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

Optimal. Leaf size=91 \[ \frac{3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}-\frac{3 b^3 (A-2 C) \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{6},\frac{11}{6},\cos ^2(c+d x)\right )}{5 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}} \]

[Out]

(-3*b^3*(A - 2*C)*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*C*Tan[c + d*x])/(d*(b*Sec[c + d*x])^(2/3))

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Rubi [A]  time = 0.115196, antiderivative size = 91, 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, 4046, 3772, 2643} \[ \frac{3 b^2 C \tan (c+d x)}{d (b \sec (c+d x))^{2/3}}-\frac{3 b^3 (A-2 C) \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{5}{6};\frac{11}{6};\cos ^2(c+d x)\right )}{5 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b^3*(A - 2*C)*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*C*Tan[c + d*x])/(d*(b*Sec[c + d*x])^(2/3))

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 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]

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

Mathematica [A]  time = 0.130123, size = 90, normalized size = 0.99 \[ -\frac{3 \sqrt{-\tan ^2(c+d x)} \cot (c+d x) (b \sec (c+d x))^{4/3} \left (2 A \cos ^2(c+d x) \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\sec ^2(c+d x)\right )-C \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\sec ^2(c+d x)\right )\right )}{4 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.313, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{2} \left ( b\sec \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}} \left ( A+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)^2*(b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{3} + A b \cos \left (d x + c\right )^{2} \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)^2*(b*sec(d*x+c))^(4/3)*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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