b*(c + d*x)^(1/3)) - (((I*b*(c + d*x))/d)^(1/3)*Gam ma[2/3, (I*b*(c + d*x))/d])/(2*b*E^(I*(a - (b*c)/d))*(c + d*x)^(1/3))))/(3 *b) + ((c + d*x)^(2/3)*Sin[a + b*x])/b
3.1.68.3.1 Defintions of rubi rules used
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c + d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d) )^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m + 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] && !IntegerQ[m]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[( -(c + d*x)^m)*(Cos[e + f*x]/f), x] + Simp[d*(m/f) Int[(c + d*x)^(m - 1)*C os[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I /2 Int[(c + d*x)^m/E^(I*(e + f*x)), x], x] - Simp[I/2 Int[(c + d*x)^m*E ^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]
\[\int \left (d x +c \right )^{\frac {2}{3}} \cos \left (b x +a \right )d x\]
Time = 0.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.92 \[ \int (c+d x)^{2/3} \cos (a+b x) \, dx=\frac {3 \, {\left (d x + c\right )}^{\frac {2}{3}} b \sin \left (b x + a\right ) + {\left (d \cos \left (-\frac {b c - a d}{d}\right ) - i \, d \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (\frac {i \, b}{d}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, \frac {i \, b d x + i \, b c}{d}\right ) + {\left (d \cos \left (-\frac {b c - a d}{d}\right ) + i \, d \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (-\frac {i \, b}{d}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, \frac {-i \, b d x - i \, b c}{d}\right )}{3 \, b^{2}} \]
1/3*(3*(d*x + c)^(2/3)*b*sin(b*x + a) + (d*cos(-(b*c - a*d)/d) - I*d*sin(- (b*c - a*d)/d))*(I*b/d)^(1/3)*gamma(2/3, (I*b*d*x + I*b*c)/d) + (d*cos(-(b *c - a*d)/d) + I*d*sin(-(b*c - a*d)/d))*(-I*b/d)^(1/3)*gamma(2/3, (-I*b*d* x - I*b*c)/d))/b^2
\[ \int (c+d x)^{2/3} \cos (a+b x) \, dx=\int \left (c + d x\right )^{\frac {2}{3}} \cos {\left (a + b x \right )}\, dx \]
Time = 0.26 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.22 \[ \int (c+d x)^{2/3} \cos (a+b x) \, dx=\frac {6 \, {\left (d x + c\right )}^{\frac {2}{3}} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {2}{3}} d \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) + {\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} d \cos \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {2}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {2}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} d \sin \left (-\frac {b c - a d}{d}\right )\right )} {\left (d x + c\right )}^{\frac {2}{3}}}{6 \, b \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {2}{3}} d} \]
1/6*(6*(d*x + c)^(2/3)*((d*x + c)*b/d)^(2/3)*d*sin(((d*x + c)*b - b*c + a* d)/d) + (((sqrt(3) + I)*gamma(2/3, I*(d*x + c)*b/d) + (sqrt(3) - I)*gamma( 2/3, -I*(d*x + c)*b/d))*d*cos(-(b*c - a*d)/d) + ((-I*sqrt(3) + 1)*gamma(2/ 3, I*(d*x + c)*b/d) + (I*sqrt(3) + 1)*gamma(2/3, -I*(d*x + c)*b/d))*d*sin( -(b*c - a*d)/d))*(d*x + c)^(2/3))/(b*((d*x + c)*b/d)^(2/3)*d)
\[ \int (c+d x)^{2/3} \cos (a+b x) \, dx=\int { {\left (d x + c\right )}^{\frac {2}{3}} \cos \left (b x + a\right ) \,d x } \]
Timed out. \[ \int (c+d x)^{2/3} \cos (a+b x) \, dx=\int \cos \left (a+b\,x\right )\,{\left (c+d\,x\right )}^{2/3} \,d x \]