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

Optimal. Leaf size=92 \[ \frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {3 (5 A+8 C) \sin (c+d x) \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )}{16 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}} \]

[Out]

3/8*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(8/3)+3/16*(5*A+8*C)*hypergeom([-1/3, 1/2],[2/3],cos(d*x+c)^2)*sin(d*x+c
)/d/(b*cos(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.10, antiderivative size = 92, normalized size of antiderivative = 1.00, 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} \[ \frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {3 (5 A+8 C) \sin (c+d x) \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )}{16 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*b^2*Sin[c + d*x])/(8*d*(b*Cos[c + d*x])^(8/3)) + (3*(5*A + 8*C)*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c +
d*x]^2]*Sin[c + d*x])/(16*d*(b*Cos[c + d*x])^(2/3)*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 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 ^3(c+d x)}{(b \cos (c+d x))^{2/3}} \, dx &=b^3 \int \frac {A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{11/3}} \, dx\\ &=\frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {1}{8} (b (5 A+8 C)) \int \frac {1}{(b \cos (c+d x))^{5/3}} \, dx\\ &=\frac {3 A b^2 \sin (c+d x)}{8 d (b \cos (c+d x))^{8/3}}+\frac {3 (5 A+8 C) \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{16 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 6.29, size = 473, normalized size = 5.14 \[ b \left (\frac {\cos ^4(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (\frac {3 \sec (c) \sec (c+d x) (5 A \sin (d x)+8 C \sin (d x))}{8 d}+\frac {3 (5 A+8 C) \csc (c) \sec (c)}{8 d}+\frac {3 A \sec (c) \sin (d x) \sec ^3(c+d x)}{4 d}+\frac {3 A \tan (c) \sec ^2(c+d x)}{4 d}\right )}{(b \cos (c+d x))^{5/3} (2 A+C \cos (2 c+2 d x)+C)}-\frac {i (5 A+8 C) \csc \left (\frac {c}{2}\right ) \sec \left (\frac {c}{2}\right ) \cos ^{\frac {11}{3}}(c+d x) \left (A \sec ^2(c+d x)+C\right ) \left (-\frac {3 i e^{-i d x} \left (2 i \sin (2 c) e^{2 i d x}+2 \cos (2 c) e^{2 i d x}+2\right )^{2/3} \, _2F_1\left (-\frac {1}{6},\frac {2}{3};\frac {5}{6};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{d \left (e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )\right )^{2/3}}-\frac {3 i e^{i d x} \left (2 i \sin (2 c) e^{2 i d x}+2 \cos (2 c) e^{2 i d x}+2\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {5}{6};\frac {11}{6};-e^{2 i d x} (\cos (c)+i \sin (c))^2\right )}{5 d \left (e^{-i d x} \left (i \sin (c) \left (-1+e^{2 i d x}\right )+\cos (c) \left (1+e^{2 i d x}\right )\right )\right )^{2/3}}\right )}{32 (b \cos (c+d x))^{5/3} (2 A+C \cos (2 c+2 d x)+C)}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

b*(((-1/32*I)*(5*A + 8*C)*Cos[c + d*x]^(11/3)*Csc[c/2]*Sec[c/2]*(C + A*Sec[c + d*x]^2)*(((-3*I)*Hypergeometric
2F1[-1/6, 2/3, 5/6, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*(2 + 2*E^((2*I)*d*x)*Cos[2*c] + (2*I)*E^((2*I)*d*x
)*Sin[2*c])^(2/3))/(d*E^(I*d*x)*(((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 + E^((2*I)*d*x))*Sin[c])/E^(I*d*x))^(2/3)
) - (((3*I)/5)*E^(I*d*x)*Hypergeometric2F1[2/3, 5/6, 11/6, -(E^((2*I)*d*x)*(Cos[c] + I*Sin[c])^2)]*(2 + 2*E^((
2*I)*d*x)*Cos[2*c] + (2*I)*E^((2*I)*d*x)*Sin[2*c])^(2/3))/(d*(((1 + E^((2*I)*d*x))*Cos[c] + I*(-1 + E^((2*I)*d
*x))*Sin[c])/E^(I*d*x))^(2/3))))/((b*Cos[c + d*x])^(5/3)*(2*A + C + C*Cos[2*c + 2*d*x])) + (Cos[c + d*x]^4*(C
+ A*Sec[c + d*x]^2)*((3*(5*A + 8*C)*Csc[c]*Sec[c])/(8*d) + (3*A*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(4*d) + (3*Sec
[c]*Sec[c + d*x]*(5*A*Sin[d*x] + 8*C*Sin[d*x]))/(8*d) + (3*A*Sec[c + d*x]^2*Tan[c])/(4*d)))/((b*Cos[c + d*x])^
(5/3)*(2*A + C + C*Cos[2*c + 2*d*x])))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )^{3}}{b \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(2/3),x)

[Out]

Timed out

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