3.52 \(\int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=154 \[ \frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}-\frac {3 b^2 (2 A+5 C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}-\frac {3 b^3 B \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}} \]

[Out]

-3/5*b^3*B*hypergeom([1/2, 5/6],[11/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(5/3)/(sin(d*x+c)^2)^(1/2)-3/
10*b^2*(2*A+5*C)*hypergeom([1/3, 1/2],[4/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(2/3)/(sin(d*x+c)^2)^(1/
2)+3/5*A*b^3*tan(d*x+c)/d/(b*sec(d*x+c))^(5/3)

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {16, 4047, 3772, 2643, 4045} \[ -\frac {3 b^2 (2 A+5 C) \sin (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right )}{10 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}-\frac {3 b^3 B \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right )}{5 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(-3*b^3*B*Hypergeometric2F1[1/2, 5/6, 11/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3)*Sqrt[Sin
[c + d*x]^2]) - (3*b^2*(2*A + 5*C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(10*d*(b*Sec
[c + d*x])^(2/3)*Sqrt[Sin[c + d*x]^2]) + (3*A*b^3*Tan[c + d*x])/(5*d*(b*Sec[c + d*x])^(5/3))

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 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rubi steps

\begin {align*} \int \cos ^3(c+d x) (b \sec (c+d x))^{4/3} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=b^3 \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/3}} \, dx\\ &=b^3 \int \frac {A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{5/3}} \, dx+\left (b^2 B\right ) \int \frac {1}{(b \sec (c+d x))^{2/3}} \, dx\\ &=\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}+\frac {1}{5} (b (2 A+5 C)) \int \sqrt [3]{b \sec (c+d x)} \, dx+\left (b^2 B \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \left (\frac {\cos (c+d x)}{b}\right )^{2/3} \, dx\\ &=-\frac {3 b B \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}+\frac {1}{5} \left (b (2 A+5 C) \sqrt [3]{\frac {\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac {1}{\sqrt [3]{\frac {\cos (c+d x)}{b}}} \, dx\\ &=-\frac {3 b (2 A+5 C) \cos (c+d x) \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{10 d \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {5}{6};\frac {11}{6};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}}+\frac {3 A b^3 \tan (c+d x)}{5 d (b \sec (c+d x))^{5/3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.19, size = 118, normalized size = 0.77 \[ -\frac {3 b \sqrt {-\tan ^2(c+d x)} \cot (c+d x) \sqrt [3]{b \sec (c+d x)} \left (2 A \cos ^2(c+d x) \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {1}{6};\sec ^2(c+d x)\right )+5 B \cos (c+d x) \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\sec ^2(c+d x)\right )-10 C \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\sec ^2(c+d x)\right )\right )}{10 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(b*Sec[c + d*x])^(4/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(-3*b*Cot[c + d*x]*(2*A*Cos[c + d*x]^2*Hypergeometric2F1[-5/6, 1/2, 1/6, Sec[c + d*x]^2] + 5*B*Cos[c + d*x]*Hy
pergeometric2F1[-1/3, 1/2, 2/3, Sec[c + d*x]^2] - 10*C*Hypergeometric2F1[1/6, 1/2, 7/6, Sec[c + d*x]^2])*(b*Se
c[c + d*x])^(1/3)*Sqrt[-Tan[c + d*x]^2])/(10*d)

________________________________________________________________________________________

fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} + B b \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + A b \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {1}{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*cos(d*x + c)^3, x)

________________________________________________________________________________________

maple [F]  time = 5.27, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{3}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \sec \left (d x +c \right )+C \left (\sec ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

int(cos(d*x+c)^3*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(b*sec(d*x+c))^(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(b*sec(d*x + c))^(4/3)*cos(d*x + c)^3, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^3\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int(cos(c + d*x)^3*(b/cos(c + d*x))^(4/3)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)

________________________________________________________________________________________

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(cos(d*x+c)**3*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________