3.74 \(\int \frac{\cos (a+b x)}{(c+d x)^{7/3}} \, dx\)

Optimal. Leaf size=182 \[ \frac{9 i b e^{i \left (a-\frac{b c}{d}\right )} \sqrt [3]{-\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{2}{3},-\frac{i b (c+d x)}{d}\right )}{8 d^2 \sqrt [3]{c+d x}}-\frac{9 i b e^{-i \left (a-\frac{b c}{d}\right )} \sqrt [3]{\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{2}{3},\frac{i b (c+d x)}{d}\right )}{8 d^2 \sqrt [3]{c+d x}}+\frac{9 b \sin (a+b x)}{4 d^2 \sqrt [3]{c+d x}}-\frac{3 \cos (a+b x)}{4 d (c+d x)^{4/3}} \]

[Out]

(-3*Cos[a + b*x])/(4*d*(c + d*x)^(4/3)) + (((9*I)/8)*b*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(1/3)*Gamma[
2/3, ((-I)*b*(c + d*x))/d])/(d^2*(c + d*x)^(1/3)) - (((9*I)/8)*b*((I*b*(c + d*x))/d)^(1/3)*Gamma[2/3, (I*b*(c
+ d*x))/d])/(d^2*E^(I*(a - (b*c)/d))*(c + d*x)^(1/3)) + (9*b*Sin[a + b*x])/(4*d^2*(c + d*x)^(1/3))

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Rubi [A]  time = 0.204047, antiderivative size = 182, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {3297, 3307, 2181} \[ \frac{9 i b e^{i \left (a-\frac{b c}{d}\right )} \sqrt [3]{-\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{2}{3},-\frac{i b (c+d x)}{d}\right )}{8 d^2 \sqrt [3]{c+d x}}-\frac{9 i b e^{-i \left (a-\frac{b c}{d}\right )} \sqrt [3]{\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{2}{3},\frac{i b (c+d x)}{d}\right )}{8 d^2 \sqrt [3]{c+d x}}+\frac{9 b \sin (a+b x)}{4 d^2 \sqrt [3]{c+d x}}-\frac{3 \cos (a+b x)}{4 d (c+d x)^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]/(c + d*x)^(7/3),x]

[Out]

(-3*Cos[a + b*x])/(4*d*(c + d*x)^(4/3)) + (((9*I)/8)*b*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(1/3)*Gamma[
2/3, ((-I)*b*(c + d*x))/d])/(d^2*(c + d*x)^(1/3)) - (((9*I)/8)*b*((I*b*(c + d*x))/d)^(1/3)*Gamma[2/3, (I*b*(c
+ d*x))/d])/(d^2*E^(I*(a - (b*c)/d))*(c + d*x)^(1/3)) + (9*b*Sin[a + b*x])/(4*d^2*(c + d*x)^(1/3))

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 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)^{7/3}} \, dx &=-\frac{3 \cos (a+b x)}{4 d (c+d x)^{4/3}}-\frac{(3 b) \int \frac{\sin (a+b x)}{(c+d x)^{4/3}} \, dx}{4 d}\\ &=-\frac{3 \cos (a+b x)}{4 d (c+d x)^{4/3}}+\frac{9 b \sin (a+b x)}{4 d^2 \sqrt [3]{c+d x}}-\frac{\left (9 b^2\right ) \int \frac{\cos (a+b x)}{\sqrt [3]{c+d x}} \, dx}{4 d^2}\\ &=-\frac{3 \cos (a+b x)}{4 d (c+d x)^{4/3}}+\frac{9 b \sin (a+b x)}{4 d^2 \sqrt [3]{c+d x}}-\frac{\left (9 b^2\right ) \int \frac{e^{-i (a+b x)}}{\sqrt [3]{c+d x}} \, dx}{8 d^2}-\frac{\left (9 b^2\right ) \int \frac{e^{i (a+b x)}}{\sqrt [3]{c+d x}} \, dx}{8 d^2}\\ &=-\frac{3 \cos (a+b x)}{4 d (c+d x)^{4/3}}+\frac{9 i b 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 )}{8 d^2 \sqrt [3]{c+d x}}-\frac{9 i b 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 )}{8 d^2 \sqrt [3]{c+d x}}+\frac{9 b \sin (a+b x)}{4 d^2 \sqrt [3]{c+d x}}\\ \end{align*}

Mathematica [A]  time = 0.053829, size = 125, normalized size = 0.69 \[ \frac{i b e^{-\frac{i (a d+b c)}{d}} \left (e^{2 i a} \sqrt [3]{-\frac{i b (c+d x)}{d}} \text{Gamma}\left (-\frac{4}{3},-\frac{i b (c+d x)}{d}\right )-e^{\frac{2 i b c}{d}} \sqrt [3]{\frac{i b (c+d x)}{d}} \text{Gamma}\left (-\frac{4}{3},\frac{i b (c+d x)}{d}\right )\right )}{2 d^2 \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]/(c + d*x)^(7/3),x]

[Out]

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

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Maple [F]  time = 0.186, size = 0, normalized size = 0. \begin{align*} \int{\cos \left ( bx+a \right ) \left ( dx+c \right ) ^{-{\frac{7}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)/(d*x+c)^(7/3),x)

[Out]

int(cos(b*x+a)/(d*x+c)^(7/3),x)

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Maxima [B]  time = 1.49109, size = 632, normalized size = 3.47 \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)^(7/3),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.86642, size = 452, normalized size = 2.48 \begin{align*} \frac{{\left (-9 i \, b d^{2} x^{2} - 18 i \, b c d x - 9 i \, b c^{2}\right )} \left (\frac{i \, b}{d}\right )^{\frac{1}{3}} e^{\left (\frac{i \, b c - i \, a d}{d}\right )} \Gamma \left (\frac{2}{3}, \frac{i \, b d x + i \, b c}{d}\right ) +{\left (9 i \, b d^{2} x^{2} + 18 i \, b c d x + 9 i \, b c^{2}\right )} \left (-\frac{i \, b}{d}\right )^{\frac{1}{3}} e^{\left (\frac{-i \, b c + i \, a d}{d}\right )} \Gamma \left (\frac{2}{3}, \frac{-i \, b d x - i \, b c}{d}\right ) - 6 \,{\left (d x + c\right )}^{\frac{2}{3}}{\left (d \cos \left (b x + a\right ) - 3 \,{\left (b d x + b c\right )} \sin \left (b x + a\right )\right )}}{8 \,{\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(7/3),x, algorithm="fricas")

[Out]

1/8*((-9*I*b*d^2*x^2 - 18*I*b*c*d*x - 9*I*b*c^2)*(I*b/d)^(1/3)*e^((I*b*c - I*a*d)/d)*gamma(2/3, (I*b*d*x + I*b
*c)/d) + (9*I*b*d^2*x^2 + 18*I*b*c*d*x + 9*I*b*c^2)*(-I*b/d)^(1/3)*e^((-I*b*c + I*a*d)/d)*gamma(2/3, (-I*b*d*x
- I*b*c)/d) - 6*(d*x + c)^(2/3)*(d*cos(b*x + a) - 3*(b*d*x + b*c)*sin(b*x + a)))/(d^4*x^2 + 2*c*d^3*x + c^2*d
^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)**(7/3),x)

[Out]

Timed out

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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{7}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(7/3),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)/(d*x + c)^(7/3), x)