3.163 \(\int \frac{(A+C \cos ^2(c+d x)) \sec ^3(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=92 \[ \frac{3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac{3 (4 A+7 C) \sin (c+d x) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

[Out]

(3*A*b^2*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(7/3)) + (3*(4*A + 7*C)*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c +
d*x]^2]*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(1/3)*Sqrt[Sin[c + d*x]^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0972061, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {16, 3012, 2643} \[ \frac{3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac{3 (4 A+7 C) \sin (c+d x) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(b*Cos[c + d*x])^(1/3),x]

[Out]

(3*A*b^2*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(7/3)) + (3*(4*A + 7*C)*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c +
d*x]^2]*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(1/3)*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 ^3(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx &=b^3 \int \frac{A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{10/3}} \, dx\\ &=\frac{3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac{1}{7} (b (4 A+7 C)) \int \frac{1}{(b \cos (c+d x))^{4/3}} \, dx\\ &=\frac{3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac{3 (4 A+7 C) \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d \sqrt [3]{b \cos (c+d x)} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [C]  time = 5.98642, size = 404, normalized size = 4.39 \[ \frac{3 b^2 \csc (c) \sec (c) e^{-i d x} \left (A+C \cos ^2(c+d x)\right ) \left (2 \cos (c) \left (e^{i d x} \sqrt [3]{e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )} ((2 A+7 C) \cos (2 c+d x)+(4 A+7 C) \cos (2 c+3 d x)+2 (5 A+7 C) \cos (d x))-2\ 2^{2/3} (4 A+7 C) \cos ^{\frac{7}{3}}(c+d x) \sqrt [3]{\cos (c+d x) (\cos (c+d x)+i \sin (c+d x))} \, _2F_1\left (-\frac{1}{3},\frac{1}{3};\frac{2}{3};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )\right )-(4 A+7 C) \sin (c) \cos (c) \csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) e^{2 i d x} \cos ^{\frac{7}{3}}(c+d x) \sqrt [3]{2 i \sin (2 c) e^{2 i d x}+2 \cos (2 c) e^{2 i d x}+2} \, _2F_1\left (\frac{1}{3},\frac{2}{3};\frac{5}{3};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )\right )}{28 d (b \cos (c+d x))^{7/3} \sqrt [3]{e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )} (2 A+C \cos (2 (c+d x))+C)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(b*Cos[c + d*x])^(1/3),x]

[Out]

(3*b^2*(A + C*Cos[c + d*x]^2)*Csc[c]*Sec[c]*(-((4*A + 7*C)*E^((2*I)*d*x)*Cos[c]*Cos[c + d*x]^(7/3)*Csc[c/2]*Hy
pergeometric2F1[1/3, 2/3, 5/3, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*Sec[c/2]*Sin[c]*(2 + 2*E^((2*I)*d*x)*Co
s[2*c] + (2*I)*E^((2*I)*d*x)*Sin[2*c])^(1/3)) + 2*Cos[c]*(E^(I*d*x)*(2*(5*A + 7*C)*Cos[d*x] + (2*A + 7*C)*Cos[
2*c + d*x] + (4*A + 7*C)*Cos[2*c + 3*d*x])*(((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*
d*x))^(1/3) - 2*2^(2/3)*(4*A + 7*C)*Cos[c + d*x]^(7/3)*Hypergeometric2F1[-1/3, 1/3, 2/3, -(E^((2*I)*d*x)*(Cos[
c] + I*Sin[c])^2)]*(Cos[c + d*x]*(Cos[c + d*x] + I*Sin[c + d*x]))^(1/3))))/(28*d*E^(I*d*x)*(b*Cos[c + d*x])^(7
/3)*(2*A + C + C*Cos[2*(c + d*x)])*(((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x))^(1
/3))

________________________________________________________________________________________

Maple [F]  time = 0.379, size = 0, normalized size = 0. \begin{align*} \int{ \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}{\frac{1}{\sqrt [3]{b\cos \left ( dx+c \right ) }}}}\, 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)^3/(b*cos(d*x+c))^(1/3),x)

[Out]

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

________________________________________________________________________________________

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 )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac{1}{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)^3/(b*cos(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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{2}{3}} \sec \left (d x + c\right )^{3}}{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)^3/(b*cos(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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 )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac{1}{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)^3/(b*cos(d*x+c))^(1/3),x, algorithm="giac")

[Out]

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