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

Optimal. Leaf size=114 \[ \frac{3 A b \sin (c+d x) \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{7/3}}+\frac{3 B \sin (c+d x) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

[Out]

(3*A*b*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(7/3)*Sqrt[Sin[c
+ d*x]^2]) + (3*B*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*d*(b*Cos[c + d*x])^(4/3)
*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.0971097, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {16, 2748, 2643} \[ \frac{3 A b \sin (c+d x) \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right )}{7 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{7/3}}+\frac{3 B \sin (c+d x) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )}{4 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*b*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(7/3)*Sqrt[Sin[c
+ d*x]^2]) + (3*B*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*d*(b*Cos[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 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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{(A+B \cos (c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx &=b^2 \int \frac{A+B \cos (c+d x)}{(b \cos (c+d x))^{10/3}} \, dx\\ &=\left (A b^2\right ) \int \frac{1}{(b \cos (c+d x))^{10/3}} \, dx+(b B) \int \frac{1}{(b \cos (c+d x))^{7/3}} \, dx\\ &=\frac{3 A b \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3} \sqrt{\sin ^2(c+d x)}}+\frac{3 B \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{4 d (b \cos (c+d x))^{4/3} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.180191, size = 89, normalized size = 0.78 \[ \frac{3 b^2 \sqrt{\sin ^2(c+d x)} \cot (c+d x) \left (4 A \, _2F_1\left (-\frac{7}{6},\frac{1}{2};-\frac{1}{6};\cos ^2(c+d x)\right )+7 B \cos (c+d x) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )\right )}{28 d (b \cos (c+d x))^{10/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*Cot[c + d*x]*(4*A*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2] + 7*B*Cos[c + d*x]*Hypergeometric2
F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(28*d*(b*Cos[c + d*x])^(10/3))

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Maple [F]  time = 0.25, size = 0, normalized size = 0. \begin{align*} \int{ \left ( A+B\cos \left ( dx+c \right ) \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2} \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^2/(b*cos(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 (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{2}{3}} \sec \left (d x + c\right )^{2}}{b^{2} \cos \left (d x + c\right )^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(2/3)*sec(d*x + c)^2/(b^2*cos(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((A+B*cos(d*x+c))*sec(d*x+c)**2/(b*cos(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 (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{2}}{\left (b \cos \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((A+B*cos(d*x+c))*sec(d*x+c)^2/(b*cos(d*x+c))^(4/3),x, algorithm="giac")

[Out]

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