3.68 \(\int (c+d x)^{2/3} \cos (a+b x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.148958, antiderivative size = 152, normalized size of antiderivative = 1., 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 = {3296, 3308, 2181} \[ \frac{d e^{i \left (a-\frac{b c}{d}\right )} \sqrt [3]{-\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{2}{3},-\frac{i b (c+d x)}{d}\right )}{3 b^2 \sqrt [3]{c+d x}}+\frac{d e^{-i \left (a-\frac{b c}{d}\right )} \sqrt [3]{\frac{i b (c+d x)}{d}} \text{Gamma}\left (\frac{2}{3},\frac{i b (c+d x)}{d}\right )}{3 b^2 \sqrt [3]{c+d x}}+\frac{(c+d x)^{2/3} \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(2/3)*Cos[a + b*x],x]

[Out]

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

Rule 3296

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 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]

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

Mathematica [A]  time = 0.118327, size = 124, normalized size = 0.82 \[ -\frac{i (c+d x)^{2/3} e^{-\frac{i (a d+b c)}{d}} \left (\frac{e^{2 i a} \text{Gamma}\left (\frac{5}{3},-\frac{i b (c+d x)}{d}\right )}{\left (-\frac{i b (c+d x)}{d}\right )^{2/3}}-\frac{e^{\frac{2 i b c}{d}} \text{Gamma}\left (\frac{5}{3},\frac{i b (c+d x)}{d}\right )}{\left (\frac{i b (c+d x)}{d}\right )^{2/3}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(2/3)*Cos[a + b*x],x]

[Out]

((-I/2)*(c + d*x)^(2/3)*((E^((2*I)*a)*Gamma[5/3, ((-I)*b*(c + d*x))/d])/(((-I)*b*(c + d*x))/d)^(2/3) - (E^(((2
*I)*b*c)/d)*Gamma[5/3, (I*b*(c + d*x))/d])/((I*b*(c + d*x))/d)^(2/3)))/(b*E^((I*(b*c + a*d))/d))

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Maple [F]  time = 0.178, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{{\frac{2}{3}}}\cos \left ( bx+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(2/3)*cos(b*x+a),x)

[Out]

int((d*x+c)^(2/3)*cos(b*x+a),x)

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Maxima [B]  time = 1.59708, size = 701, normalized size = 4.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(2/3)*cos(b*x+a),x, algorithm="maxima")

[Out]

1/6*(6*(d*x + c)^(2/3)*d*((d*x + c)*abs(b)/abs(d))^(2/3)*sin(((d*x + c)*b - b*c + a*d)/d) + (((I*gamma(2/3, I*
(d*x + c)*b/d) - I*gamma(2/3, -I*(d*x + c)*b/d))*cos(1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2)))
+ (I*gamma(2/3, I*(d*x + c)*b/d) - I*gamma(2/3, -I*(d*x + c)*b/d))*cos(-1/3*pi + 2/3*arctan2(0, b) + 2/3*arct
an2(0, d/sqrt(d^2))) + (gamma(2/3, I*(d*x + c)*b/d) + gamma(2/3, -I*(d*x + c)*b/d))*sin(1/3*pi + 2/3*arctan2(0
, b) + 2/3*arctan2(0, d/sqrt(d^2))) - (gamma(2/3, I*(d*x + c)*b/d) + gamma(2/3, -I*(d*x + c)*b/d))*sin(-1/3*pi
+ 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))))*d*cos(-(b*c - a*d)/d) + ((gamma(2/3, I*(d*x + c)*b/d) + g
amma(2/3, -I*(d*x + c)*b/d))*cos(1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))) + (gamma(2/3, I*(d*
x + c)*b/d) + gamma(2/3, -I*(d*x + c)*b/d))*cos(-1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(0, d/sqrt(d^2))) + (
-I*gamma(2/3, I*(d*x + c)*b/d) + I*gamma(2/3, -I*(d*x + c)*b/d))*sin(1/3*pi + 2/3*arctan2(0, b) + 2/3*arctan2(
0, d/sqrt(d^2))) + (I*gamma(2/3, I*(d*x + c)*b/d) - I*gamma(2/3, -I*(d*x + c)*b/d))*sin(-1/3*pi + 2/3*arctan2(
0, b) + 2/3*arctan2(0, d/sqrt(d^2))))*d*sin(-(b*c - a*d)/d))*(d*x + c)^(2/3))/(b*d*((d*x + c)*abs(b)/abs(d))^(
2/3))

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Fricas [A]  time = 1.72673, size = 258, normalized size = 1.7 \begin{align*} \frac{d \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 ) + d \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 ) + 3 \,{\left (d x + c\right )}^{\frac{2}{3}} b \sin \left (b x + a\right )}{3 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(2/3)*cos(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c + d x\right )^{\frac{2}{3}} \cos{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**(2/3)*cos(b*x+a),x)

[Out]

Integral((c + d*x)**(2/3)*cos(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{\frac{2}{3}} \cos \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(2/3)*cos(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^(2/3)*cos(b*x + a), x)