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

Optimal. Leaf size=90 \[ \frac {3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}} \]

[Out]

3*A*sin(d*x+c)/d/(b*cos(d*x+c))^(1/3)+3/5*(2*A-C)*(b*cos(d*x+c))^(5/3)*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^
2)*sin(d*x+c)/b^2/d/(sin(d*x+c)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 90, normalized size of antiderivative = 1.00, 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} \[ \frac {3 (2 A-C) \sin (c+d x) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right )}{5 b^2 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

\begin {align*} \int \frac {\left (A+C \cos ^2(c+d x)\right ) \sec (c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx &=b \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\\ &=\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}-\frac {(2 A-C) \int (b \cos (c+d x))^{2/3} \, dx}{b}\\ &=\frac {3 A \sin (c+d x)}{d \sqrt [3]{b \cos (c+d x)}}+\frac {3 (2 A-C) (b \cos (c+d x))^{5/3} \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 b^2 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 3.87, size = 283, normalized size = 3.14 \[ -\frac {3 \csc (c) e^{-i d x} \sqrt [3]{\cos (c+d x)} (\cos (d x)+i \sin (d x)) \left (2 (2 A-C) (\cos (d x)-i \sin (d x)) \sqrt [3]{i \sin (2 (c+d x))+\cos (2 (c+d x))+1} \, _2F_1\left (-\frac {1}{3},\frac {1}{3};\frac {2}{3};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )+(2 A-C) (\cos (d x)+i \sin (d x)) \sqrt [3]{i \sin (2 (c+d x))+\cos (2 (c+d x))+1} \, _2F_1\left (\frac {1}{3},\frac {2}{3};\frac {5}{3};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )-8 A \cos (d x)+2 C \cos (2 c+d x)+2 C \cos (d x)\right )}{4\ 2^{2/3} d \sqrt [3]{b \cos (c+d x)} \sqrt [3]{e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )}{b \cos \left (d x + c\right )}, x\right ) \]

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))^(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)/(b*cos(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}}\,{d x} \]

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))^(1/3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

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))^(1/3),x)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}}\,{d x} \]

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))^(1/3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\cos \left (c+d\,x\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{1/3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \cos ^{2}{\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\sqrt [3]{b \cos {\left (c + d x \right )}}}\, dx \]

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))**(1/3),x)

[Out]

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

________________________________________________________________________________________