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

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

[Out]

-3/4*cos(b*x+a)/d/(d*x+c)^(4/3)+9/8*I*b*exp(I*(a-b*c/d))*(-I*b*(d*x+c)/d)^(1/3)*GAMMA(2/3,-I*b*(d*x+c)/d)/d^2/
(d*x+c)^(1/3)-9/8*I*b*(I*b*(d*x+c)/d)^(1/3)*GAMMA(2/3,I*b*(d*x+c)/d)/d^2/exp(I*(a-b*c/d))/(d*x+c)^(1/3)+9/4*b*
sin(b*x+a)/d^2/(d*x+c)^(1/3)

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Rubi [A]  time = 0.20, antiderivative size = 182, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.05, size = 125, normalized size = 0.69 \[ \frac {i b e^{-\frac {i (a d+b c)}{d}} \left (e^{2 i a} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (-\frac {4}{3},-\frac {i b (c+d x)}{d}\right )-e^{\frac {2 i b c}{d}} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (-\frac {4}{3},\frac {i b (c+d x)}{d}\right )\right )}{2 d^2 \sqrt [3]{c+d x}} \]

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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fricas [A]  time = 0.91, size = 183, normalized size = 1.01 \[ \frac {{\left (-9 i \, b d^{2} x^{2} - 18 i \, b c d x - 9 i \, b c^{2}\right )} \left (\frac {i \, b}{d}\right )^{\frac {1}{3}} e^{\left (\frac {i \, b c - i \, a d}{d}\right )} \Gamma \left (\frac {2}{3}, \frac {i \, b d x + i \, b c}{d}\right ) + {\left (9 i \, b d^{2} x^{2} + 18 i \, b c d x + 9 i \, b c^{2}\right )} \left (-\frac {i \, b}{d}\right )^{\frac {1}{3}} e^{\left (\frac {-i \, b c + i \, a d}{d}\right )} \Gamma \left (\frac {2}{3}, \frac {-i \, b d x - i \, b c}{d}\right ) - 6 \, {\left (d x + c\right )}^{\frac {2}{3}} {\left (d \cos \left (b x + a\right ) - 3 \, {\left (b d x + b c\right )} \sin \left (b x + a\right )\right )}}{8 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {7}{3}}}\,{d x} \]

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)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x +a \right )}{\left (d x +c \right )^{\frac {7}{3}}}\, dx \]

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 [A]  time = 1.82, size = 137, normalized size = 0.75 \[ -\frac {{\left ({\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (-\frac {4}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (-\frac {4}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (-\frac {4}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {4}{3}}}{4 \, {\left (d x + c\right )}^{\frac {4}{3}} d} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\cos \left (a+b\,x\right )}{{\left (c+d\,x\right )}^{7/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)/(c + d*x)^(7/3),x)

[Out]

int(cos(a + b*x)/(c + d*x)^(7/3), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos {\left (a + b x \right )}}{\left (c + d x\right )^{\frac {7}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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