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

Optimal. Leaf size=146 \[ -\frac{3 b^2 (4 A+C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{3},\frac{1}{2},\frac{4}{3},\cos ^2(c+d x)\right )}{8 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac{3 b B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{2},\frac{5}{6},\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)}}+\frac{3 b C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 d} \]

[Out]

(-3*b^2*(4*A + C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*(b*Sec[c + d*x])^(2/3)*S
qrt[Sin[c + d*x]^2]) + (3*b*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1/3)*Sin[c +
d*x])/(d*Sqrt[Sin[c + d*x]^2]) + (3*b*C*(b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*d)

________________________________________________________________________________________

Rubi [A]  time = 0.158827, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {16, 4047, 3772, 2643, 4046} \[ -\frac{3 b^2 (4 A+C) \sin (c+d x) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};\cos ^2(c+d x)\right )}{8 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{2/3}}+\frac{3 b B \sin (c+d x) \sqrt [3]{b \sec (c+d x)} \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)}}+\frac{3 b C \tan (c+d x) \sqrt [3]{b \sec (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^2*(4*A + C)*Hypergeometric2F1[1/3, 1/2, 4/3, Cos[c + d*x]^2]*Sin[c + d*x])/(8*d*(b*Sec[c + d*x])^(2/3)*S
qrt[Sin[c + d*x]^2]) + (3*b*B*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(1/3)*Sin[c +
d*x])/(d*Sqrt[Sin[c + d*x]^2]) + (3*b*C*(b*Sec[c + d*x])^(1/3)*Tan[c + d*x])/(4*d)

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 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]

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 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 4046

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

Rubi steps

\begin{align*} \int \cos (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 \int \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\\ &=b \int \sqrt [3]{b \sec (c+d x)} \left (A+C \sec ^2(c+d x)\right ) \, dx+B \int (b \sec (c+d x))^{4/3} \, dx\\ &=\frac{3 b C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d}+\frac{1}{4} (b (4 A+C)) \int \sqrt [3]{b \sec (c+d x)} \, dx+\left (B \sqrt [3]{\frac{\cos (c+d x)}{b}} \sqrt [3]{b \sec (c+d x)}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{4/3}} \, dx\\ &=\frac{3 b B \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d \sqrt{\sin ^2(c+d x)}}+\frac{3 b C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d}+\frac{1}{4} \left (b (4 A+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 B \, _2F_1\left (-\frac{1}{6},\frac{1}{2};\frac{5}{6};\cos ^2(c+d x)\right ) \sqrt [3]{b \sec (c+d x)} \sin (c+d x)}{d \sqrt{\sin ^2(c+d x)}}-\frac{3 b (4 A+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)}{8 d \sqrt{\sin ^2(c+d x)}}+\frac{3 b C \sqrt [3]{b \sec (c+d x)} \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [C]  time = 2.35104, size = 303, normalized size = 2.08 \[ \frac{3 b \sqrt [3]{b \sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\sqrt [3]{\sec (c+d x)} (4 B \csc (c) \cos (d x)+C \tan (c+d x))-\frac{i \sqrt [3]{2} e^{-i (c+d x)} \sqrt [3]{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\left (-1+e^{2 i c}\right ) (4 A+C) e^{i (c+d x)} \sqrt [3]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{6},\frac{1}{3},\frac{7}{6},-e^{2 i (c+d x)}\right )+4 B \left (-1+e^{2 i c}\right ) \sqrt [3]{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{3},\frac{2}{3},-e^{2 i (c+d x)}\right )+4 B \left (1+e^{2 i (c+d x)}\right )\right )}{-1+e^{2 i c}}\right )}{2 d \sec ^{\frac{7}{3}}(c+d x) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(b*Sec[c + d*x])^(1/3)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(((-I)*2^(1/3)*(E^(I*(c + d*x))/(1 + E^((2
*I)*(c + d*x))))^(1/3)*(4*B*(1 + E^((2*I)*(c + d*x))) + 4*B*(-1 + E^((2*I)*c))*(1 + E^((2*I)*(c + d*x)))^(1/3)
*Hypergeometric2F1[-1/3, 1/3, 2/3, -E^((2*I)*(c + d*x))] + (4*A + C)*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*(1 + E
^((2*I)*(c + d*x)))^(1/3)*Hypergeometric2F1[1/6, 1/3, 7/6, -E^((2*I)*(c + d*x))]))/(E^(I*(c + d*x))*(-1 + E^((
2*I)*c))) + Sec[c + d*x]^(1/3)*(4*B*Cos[d*x]*Csc[c] + C*Tan[c + d*x])))/(2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*C
os[2*(c + d*x)])*Sec[c + d*x]^(7/3))

________________________________________________________________________________________

Maple [F]  time = 0.24, size = 0, normalized size = 0. \begin{align*} \int \cos \left ( dx+c \right ) \left ( b\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}} \left ( A+B\sec \left ( dx+c \right ) +C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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)*sec(d*x + c)^3 + B*b*cos(d*x + c)*sec(d*x + c)^2 + A*b*cos(d*x + c)*sec(d*x + c))*(
b*sec(d*x + c))^(1/3), 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(cos(d*x+c)*(b*sec(d*x+c))**(4/3)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(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), x)