3.73 \(\int \frac{\cos (a+b x)}{(c+d x)^{5/3}} \, dx\)

Optimal. Leaf size=153 \[ \frac{3 e^{i \left (a-\frac{b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )}{4 d (c+d x)^{2/3}}+\frac{3 e^{-i \left (a-\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )}{4 d (c+d x)^{2/3}}-\frac{3 \cos (a+b x)}{2 d (c+d x)^{2/3}} \]

[Out]

(-3*Cos[a + b*x])/(2*d*(c + d*x)^(2/3)) + (3*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(2/3)*Gamma[1/3, ((-I)
*b*(c + d*x))/d])/(4*d*(c + d*x)^(2/3)) + (3*((I*b*(c + d*x))/d)^(2/3)*Gamma[1/3, (I*b*(c + d*x))/d])/(4*d*E^(
I*(a - (b*c)/d))*(c + d*x)^(2/3))

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Rubi [A]  time = 0.154681, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3297, 3308, 2181} \[ \frac{3 e^{i \left (a-\frac{b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )}{4 d (c+d x)^{2/3}}+\frac{3 e^{-i \left (a-\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )}{4 d (c+d x)^{2/3}}-\frac{3 \cos (a+b x)}{2 d (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]/(c + d*x)^(5/3),x]

[Out]

(-3*Cos[a + b*x])/(2*d*(c + d*x)^(2/3)) + (3*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(2/3)*Gamma[1/3, ((-I)
*b*(c + d*x))/d])/(4*d*(c + d*x)^(2/3)) + (3*((I*b*(c + d*x))/d)^(2/3)*Gamma[1/3, (I*b*(c + d*x))/d])/(4*d*E^(
I*(a - (b*c)/d))*(c + d*x)^(2/3))

Rule 3297

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] - Dist[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]

Rule 3308

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

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)^{5/3}} \, dx &=-\frac{3 \cos (a+b x)}{2 d (c+d x)^{2/3}}-\frac{(3 b) \int \frac{\sin (a+b x)}{(c+d x)^{2/3}} \, dx}{2 d}\\ &=-\frac{3 \cos (a+b x)}{2 d (c+d x)^{2/3}}-\frac{(3 i b) \int \frac{e^{-i (a+b x)}}{(c+d x)^{2/3}} \, dx}{4 d}+\frac{(3 i b) \int \frac{e^{i (a+b x)}}{(c+d x)^{2/3}} \, dx}{4 d}\\ &=-\frac{3 \cos (a+b x)}{2 d (c+d x)^{2/3}}+\frac{3 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 )}{4 d (c+d x)^{2/3}}+\frac{3 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 )}{4 d (c+d x)^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0569638, size = 121, normalized size = 0.79 \[ -\frac{e^{-\frac{i (a d+b c)}{d}} \left (e^{2 i a} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},-\frac{i b (c+d x)}{d}\right )+e^{\frac{2 i b c}{d}} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (-\frac{2}{3},\frac{i b (c+d x)}{d}\right )\right )}{2 d (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]/(c + d*x)^(5/3),x]

[Out]

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

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Maple [F]  time = 0.183, size = 0, normalized size = 0. \begin{align*} \int{\cos \left ( bx+a \right ) \left ( dx+c \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)/(d*x+c)^(5/3),x)

[Out]

int(cos(b*x+a)/(d*x+c)^(5/3),x)

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Maxima [B]  time = 1.55815, size = 632, normalized size = 4.13 \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)^(5/3),x, algorithm="maxima")

[Out]

-1/4*(((gamma(-2/3, I*(d*x + c)*b/d) + gamma(-2/3, -I*(d*x + c)*b/d))*cos(1/3*pi + 2/3*arctan2(0, b) + 2/3*arc
tan2(0, d/sqrt(d^2))) + (gamma(-2/3, I*(d*x + c)*b/d) + gamma(-2/3, -I*(d*x + c)*b/d))*cos(-1/3*pi + 2/3*arcta
n2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))) + (I*gamma(-2/3, I*(d*x + c)*b/d) - I*gamma(-2/3, -I*(d*x + c)*b/d))*s
in(1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))) + (-I*gamma(-2/3, I*(d*x + c)*b/d) + I*gamma(-2/3
, -I*(d*x + c)*b/d))*sin(-1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))))*cos(-(b*c - a*d)/d) + ((-
I*gamma(-2/3, I*(d*x + c)*b/d) + I*gamma(-2/3, -I*(d*x + c)*b/d))*cos(1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2
(0, d/sqrt(d^2))) + (-I*gamma(-2/3, I*(d*x + c)*b/d) + I*gamma(-2/3, -I*(d*x + c)*b/d))*cos(-1/3*pi + 2/3*arct
an2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))) + (gamma(-2/3, I*(d*x + c)*b/d) + gamma(-2/3, -I*(d*x + c)*b/d))*sin(
1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))) - (gamma(-2/3, I*(d*x + c)*b/d) + gamma(-2/3, -I*(d*
x + c)*b/d))*sin(-1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))))*sin(-(b*c - a*d)/d))*((d*x + c)*a
bs(b)/abs(d))^(2/3)/((d*x + c)^(2/3)*d)

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Fricas [A]  time = 1.73329, size = 290, normalized size = 1.9 \begin{align*} \frac{3 \,{\left ({\left (d x + c\right )} \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 ) +{\left (d x + c\right )} \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 \,{\left (d x + c\right )}^{\frac{1}{3}} \cos \left (b x + a\right )\right )}}{4 \,{\left (d^{2} x + c d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(5/3),x, algorithm="fricas")

[Out]

3/4*((d*x + c)*(I*b/d)^(2/3)*e^((I*b*c - I*a*d)/d)*gamma(1/3, (I*b*d*x + I*b*c)/d) + (d*x + c)*(-I*b/d)^(2/3)*
e^((-I*b*c + I*a*d)/d)*gamma(1/3, (-I*b*d*x - I*b*c)/d) - 2*(d*x + c)^(1/3)*cos(b*x + a))/(d^2*x + c*d)

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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{5}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)**(5/3),x)

[Out]

Integral(cos(a + b*x)/(c + d*x)**(5/3), 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{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(5/3),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)/(d*x + c)^(5/3), x)