3.4.62 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx\) [362]

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

[Out]

3/7*A*b*sin(d*x+c)/d/(b*cos(d*x+c))^(7/3)+3/4*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)/b/d/(b*c
os(d*x+c))^(1/3)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.13, 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, 3100, 2827, 2722} \begin {gather*} \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 b d \sqrt {\sin ^2(c+d x)} \sqrt [3]{b \cos (c+d x)}}+\frac {3 A b \sin (c+d x)}{7 d (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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*A*b*Sin[c + d*x])/(7*d*(b*Cos[c + d*x])^(7/3)) + (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]) + (3*(4*A + 7*C)*Hypergeometric2F1[-1/6, 1/2, 5/6
, Cos[c + d*x]^2]*Sin[c + d*x])/(7*b*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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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 3100

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 ^2(c+d x)}{(b \cos (c+d x))^{4/3}} \, dx &=b^2 \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 \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\frac {3 \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}{7 b}\\ &=\frac {3 A b \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+(b B) \int \frac {1}{(b \cos (c+d x))^{7/3}} \, dx+\frac {1}{7} (4 A+7 C) \int \frac {1}{(b \cos (c+d x))^{4/3}} \, dx\\ &=\frac {3 A b \sin (c+d x)}{7 d (b \cos (c+d x))^{7/3}}+\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)}}+\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 b d \sqrt [3]{b \cos (c+d x)} \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 118, normalized size = 0.79 \begin {gather*} \frac {3 b^2 \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 \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 ) \sqrt {\sin ^2(c+d x)}}{28 d (b \cos (c+d x))^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*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*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])^(10/3))

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Maple [F]
time = 0.35, size = 0, normalized size = 0.00 \[\int \frac {\left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{2}\left (d x +c \right )\right )}{\left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}}}\, dx\]

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)^2/(b*cos(d*x+c))^(4/3),x)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

[Out]

integrate((C*cos(d*x + c)^2 + 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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^2/(b*cos(d*x+c))^(4/3),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)^2/(b^2*cos(d*x + c)^2), x
)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)**2/(b*cos(d*x+c))**(4/3),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^2/(b*cos(d*x+c))^(4/3),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\cos \left (c+d\,x\right )}^2\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}} \,d x \end {gather*}

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)^2*(b*cos(c + d*x))^(4/3)),x)

[Out]

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

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