3.1.69 \(\int \sqrt [3]{c+d x} \cos (a+b x) \, dx\) [69]

Optimal. Leaf size=152 \[ \frac {d e^{i \left (a-\frac {b c}{d}\right )} \left (-\frac {i b (c+d x)}{d}\right )^{2/3} \text {Gamma}\left (\frac {1}{3},-\frac {i b (c+d x)}{d}\right )}{6 b^2 (c+d x)^{2/3}}+\frac {d e^{-i \left (a-\frac {b c}{d}\right )} \left (\frac {i b (c+d x)}{d}\right )^{2/3} \text {Gamma}\left (\frac {1}{3},\frac {i b (c+d x)}{d}\right )}{6 b^2 (c+d x)^{2/3}}+\frac {\sqrt [3]{c+d x} \sin (a+b x)}{b} \]

[Out]

1/6*d*exp(I*(a-b*c/d))*(-I*b*(d*x+c)/d)^(2/3)*GAMMA(1/3,-I*b*(d*x+c)/d)/b^2/(d*x+c)^(2/3)+1/6*d*(I*b*(d*x+c)/d
)^(2/3)*GAMMA(1/3,I*b*(d*x+c)/d)/b^2/exp(I*(a-b*c/d))/(d*x+c)^(2/3)+(d*x+c)^(1/3)*sin(b*x+a)/b

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 152, 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 = {3377, 3389, 2212} \begin {gather*} \frac {d 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 )}{6 b^2 (c+d x)^{2/3}}+\frac {d 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 )}{6 b^2 (c+d x)^{2/3}}+\frac {\sqrt [3]{c+d x} \sin (a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(1/3)*Cos[a + b*x],x]

[Out]

(d*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(2/3)*Gamma[1/3, ((-I)*b*(c + d*x))/d])/(6*b^2*(c + d*x)^(2/3))
+ (d*((I*b*(c + d*x))/d)^(2/3)*Gamma[1/3, (I*b*(c + d*x))/d])/(6*b^2*E^(I*(a - (b*c)/d))*(c + d*x)^(2/3)) + ((
c + d*x)^(1/3)*Sin[a + b*x])/b

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3389

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 \sqrt [3]{c+d x} \cos (a+b x) \, dx &=\frac {\sqrt [3]{c+d x} \sin (a+b x)}{b}-\frac {d \int \frac {\sin (a+b x)}{(c+d x)^{2/3}} \, dx}{3 b}\\ &=\frac {\sqrt [3]{c+d x} \sin (a+b x)}{b}-\frac {(i d) \int \frac {e^{-i (a+b x)}}{(c+d x)^{2/3}} \, dx}{6 b}+\frac {(i d) \int \frac {e^{i (a+b x)}}{(c+d x)^{2/3}} \, dx}{6 b}\\ &=\frac {d 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 )}{6 b^2 (c+d x)^{2/3}}+\frac {d 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 )}{6 b^2 (c+d x)^{2/3}}+\frac {\sqrt [3]{c+d x} \sin (a+b x)}{b}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.06, size = 124, normalized size = 0.82 \begin {gather*} -\frac {i e^{-\frac {i (b c+a d)}{d}} \sqrt [3]{c+d x} \left (\frac {e^{2 i a} \text {Gamma}\left (\frac {4}{3},-\frac {i b (c+d x)}{d}\right )}{\sqrt [3]{-\frac {i b (c+d x)}{d}}}-\frac {e^{\frac {2 i b c}{d}} \text {Gamma}\left (\frac {4}{3},\frac {i b (c+d x)}{d}\right )}{\sqrt [3]{\frac {i b (c+d x)}{d}}}\right )}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{\frac {1}{3}} \cos \left (b x +a \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.35, size = 186, normalized size = 1.22 \begin {gather*} \frac {12 \, {\left (d x + c\right )}^{\frac {1}{3}} \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {1}{3}} d \sin \left (\frac {{\left (d x + c\right )} b - b c + a d}{d}\right ) + {\left ({\left ({\left (i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {1}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (-i \, \sqrt {3} + 1\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} d \cos \left (-\frac {b c - a d}{d}\right ) + {\left ({\left (\sqrt {3} - i\right )} \Gamma \left (\frac {1}{3}, \frac {i \, {\left (d x + c\right )} b}{d}\right ) + {\left (\sqrt {3} + i\right )} \Gamma \left (\frac {1}{3}, -\frac {i \, {\left (d x + c\right )} b}{d}\right )\right )} d \sin \left (-\frac {b c - a d}{d}\right )\right )} {\left (d x + c\right )}^{\frac {1}{3}}}{12 \, b \left (\frac {{\left (d x + c\right )} b}{d}\right )^{\frac {1}{3}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(12*(d*x + c)^(1/3)*((d*x + c)*b/d)^(1/3)*d*sin(((d*x + c)*b - b*c + a*d)/d) + (((I*sqrt(3) + 1)*gamma(1/
3, I*(d*x + c)*b/d) + (-I*sqrt(3) + 1)*gamma(1/3, -I*(d*x + c)*b/d))*d*cos(-(b*c - a*d)/d) + ((sqrt(3) - I)*ga
mma(1/3, I*(d*x + c)*b/d) + (sqrt(3) + I)*gamma(1/3, -I*(d*x + c)*b/d))*d*sin(-(b*c - a*d)/d))*(d*x + c)^(1/3)
)/(b*((d*x + c)*b/d)^(1/3)*d)

________________________________________________________________________________________

Fricas [A]
time = 0.11, size = 102, normalized size = 0.67 \begin {gather*} \frac {d \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 ) + d \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 ) + 6 \, {\left (d x + c\right )}^{\frac {1}{3}} b \sin \left (b x + a\right )}{6 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________