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

Optimal. Leaf size=154 \[ -\frac{3 b^2 (4 A+7 C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(c+d x)\right )}{28 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}-\frac{3 b^3 B \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{7}{6},\frac{13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}+\frac{3 A b^3 \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}} \]

[Out]

(-3*b^3*B*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin
[c + d*x]^2]) - (3*b^2*(4*A + 7*C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(28*d*(b*Sec
[c + d*x])^(4/3)*Sqrt[Sin[c + d*x]^2]) + (3*A*b^3*Tan[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3))

________________________________________________________________________________________

Rubi [A]  time = 0.192032, antiderivative size = 154, normalized size of antiderivative = 1., 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} \[ -\frac{3 b^2 (4 A+7 C) \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{28 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}}+\frac{3 A b^3 \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}-\frac{3 b^3 B \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^3*B*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3)*Sqrt[Sin
[c + d*x]^2]) - (3*b^2*(4*A + 7*C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(28*d*(b*Sec
[c + d*x])^(4/3)*Sqrt[Sin[c + d*x]^2]) + (3*A*b^3*Tan[c + d*x])/(7*d*(b*Sec[c + d*x])^(7/3))

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 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 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]

Rubi steps

\begin{align*} \int \cos ^3(c+d x) (b \sec (c+d x))^{2/3} \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))^{7/3}} \, dx\\ &=b^3 \int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{7/3}} \, dx+\left (b^2 B\right ) \int \frac{1}{(b \sec (c+d x))^{4/3}} \, dx\\ &=\frac{3 A b^3 \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}+\frac{1}{7} (b (4 A+7 C)) \int \frac{1}{\sqrt [3]{b \sec (c+d x)}} \, dx+\left (b^2 B \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \left (\frac{\cos (c+d x)}{b}\right )^{4/3} \, dx\\ &=-\frac{3 B \cos ^3(c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{7 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A b^3 \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}+\frac{1}{7} \left (b (4 A+7 C) \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \sqrt [3]{\frac{\cos (c+d x)}{b}} \, dx\\ &=-\frac{3 (4 A+7 C) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{28 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \cos ^3(c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{7 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A b^3 \tan (c+d x)}{7 d (b \sec (c+d x))^{7/3}}\\ \end{align*}

Mathematica [A]  time = 0.224726, size = 118, normalized size = 0.77 \[ -\frac{3 b \sqrt{-\tan ^2(c+d x)} \cot (c+d x) \left (4 A \cos ^2(c+d x) \text{Hypergeometric2F1}\left (-\frac{7}{6},\frac{1}{2},-\frac{1}{6},\sec ^2(c+d x)\right )+7 B \cos (c+d x) \text{Hypergeometric2F1}\left (-\frac{2}{3},\frac{1}{2},\frac{1}{3},\sec ^2(c+d x)\right )+28 C \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{2},\frac{5}{6},\sec ^2(c+d x)\right )\right )}{28 d \sqrt [3]{b \sec (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-3*b*Cot[c + d*x]*(4*A*Cos[c + d*x]^2*Hypergeometric2F1[-7/6, 1/2, -1/6, Sec[c + d*x]^2] + 7*B*Cos[c + d*x]*H
ypergeometric2F1[-2/3, 1/2, 1/3, Sec[c + d*x]^2] + 28*C*Hypergeometric2F1[-1/6, 1/2, 5/6, Sec[c + d*x]^2])*Sqr
t[-Tan[c + d*x]^2])/(28*d*(b*Sec[c + d*x])^(1/3))

________________________________________________________________________________________

Maple [F]  time = 0.638, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{3} \left ( b\sec \left ( dx+c \right ) \right ) ^{{\frac{2}{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)^3*(b*sec(d*x+c))^(2/3)*(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))^(2/3)*(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{2}{3}} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^3*sec(d*x + c)^2 + B*cos(d*x + c)^3*sec(d*x + c) + A*cos(d*x + c)^3)*(b*sec(d*x + c))
^(2/3), 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)**3*(b*sec(d*x+c))**(2/3)*(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{2}{3}} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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