3.67 \(\int (c+d x)^{4/3} \cos (a+b x) \, dx\)

Optimal. Leaf size=183 \[ \frac{2 i d^2 e^{i \left (a-\frac{b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )}{9 b^3 (c+d x)^{2/3}}-\frac{2 i d^2 e^{-i \left (a-\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )}{9 b^3 (c+d x)^{2/3}}+\frac{4 d \sqrt [3]{c+d x} \cos (a+b x)}{3 b^2}+\frac{(c+d x)^{4/3} \sin (a+b x)}{b} \]

[Out]

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

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Rubi [A]  time = 0.2404, antiderivative size = 183, 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 = {3296, 3307, 2181} \[ \frac{2 i d^2 e^{i \left (a-\frac{b c}{d}\right )} \left (-\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},-\frac{i b (c+d x)}{d}\right )}{9 b^3 (c+d x)^{2/3}}-\frac{2 i d^2 e^{-i \left (a-\frac{b c}{d}\right )} \left (\frac{i b (c+d x)}{d}\right )^{2/3} \text{Gamma}\left (\frac{1}{3},\frac{i b (c+d x)}{d}\right )}{9 b^3 (c+d x)^{2/3}}+\frac{4 d \sqrt [3]{c+d x} \cos (a+b x)}{3 b^2}+\frac{(c+d x)^{4/3} \sin (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^(4/3)*Cos[a + b*x],x]

[Out]

(4*d*(c + d*x)^(1/3)*Cos[a + b*x])/(3*b^2) + (((2*I)/9)*d^2*E^(I*(a - (b*c)/d))*(((-I)*b*(c + d*x))/d)^(2/3)*G
amma[1/3, ((-I)*b*(c + d*x))/d])/(b^3*(c + d*x)^(2/3)) - (((2*I)/9)*d^2*((I*b*(c + d*x))/d)^(2/3)*Gamma[1/3, (
I*b*(c + d*x))/d])/(b^3*E^(I*(a - (b*c)/d))*(c + d*x)^(2/3)) + ((c + d*x)^(4/3)*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 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 (c+d x)^{4/3} \cos (a+b x) \, dx &=\frac{(c+d x)^{4/3} \sin (a+b x)}{b}-\frac{(4 d) \int \sqrt [3]{c+d x} \sin (a+b x) \, dx}{3 b}\\ &=\frac{4 d \sqrt [3]{c+d x} \cos (a+b x)}{3 b^2}+\frac{(c+d x)^{4/3} \sin (a+b x)}{b}-\frac{\left (4 d^2\right ) \int \frac{\cos (a+b x)}{(c+d x)^{2/3}} \, dx}{9 b^2}\\ &=\frac{4 d \sqrt [3]{c+d x} \cos (a+b x)}{3 b^2}+\frac{(c+d x)^{4/3} \sin (a+b x)}{b}-\frac{\left (2 d^2\right ) \int \frac{e^{-i (a+b x)}}{(c+d x)^{2/3}} \, dx}{9 b^2}-\frac{\left (2 d^2\right ) \int \frac{e^{i (a+b x)}}{(c+d x)^{2/3}} \, dx}{9 b^2}\\ &=\frac{4 d \sqrt [3]{c+d x} \cos (a+b x)}{3 b^2}+\frac{2 i d^2 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 )}{9 b^3 (c+d x)^{2/3}}-\frac{2 i d^2 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 )}{9 b^3 (c+d x)^{2/3}}+\frac{(c+d x)^{4/3} \sin (a+b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.107263, size = 122, normalized size = 0.67 \[ \frac{d \sqrt [3]{c+d x} e^{-\frac{i (a d+b c)}{d}} \left (\frac{e^{2 i a} \text{Gamma}\left (\frac{7}{3},-\frac{i b (c+d x)}{d}\right )}{\sqrt [3]{-\frac{i b (c+d x)}{d}}}+\frac{e^{\frac{2 i b c}{d}} \text{Gamma}\left (\frac{7}{3},\frac{i b (c+d x)}{d}\right )}{\sqrt [3]{\frac{i b (c+d x)}{d}}}\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^(4/3)*Cos[a + b*x],x]

