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

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

[Out]

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

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Rubi [A]  time = 0.15, antiderivative size = 153, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.05, size = 121, normalized size = 0.79 \[ -\frac {e^{-\frac {i (a d+b c)}{d}} \left (e^{2 i a} \left (-\frac {i b (c+d x)}{d}\right )^{2/3} \Gamma \left (-\frac {2}{3},-\frac {i b (c+d x)}{d}\right )+e^{\frac {2 i b c}{d}} \left (\frac {i b (c+d x)}{d}\right )^{2/3} \Gamma \left (-\frac {2}{3},\frac {i b (c+d x)}{d}\right )\right )}{2 d (c+d x)^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.90, size = 117, normalized size = 0.76 \[ \frac {3 \, {\left ({\left (d x + c\right )} \left (\frac {i \, b}{d}\right )^{\frac {2}{3}} e^{\left (\frac {i \, b c - i \, a d}{d}\right )} \Gamma \left (\frac {1}{3}, \frac {i \, b d x + i \, b c}{d}\right ) + {\left (d x + c\right )} \left (-\frac {i \, b}{d}\right )^{\frac {2}{3}} e^{\left (\frac {-i \, b c + i \, a d}{d}\right )} \Gamma \left (\frac {1}{3}, \frac {-i \, b d x - i \, b c}{d}\right ) - 2 \, {\left (d x + c\right )}^{\frac {1}{3}} \cos \left (b x + a\right )\right )}}{4 \, {\left (d^{2} x + c d\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {5}{3}}}\,{d x} \]

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)

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x +a \right )}{\left (d x +c \right )^{\frac {5}{3}}}\, dx \]

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 [A]  time = 1.96, size = 138, normalized size = 0.90 \[ \frac {{\left ({\left ({\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {2}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {2}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) - {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {2}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {2}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {2}{3}}}{4 \, {\left (d x + c\right )}^{\frac {2}{3}} d} \]

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*(((-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))*cos(-(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))*sin(-(
b*c - a*d)/d))*((d*x + c)*b/d)^(2/3)/((d*x + c)^(2/3)*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x\right )}{{\left (c+d\,x\right )}^{5/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)/(c + d*x)^(5/3),x)

[Out]

int(cos(a + b*x)/(c + d*x)^(5/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {5}{3}}}\, dx \]

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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