Integrand size = 16, antiderivative size = 151 \[ \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx=-\frac {3 \cos (a+b x)}{d \sqrt [3]{c+d x}}+\frac {3 e^{i \left (a-\frac {b c}{d}\right )} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},-\frac {i b (c+d x)}{d}\right )}{2 d \sqrt [3]{c+d x}}+\frac {3 e^{-i \left (a-\frac {b c}{d}\right )} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},\frac {i b (c+d x)}{d}\right )}{2 d \sqrt [3]{c+d x}} \]
-3*cos(b*x+a)/d/(d*x+c)^(1/3)+3/2*exp(I*(a-b*c/d))*(-I*b*(d*x+c)/d)^(1/3)* GAMMA(2/3,-I*b*(d*x+c)/d)/d/(d*x+c)^(1/3)+3/2*(I*b*(d*x+c)/d)^(1/3)*GAMMA( 2/3,I*b*(d*x+c)/d)/d/exp(I*(a-b*c/d))/(d*x+c)^(1/3)
Time = 0.05 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.80 \[ \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx=-\frac {e^{-\frac {i (b c+a d)}{d}} \left (e^{2 i a} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (-\frac {1}{3},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (-\frac {1}{3},\frac {i b (c+d x)}{d}\right )\right )}{2 d \sqrt [3]{c+d x}} \]
-1/2*(E^((2*I)*a)*(((-I)*b*(c + d*x))/d)^(1/3)*Gamma[-1/3, ((-I)*b*(c + d* x))/d] + E^(((2*I)*b*c)/d)*((I*b*(c + d*x))/d)^(1/3)*Gamma[-1/3, (I*b*(c + d*x))/d])/(d*E^((I*(b*c + a*d))/d)*(c + d*x)^(1/3))
Time = 0.42 (sec) , antiderivative size = 158, 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, 3778, 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.
\(\displaystyle \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (a+b x+\frac {\pi }{2}\right )}{(c+d x)^{4/3}}dx\) |
\(\Big \downarrow \) 3778 |
\(\displaystyle \frac {3 b \int -\frac {\sin (a+b x)}{\sqrt [3]{c+d x}}dx}{d}-\frac {3 \cos (a+b x)}{d \sqrt [3]{c+d x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {3 b \int \frac {\sin (a+b x)}{\sqrt [3]{c+d x}}dx}{d}-\frac {3 \cos (a+b x)}{d \sqrt [3]{c+d x}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {3 b \int \frac {\sin (a+b x)}{\sqrt [3]{c+d x}}dx}{d}-\frac {3 \cos (a+b x)}{d \sqrt [3]{c+d x}}\) |
\(\Big \downarrow \) 3789 |
\(\displaystyle -\frac {3 \cos (a+b x)}{d \sqrt [3]{c+d x}}-\frac {3 b \left (\frac {1}{2} i \int \frac {e^{-i (a+b x)}}{\sqrt [3]{c+d x}}dx-\frac {1}{2} i \int \frac {e^{i (a+b x)}}{\sqrt [3]{c+d x}}dx\right )}{d}\) |
\(\Big \downarrow \) 2612 |
\(\displaystyle -\frac {3 \cos (a+b x)}{d \sqrt [3]{c+d x}}-\frac {3 b \left (-\frac {e^{i \left (a-\frac {b c}{d}\right )} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},-\frac {i b (c+d x)}{d}\right )}{2 b \sqrt [3]{c+d x}}-\frac {e^{-i \left (a-\frac {b c}{d}\right )} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},\frac {i b (c+d x)}{d}\right )}{2 b \sqrt [3]{c+d x}}\right )}{d}\) |
(-3*Cos[a + b*x])/(d*(c + d*x)^(1/3)) - (3*b*(-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)*Gamma[2/3, (I*b*(c + d*x))/d])/(2*b*E^ (I*(a - (b*c)/d))*(c + d*x)^(1/3))))/d
3.1.72.3.1 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 + 1)*(Sin[e + f*x]/(d*(m + 1))), x] - Simp[f/(d*(m + 1)) Int[( c + d*x)^(m + 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && LtQ[m, - 1]
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]
\[\int \frac {\cos \left (b x +a \right )}{\left (d x +c \right )^{\frac {4}{3}}}d x\]
Time = 0.10 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx=\frac {3 \, {\left ({\left ({\left (d x + c\right )} \cos \left (-\frac {b c - a d}{d}\right ) - {\left (i \, d x + i \, c\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (\frac {i \, b}{d}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, \frac {i \, b d x + i \, b c}{d}\right ) + {\left ({\left (d x + c\right )} \cos \left (-\frac {b c - a d}{d}\right ) - {\left (-i \, d x - i \, c\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (-\frac {i \, b}{d}\right )^{\frac {1}{3}} \Gamma \left (\frac {2}{3}, \frac {-i \, b d x - i \, b c}{d}\right ) - 2 \, {\left (d x + c\right )}^{\frac {2}{3}} \cos \left (b x + a\right )\right )}}{2 \, {\left (d^{2} x + c d\right )}} \]
3/2*(((d*x + c)*cos(-(b*c - a*d)/d) - (I*d*x + I*c)*sin(-(b*c - a*d)/d))*( I*b/d)^(1/3)*gamma(2/3, (I*b*d*x + I*b*c)/d) + ((d*x + c)*cos(-(b*c - a*d) /d) - (-I*d*x - I*c)*sin(-(b*c - a*d)/d))*(-I*b/d)^(1/3)*gamma(2/3, (-I*b* d*x - I*b*c)/d) - 2*(d*x + c)^(2/3)*cos(b*x + a))/(d^2*x + c*d)
\[ \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx=\int \frac {\cos {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.91 \[ \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx=-\frac {{\left ({\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {1}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {1}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - {\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {1}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {1}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {1}{3}}}{4 \, {\left (d x + c\right )}^{\frac {1}{3}} d} \]
-1/4*(((sqrt(3) + I)*gamma(-1/3, I*(d*x + c)*b/d) + (sqrt(3) - I)*gamma(-1 /3, -I*(d*x + c)*b/d))*cos(-(b*c - a*d)/d) - ((I*sqrt(3) - 1)*gamma(-1/3, I*(d*x + c)*b/d) + (-I*sqrt(3) - 1)*gamma(-1/3, -I*(d*x + c)*b/d))*sin(-(b *c - a*d)/d))*((d*x + c)*b/d)^(1/3)/((d*x + c)^(1/3)*d)
\[ \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx=\int { \frac {\cos \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {\cos (a+b x)}{(c+d x)^{4/3}} \, dx=\int \frac {\cos \left (a+b\,x\right )}{{\left (c+d\,x\right )}^{4/3}} \,d x \]