3.71 \(\int \frac{\cos (a+b x)}{(c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=135 \[ \frac{i e^{-i \left (a-\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )}{2 b (c+d x)^{2/3}}-\frac{i e^{i \left (a-\frac{b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )}{2 b (c+d x)^{2/3}} \]

[Out]

((-I/2)*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(2/3)*Gamma[1/3, ((-I)*b*(c + d*x))/d])/(b*(c + d*x)^(2/3))
+ ((I/2)*((I*b*(c + d*x))/d)^(2/3)*Gamma[1/3, (I*b*(c + d*x))/d])/(b*E^(I*(a - (b*c)/d))*(c + d*x)^(2/3))

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Rubi [A]  time = 0.119584, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3307, 2181} \[ \frac{i e^{-i \left (a-\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )}{2 b (c+d x)^{2/3}}-\frac{i e^{i \left (a-\frac{b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )}{2 b (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]/(c + d*x)^(2/3),x]

[Out]

((-I/2)*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(2/3)*Gamma[1/3, ((-I)*b*(c + d*x))/d])/(b*(c + d*x)^(2/3))
+ ((I/2)*((I*b*(c + d*x))/d)^(2/3)*Gamma[1/3, (I*b*(c + d*x))/d])/(b*E^(I*(a - (b*c)/d))*(c + d*x)^(2/3))

Rule 3307

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int \frac{\cos (a+b x)}{(c+d x)^{2/3}} \, dx &=\frac{1}{2} \int \frac{e^{-i (a+b x)}}{(c+d x)^{2/3}} \, dx+\frac{1}{2} \int \frac{e^{i (a+b x)}}{(c+d x)^{2/3}} \, dx\\ &=-\frac{i e^{i \left (a-\frac{b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \Gamma \left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )}{2 b (c+d x)^{2/3}}+\frac{i e^{-i \left (a-\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \Gamma \left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )}{2 b (c+d x)^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0638611, size = 124, normalized size = 0.92 \[ \frac{i e^{-\frac{i (a d+b c)}{d}} \left (e^{\frac{2 i b c}{d}} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )-e^{2 i a} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )\right )}{2 b (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]/(c + d*x)^(2/3),x]

[Out]

((I/2)*(-(E^((2*I)*a)*(((-I)*b*(c + d*x))/d)^(2/3)*Gamma[1/3, ((-I)*b*(c + d*x))/d]) + E^(((2*I)*b*c)/d)*((I*b
*(c + d*x))/d)^(2/3)*Gamma[1/3, (I*b*(c + d*x))/d]))/(b*E^((I*(b*c + a*d))/d)*(c + d*x)^(2/3))

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Maple [F]  time = 0.165, size = 0, normalized size = 0. \begin{align*} \int{\cos \left ( bx+a \right ) \left ( dx+c \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)/(d*x+c)^(2/3),x)

[Out]

int(cos(b*x+a)/(d*x+c)^(2/3),x)

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Maxima [B]  time = 1.61575, size = 636, normalized size = 4.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(2/3),x, algorithm="maxima")

[Out]

-1/4*(d*x + c)^(1/3)*(((gamma(1/3, I*(d*x + c)*b/d) + gamma(1/3, -I*(d*x + c)*b/d))*cos(1/6*pi + 1/3*arctan2(0
, b) + 1/3*arctan2(0, d/sqrt(d^2))) + (gamma(1/3, I*(d*x + c)*b/d) + gamma(1/3, -I*(d*x + c)*b/d))*cos(-1/6*pi
+ 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) - (I*gamma(1/3, I*(d*x + c)*b/d) - I*gamma(1/3, -I*(d*x +
c)*b/d))*sin(1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) - (-I*gamma(1/3, I*(d*x + c)*b/d) + I*g
amma(1/3, -I*(d*x + c)*b/d))*sin(-1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))))*cos(-(b*c - a*d)/
d) - ((I*gamma(1/3, I*(d*x + c)*b/d) - I*gamma(1/3, -I*(d*x + c)*b/d))*cos(1/6*pi + 1/3*arctan2(0, b) + 1/3*ar
ctan2(0, d/sqrt(d^2))) + (I*gamma(1/3, I*(d*x + c)*b/d) - I*gamma(1/3, -I*(d*x + c)*b/d))*cos(-1/6*pi + 1/3*ar
ctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) + (gamma(1/3, I*(d*x + c)*b/d) + gamma(1/3, -I*(d*x + c)*b/d))*sin(
1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) - (gamma(1/3, I*(d*x + c)*b/d) + gamma(1/3, -I*(d*x
+ c)*b/d))*sin(-1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))))*sin(-(b*c - a*d)/d))/(d*((d*x + c)*
abs(b)/abs(d))^(1/3))

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Fricas [A]  time = 1.66481, size = 208, normalized size = 1.54 \begin{align*} \frac{i \, \left (\frac{i \, b}{d}\right )^{\frac{2}{3}} e^{\left (\frac{i \, b c - i \, a d}{d}\right )} \Gamma \left (\frac{1}{3}, \frac{i \, b d x + i \, b c}{d}\right ) - i \, \left (-\frac{i \, b}{d}\right )^{\frac{2}{3}} e^{\left (\frac{-i \, b c + i \, a d}{d}\right )} \Gamma \left (\frac{1}{3}, \frac{-i \, b d x - i \, b c}{d}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(2/3),x, algorithm="fricas")

[Out]

1/2*(I*(I*b/d)^(2/3)*e^((I*b*c - I*a*d)/d)*gamma(1/3, (I*b*d*x + I*b*c)/d) - I*(-I*b/d)^(2/3)*e^((-I*b*c + I*a
*d)/d)*gamma(1/3, (-I*b*d*x - I*b*c)/d))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (a + b x \right )}}{\left (c + d x\right )^{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)**(2/3),x)

[Out]

Integral(cos(a + b*x)/(c + d*x)**(2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (b x + a\right )}{{\left (d x + c\right )}^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(2/3),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)/(d*x + c)^(2/3), x)