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

Optimal. Leaf size=116 \[ \frac{3 A \sin (c+d x) (b \sec (c+d x))^{2/3} \text{Hypergeometric2F1}\left (-\frac{1}{3},\frac{1}{2},\frac{2}{3},\cos ^2(c+d x)\right )}{2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{5/3} \text{Hypergeometric2F1}\left (-\frac{5}{6},\frac{1}{2},\frac{1}{6},\cos ^2(c+d x)\right )}{5 b d \sqrt{\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.0984483, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {16, 3787, 3772, 2643} \[ \frac{3 A \sin (c+d x) (b \sec (c+d x))^{2/3} \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right )}{2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \sin (c+d x) (b \sec (c+d x))^{5/3} \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right )}{5 b d \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(2/3)*Sin[c + d*x])/(2*d*Sqrt[Sin[c +
d*x]^2]) + (3*B*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2]*(b*Sec[c + d*x])^(5/3)*Sin[c + d*x])/(5*b*d*
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 \sec (c+d x) (b \sec (c+d x))^{2/3} (A+B \sec (c+d x)) \, dx &=\frac{\int (b \sec (c+d x))^{5/3} (A+B \sec (c+d x)) \, dx}{b}\\ &=\frac{A \int (b \sec (c+d x))^{5/3} \, dx}{b}+\frac{B \int (b \sec (c+d x))^{8/3} \, dx}{b^2}\\ &=\frac{\left (A \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{5/3}} \, dx}{b}+\frac{\left (B \left (\frac{\cos (c+d x)}{b}\right )^{2/3} (b \sec (c+d x))^{2/3}\right ) \int \frac{1}{\left (\frac{\cos (c+d x)}{b}\right )^{8/3}} \, dx}{b^2}\\ &=\frac{3 A \, _2F_1\left (-\frac{1}{3},\frac{1}{2};\frac{2}{3};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{2/3} \sin (c+d x)}{2 d \sqrt{\sin ^2(c+d x)}}+\frac{3 B \, _2F_1\left (-\frac{5}{6},\frac{1}{2};\frac{1}{6};\cos ^2(c+d x)\right ) (b \sec (c+d x))^{5/3} \sin (c+d x)}{5 b d \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.117802, size = 91, normalized size = 0.78 \[ \frac{3 \sqrt{-\tan ^2(c+d x)} \csc (c+d x) (b \sec (c+d x))^{5/3} \left (8 A \cos (c+d x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{5}{6},\frac{11}{6},\sec ^2(c+d x)\right )+5 B \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{4}{3},\frac{7}{3},\sec ^2(c+d x)\right )\right )}{40 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Csc[c + d*x]*(8*A*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/6, 11/6, Sec[c + d*x]^2] + 5*B*Hypergeometric2F1[1/
2, 4/3, 7/3, Sec[c + d*x]^2])*(b*Sec[c + d*x])^(5/3)*Sqrt[-Tan[c + d*x]^2])/(40*b*d)

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Maple [F]  time = 0.105, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \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(sec(d*x+c)*(b*sec(d*x+c))^(2/3)*(A+B*sec(d*x+c)),x)

[Out]

int(sec(d*x+c)*(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}} \sec \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)*sec(d*x + c), x)

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(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)*sec(d*x + c), x)