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

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

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 2748, 2643} \[ \frac {3 A \sin (c+d x) \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac {3 B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rubi steps

\begin {align*} \int \frac {(A+B \cos (c+d x)) \sec (c+d x)}{(b \cos (c+d x))^{2/3}} \, dx &=b \int \frac {A+B \cos (c+d x)}{(b \cos (c+d x))^{5/3}} \, dx\\ &=(A b) \int \frac {1}{(b \cos (c+d x))^{5/3}} \, dx+B \int \frac {1}{(b \cos (c+d x))^{2/3}} \, dx\\ &=\frac {3 A \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{b d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 85, normalized size = 0.75 \[ \frac {3 \sqrt {\sin ^2(c+d x)} \csc (c+d x) \left (A \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )-2 B \cos (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )\right )}{2 d (b \cos (c+d x))^{2/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}} \sec \left (d x + c\right )}{b \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^(1/3)*sec(d*x + c)/(b*cos(d*x + c)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.20, size = 0, normalized size = 0.00 \[ \int \frac {\left (A +B \cos \left (d x +c \right )\right ) \sec \left (d x +c \right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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