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

Optimal. Leaf size=92 \[ \frac{3 A b^2 \sin (c+d x)}{10 d (b \cos (c+d x))^{10/3}}+\frac{3 (7 A+10 C) \sin (c+d x) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )}{40 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

[Out]

(3*A*b^2*Sin[c + d*x])/(10*d*(b*Cos[c + d*x])^(10/3)) + (3*(7*A + 10*C)*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[
c + d*x]^2]*Sin[c + d*x])/(40*d*(b*Cos[c + d*x])^(4/3)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.104965, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {16, 3012, 2643} \[ \frac{3 A b^2 \sin (c+d x)}{10 d (b \cos (c+d x))^{10/3}}+\frac{3 (7 A+10 C) \sin (c+d x) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )}{40 d \sqrt{\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*b^2*Sin[c + d*x])/(10*d*(b*Cos[c + d*x])^(10/3)) + (3*(7*A + 10*C)*Hypergeometric2F1[-2/3, 1/2, 1/3, Cos[
c + d*x]^2]*Sin[c + d*x])/(40*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 3012

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*Cos[e
+ f*x]*(b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1)), Int[(b*Sin[e
+ f*x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]

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{\left (A+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx &=b^3 \int \frac{A+C \cos ^2(c+d x)}{(b \cos (c+d x))^{13/3}} \, dx\\ &=\frac{3 A b^2 \sin (c+d x)}{10 d (b \cos (c+d x))^{10/3}}+\frac{1}{10} (b (7 A+10 C)) \int \frac{1}{(b \cos (c+d x))^{7/3}} \, dx\\ &=\frac{3 A b^2 \sin (c+d x)}{10 d (b \cos (c+d x))^{10/3}}+\frac{3 (7 A+10 C) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{40 d (b \cos (c+d x))^{4/3} \sqrt{\sin ^2(c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.209038, size = 91, normalized size = 0.99 \[ \frac{3 b^2 \sqrt{\sin ^2(c+d x)} \csc (c+d x) \left (2 A \, _2F_1\left (-\frac{5}{3},\frac{1}{2};-\frac{2}{3};\cos ^2(c+d x)\right )+5 C \cos ^2(c+d x) \, _2F_1\left (-\frac{2}{3},\frac{1}{2};\frac{1}{3};\cos ^2(c+d x)\right )\right )}{20 d (b \cos (c+d x))^{10/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b^2*Csc[c + d*x]*(2*A*Hypergeometric2F1[-5/3, 1/2, -2/3, Cos[c + d*x]^2] + 5*C*Cos[c + d*x]^2*Hypergeometri
c2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2])*Sqrt[Sin[c + d*x]^2])/(20*d*(b*Cos[c + d*x])^(10/3))

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Maple [F]  time = 0.393, size = 0, normalized size = 0. \begin{align*} \int{ \left ( A+C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3} \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+C*cos(d*x+c)^2)*sec(d*x+c)^3/(b*cos(d*x+c))^(4/3),x)

[Out]

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

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

[Out]

Timed out

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

[Out]

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

[Out]

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