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 \[ -\frac {3 A b^3 \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{7/3}}-\frac {3 b^2 B \sin (c+d x) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \sec (c+d x))^{4/3}} \]

[Out]

-3/7*A*b^3*hypergeom([1/2, 7/6],[13/6],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(7/3)/(sin(d*x+c)^2)^(1/2)-3/
4*b^2*B*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*sin(d*x+c)/d/(b*sec(d*x+c))^(4/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 119, normalized size of antiderivative = 1.00, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.14, size = 88, normalized size = 0.74 \[ -\frac {3 b \sqrt {-\tan ^2(c+d x)} \cot (c+d x) \left (A \cos (c+d x) \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\sec ^2(c+d x)\right )+4 B \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\sec ^2(c+d x)\right )\right )}{4 d \sqrt [3]{b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{2}\,{d x} \]

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)

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maple [F]  time = 2.66, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (d x +c \right )\right ) \left (b \sec \left (d x +c \right )\right )^{\frac {2}{3}} \left (A +B \sec \left (d x +c \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} \left (b \sec \left (d x + c\right )\right )^{\frac {2}{3}} \cos \left (d x + c\right )^{2}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{2/3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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