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

Optimal. Leaf size=119 \[ -\frac{3 A b^3 \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{7}{6},\frac{13}{6},\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac{3 b^2 B \sin (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{2}{3},\frac{5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

[Out]

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

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Rubi [A]  time = 0.121113, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {16, 3787, 3772, 2643} \[ -\frac{3 A b^3 \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac{3 b^2 B \sin (c+d x) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, 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]

Rubi steps

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

Mathematica [A]  time = 0.137663, size = 88, normalized size = 0.74 \[ -\frac{3 b \sqrt{-\tan ^2(c+d x)} \cot (c+d x) \left (A \cos (c+d x) \text{Hypergeometric2F1}\left (-\frac{2}{3},\frac{1}{2},\frac{1}{3},\sec ^2(c+d x)\right )+4 B \text{Hypergeometric2F1}\left (-\frac{1}{6},\frac{1}{2},\frac{5}{6},\sec ^2(c+d x)\right )\right )}{4 d \sqrt [3]{b \sec (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]  time = 0.358, 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{2}{3}}} \left ( A+B\sec \left ( dx+c \right ) \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))^(2/3)*(A+B*sec(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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 )^{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))^(2/3)*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c))^(2/3)*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 (B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\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)^2*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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 )^{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))^(2/3)*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

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