3.70 \(\int \frac {\cos (a+b x)}{\sqrt [3]{c+d x}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 135, normalized size of antiderivative = 1.00, 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} \[ \frac {i e^{-i \left (a-\frac {b c}{d}\right )} \sqrt [3]{\frac {i b (c+d x)}{d}} \text {Gamma}\left (\frac {2}{3},\frac {i b (c+d x)}{d}\right )}{2 b \sqrt [3]{c+d x}}-\frac {i e^{i \left (a-\frac {b c}{d}\right )} \sqrt [3]{-\frac {i b (c+d x)}{d}} \text {Gamma}\left (\frac {2}{3},-\frac {i b (c+d x)}{d}\right )}{2 b \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]/(c + d*x)^(1/3),x]

[Out]

((-I/2)*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(1/3)*Gamma[2/3, ((-I)*b*(c + d*x))/d])/(b*(c + d*x)^(1/3))
+ ((I/2)*((I*b*(c + d*x))/d)^(1/3)*Gamma[2/3, (I*b*(c + d*x))/d])/(b*E^(I*(a - (b*c)/d))*(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 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)}{\sqrt [3]{c+d x}} \, dx &=\frac {1}{2} \int \frac {e^{-i (a+b x)}}{\sqrt [3]{c+d x}} \, dx+\frac {1}{2} \int \frac {e^{i (a+b x)}}{\sqrt [3]{c+d x}} \, dx\\ &=-\frac {i 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 )}{2 b \sqrt [3]{c+d x}}+\frac {i 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 )}{2 b \sqrt [3]{c+d x}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 124, normalized size = 0.92 \[ \frac {i e^{-\frac {i (a d+b c)}{d}} \left (e^{\frac {2 i b c}{d}} \sqrt [3]{\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},\frac {i b (c+d x)}{d}\right )-e^{2 i a} \sqrt [3]{-\frac {i b (c+d x)}{d}} \Gamma \left (\frac {2}{3},-\frac {i b (c+d x)}{d}\right )\right )}{2 b \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[a + b*x]/(c + d*x)^(1/3),x]

[Out]

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

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fricas [A]  time = 0.98, size = 86, normalized size = 0.64 \[ \frac {i \, \left (\frac {i \, b}{d}\right )^{\frac {1}{3}} e^{\left (\frac {i \, b c - i \, a d}{d}\right )} \Gamma \left (\frac {2}{3}, \frac {i \, b d x + i \, b c}{d}\right ) - i \, \left (-\frac {i \, b}{d}\right )^{\frac {1}{3}} e^{\left (\frac {-i \, b c + i \, a d}{d}\right )} \Gamma \left (\frac {2}{3}, \frac {-i \, b d x - i \, b c}{d}\right )}{2 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(1/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x + a\right )}{{\left (d x + c\right )}^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(1/3),x, algorithm="giac")

[Out]

integrate(cos(b*x + a)/(d*x + c)^(1/3), x)

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (b x +a \right )}{\left (d x +c \right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(b*x+a)/(d*x+c)^(1/3),x)

[Out]

int(cos(b*x+a)/(d*x+c)^(1/3),x)

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maxima [A]  time = 2.07, size = 137, normalized size = 1.01 \[ \frac {{\left (d x + c\right )}^{\frac {2}{3}} {\left ({\left ({\left (i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (-i \, \sqrt {3} - 1\right )} \Gamma \left (\frac {2}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \cos \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (\sqrt {3} + i\right )} \Gamma \left (\frac {2}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (\sqrt {3} - i\right )} \Gamma \left (\frac {2}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} \sin \left (-\frac {b c - a d}{d}\right )\right )}}{4 \, \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {2}{3}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)^(1/3),x, algorithm="maxima")

[Out]

1/4*(d*x + c)^(2/3)*(((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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)/(c + d*x)^(1/3),x)

[Out]

int(cos(a + b*x)/(c + d*x)^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)/(d*x+c)**(1/3),x)

[Out]

Integral(cos(a + b*x)/(c + d*x)**(1/3), x)

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