[Out]

(d*(c + d*x)^(1/3)*((E^((2*I)*a)*Gamma[7/3, ((-I)*b*(c + d*x))/d])/(((-I)*b*(c + d*x))/d)^(1/3) + (E^(((2*I)*b
*c)/d)*Gamma[7/3, (I*b*(c + d*x))/d])/((I*b*(c + d*x))/d)^(1/3)))/(2*b^2*E^((I*(b*c + a*d))/d))

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Maple [F]  time = 0.191, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) ^{{\frac{4}{3}}}\cos \left ( bx+a \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^(4/3)*cos(b*x+a),x)

[Out]

int((d*x+c)^(4/3)*cos(b*x+a),x)

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Maxima [B]  time = 1.5645, size = 770, normalized size = 4.21 \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)^(4/3)*cos(b*x+a),x, algorithm="maxima")

[Out]

1/9*(9*(d*x + c)^(4/3)*b*d*((d*x + c)*abs(b)/abs(d))^(1/3)*sin(((d*x + c)*b - b*c + a*d)/d) + 12*(d*x + c)^(1/
3)*d^2*((d*x + c)*abs(b)/abs(d))^(1/3)*cos(((d*x + c)*b - b*c + a*d)/d) + (((gamma(1/3, I*(d*x + c)*b/d) + gam
ma(1/3, -I*(d*x + c)*b/d))*cos(1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) + (gamma(1/3, I*(d*x
+ c)*b/d) + gamma(1/3, -I*(d*x + c)*b/d))*cos(-1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) + (-I
*gamma(1/3, I*(d*x + c)*b/d) + I*gamma(1/3, -I*(d*x + c)*b/d))*sin(1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0,
d/sqrt(d^2))) + (I*gamma(1/3, I*(d*x + c)*b/d) - I*gamma(1/3, -I*(d*x + c)*b/d))*sin(-1/6*pi + 1/3*arctan2(0,
b) + 1/3*arctan2(0, d/sqrt(d^2))))*d^2*cos(-(b*c - a*d)/d) + ((-I*gamma(1/3, I*(d*x + c)*b/d) + I*gamma(1/3,
-I*(d*x + c)*b/d))*cos(1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) + (-I*gamma(1/3, I*(d*x + c)*
b/d) + I*gamma(1/3, -I*(d*x + c)*b/d))*cos(-1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(d^2))) - (gamma
(1/3, I*(d*x + c)*b/d) + gamma(1/3, -I*(d*x + c)*b/d))*sin(1/6*pi + 1/3*arctan2(0, b) + 1/3*arctan2(0, d/sqrt(
d^2))) + (gamma(1/3, I*(d*x + c)*b/d) + gamma(1/3, -I*(d*x + c)*b/d))*sin(-1/6*pi + 1/3*arctan2(0, b) + 1/3*ar
ctan2(0, d/sqrt(d^2))))*d^2*sin(-(b*c - a*d)/d))*(d*x + c)^(1/3))/(b^2*d*((d*x + c)*abs(b)/abs(d))^(1/3))

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Fricas [A]  time = 1.75519, size = 331, normalized size = 1.81 \begin{align*} \frac{-2 i \, d^{2} \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 ) + 2 i \, d^{2} \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 ) + 3 \,{\left (4 \, b d \cos \left (b x + a\right ) + 3 \,{\left (b^{2} d x + b^{2} c\right )} \sin \left (b x + a\right )\right )}{\left (d x + c\right )}^{\frac{1}{3}}}{9 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^(4/3)*cos(b*x+a),x, algorithm="fricas")

[Out]

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

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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((d*x+c)**(4/3)*cos(b*x+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )}^{\frac{4}{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)^(4/3)*cos(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^(4/3)*cos(b*x + a), x)