3.167 \(\int \frac{(A+C \cos ^2(c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx\)

Optimal. Leaf size=90 \[ \frac{3 (A-2 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{8 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}} \]

[Out]

(3*A*Sin[c + d*x])/(2*d*(b*Cos[c + d*x])^(2/3)) + (3*(A - 2*C)*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2, 2
/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b^2*d*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.0813656, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {16, 3012, 2643} \[ \frac{3 (A-2 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{8 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x])/(b*Cos[c + d*x])^(2/3),x]

[Out]

(3*A*Sin[c + d*x])/(2*d*(b*Cos[c + d*x])^(2/3)) + (3*(A - 2*C)*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2, 2
/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*b^2*d*Sqrt[Sin[c + d*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rubi steps

\begin{align*} \int \frac{\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx &=b \int \frac{A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{5/3}} \, dx\\ &=\frac{3 A \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}}-\frac{(A-2 C) \int \sqrt [3]{b \cos (c+d x)} \, dx}{2 b}\\ &=\frac{3 A \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}}+\frac{3 (A-2 C) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{8 b^2 d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 3.64923, size = 277, normalized size = 3.08 \[ -\frac{3 \csc (c) e^{-i d x} \cos ^{\frac{2}{3}}(c+d x) (\cos (d x)+i \sin (d x)) \left (5 (A-2 C) (\cos (d x)-i \sin (d x)) (i \sin (2 (c+d x))+\cos (2 (c+d x))+1)^{2/3} \, _2F_1\left (-\frac{1}{6},\frac{2}{3};\frac{5}{6};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )+(A-2 C) (\cos (d x)+i \sin (d x)) (i \sin (2 (c+d x))+\cos (2 (c+d x))+1)^{2/3} \, _2F_1\left (\frac{2}{3},\frac{5}{6};\frac{11}{6};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )+10 ((C-A) \cos (d x)+C \cos (2 c+d x))\right )}{10 \sqrt [3]{2} d (b \cos (c+d x))^{2/3} \left (e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Cos[c + d*x]^(2/3)*Csc[c]*(Cos[d*x] + I*Sin[d*x])*(10*((-A + C)*Cos[d*x] + C*Cos[2*c + d*x]) + 5*(A - 2*C)
*Hypergeometric2F1[-1/6, 2/3, 5/6, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*(Cos[d*x] - I*Sin[d*x])*(1 + Cos[2*
(c + d*x)] + I*Sin[2*(c + d*x)])^(2/3) + (A - 2*C)*Hypergeometric2F1[2/3, 5/6, 11/6, -(E^((2*I)*d*x)*(Cos[c] +
I*Sin[c])^2)]*(Cos[d*x] + I*Sin[d*x])*(1 + Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])^(2/3)))/(10*2^(1/3)*d*E^(I*
d*x)*(b*Cos[c + d*x])^(2/3)*(((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x))^(2/3))

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Maple [F]  time = 0.329, size = 0, normalized size = 0. \begin{align*} \int{ \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sec \left ( dx+c \right ) \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(b*cos(d*x+c))^(2/3),x)

[Out]

int((A+C*cos(d*x+c)^2)*sec(d*x+c)/(b*cos(d*x+c))^(2/3),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)/(b*cos(d*x + c))^(2/3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}} \sec \left (d x + c\right )}{b \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(1/3)*sec(d*x + c)/(b*cos(d*x + c)), x)

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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((A+C*cos(d*x+c)**2)*sec(d*x+c)/(b*cos(d*x+c))**(2/3),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*sec(d*x + c)/(b*cos(d*x + c))^(2/3), x)