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

Optimal. Leaf size=93 \[ \frac{3 A b^2 \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}-\frac{3 b (7 A+10 C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{7}{6},\frac{13}{6},\cos ^2(c+d x)\right )}{70 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}} \]

[Out]

(-3*b*(7*A + 10*C)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(70*d*(b*Sec[c + d*x])^(7/3
)*Sqrt[Sin[c + d*x]^2]) + (3*A*b^2*Tan[c + d*x])/(10*d*(b*Sec[c + d*x])^(10/3))

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Rubi [A]  time = 0.109126, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {16, 4045, 3772, 2643} \[ \frac{3 A b^2 \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}-\frac{3 b (7 A+10 C) \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right )}{70 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(7*A + 10*C)*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(70*d*(b*Sec[c + d*x])^(7/3
)*Sqrt[Sin[c + d*x]^2]) + (3*A*b^2*Tan[c + d*x])/(10*d*(b*Sec[c + d*x])^(10/3))

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(b \sec (c+d x))^{4/3}} \, dx &=b^2 \int \frac{A+C \sec ^2(c+d x)}{(b \sec (c+d x))^{10/3}} \, dx\\ &=\frac{3 A b^2 \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}+\frac{1}{10} (7 A+10 C) \int \frac{1}{(b \sec (c+d x))^{4/3}} \, dx\\ &=\frac{3 A b^2 \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}+\frac{1}{10} \left ((7 A+10 C) \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 (7 A+10 C) \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)}{70 b^2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 A b^2 \tan (c+d x)}{10 d (b \sec (c+d x))^{10/3}}\\ \end{align*}

Mathematica [A]  time = 0.67149, size = 96, normalized size = 1.03 \[ \frac{\tan (c+d x) \left ((7 A+10 C) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{6},\frac{3}{2},\sin ^2(c+d x)\right )+3 \sqrt [6]{\cos ^2(c+d x)} (2 A \cos (2 (c+d x))+9 A+10 C)\right )}{40 d \sqrt [6]{\cos ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*(Cos[c + d*x]^2)^(1/6)*(9*A + 10*C + 2*A*Cos[2*(c + d*x)]) + (7*A + 10*C)*Hypergeometric2F1[1/2, 5/6, 3/2,
Sin[c + d*x]^2])*Tan[c + d*x])/(40*d*(Cos[c + d*x]^2)^(1/6)*(b*Sec[c + d*x])^(4/3))

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Maple [F]  time = 0.305, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( A+C \left ( \sec \left ( dx+c \right ) \right ) ^{2} \right ) \left ( b\sec \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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