3.356 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=149 \[ \frac {3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac {3 (4 A+7 C) \sin (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}+\frac {3 b B \sin (c+d x) \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {16, 3021, 2748, 2643} \[ \frac {3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac {3 (4 A+7 C) \sin (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )}{7 d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}+\frac {3 b B \sin (c+d x) \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{4/3}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(b*Cos[c + d*x])^(1/3),x]

[Out]

(3*A*b^2*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(7/3)) + (3*b*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]) + (3*(4*A + 7*C)*Hypergeometric2F1[-1/6, 1/2,
5/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(1/3)*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 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 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{\sqrt [3]{b \cos (c+d x)}} \, dx &=b^3 \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{10/3}} \, dx\\ &=\frac {3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac {3}{7} \int \frac {\frac {7 b^2 B}{3}+\frac {1}{3} b^2 (4 A+7 C) \cos (c+d x)}{(b \cos (c+d x))^{7/3}} \, dx\\ &=\frac {3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\left (b^2 B\right ) \int \frac {1}{(b \cos (c+d x))^{7/3}} \, dx+\frac {1}{7} (b (4 A+7 C)) \int \frac {1}{(b \cos (c+d x))^{4/3}} \, dx\\ &=\frac {3 A b^2 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac {3 b 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)}}+\frac {3 (4 A+7 C) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.27, size = 118, normalized size = 0.79 \[ \frac {3 b^2 \sqrt {\sin ^2(c+d x)} \csc (c+d x) \left (4 A \, _2F_1\left (-\frac {7}{6},\frac {1}{2};-\frac {1}{6};\cos ^2(c+d x)\right )+7 \cos (c+d x) \left (B \, _2F_1\left (-\frac {2}{3},\frac {1}{2};\frac {1}{3};\cos ^2(c+d x)\right )+4 C \cos (c+d x) \, _2F_1\left (-\frac {1}{6},\frac {1}{2};\frac {5}{6};\cos ^2(c+d x)\right )\right )\right )}{28 d (b \cos (c+d x))^{7/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3)/(b*Cos[c + d*x])^(1/3),x]

[Out]

(3*b^2*Csc[c + d*x]*(4*A*Hypergeometric2F1[-7/6, 1/2, -1/6, Cos[c + d*x]^2] + 7*Cos[c + d*x]*(B*Hypergeometric
2F1[-2/3, 1/2, 1/3, Cos[c + d*x]^2] + 4*C*Cos[c + d*x]*Hypergeometric2F1[-1/6, 1/2, 5/6, Cos[c + d*x]^2]))*Sqr
t[Sin[c + d*x]^2])/(28*d*(b*Cos[c + d*x])^(7/3))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {2}{3}} \sec \left (d x + c\right )^{3}}{b \cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(b*cos(d*x+c))^(1/3),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(b*cos(d*x+c))^(1/3),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^3/(b*cos(d*x + c))^(1/3), x)

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \frac {\left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{3}\left (d x +c \right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {1}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(b*cos(d*x+c))^(1/3),x)

[Out]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(b*cos(d*x+c))^(1/3),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\left (b \cos \left (d x + c\right )\right )^{\frac {1}{3}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3/(b*cos(d*x+c))^(1/3),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*sec(d*x + c)^3/(b*cos(d*x + c))^(1/3), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(b*cos(c + d*x))^(1/3)),x)

[Out]

int((A + B*cos(c + d*x) + C*cos(c + d*x)^2)/(cos(c + d*x)^3*(b*cos(c + d*x))^(1/3)), 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((A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**3/(b*cos(d*x+c))**(1/3),x)

[Out]

Timed out

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