3.4.47 \(\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [347]

Optimal. Leaf size=148 \[ \frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}-\frac {3 (7 A+4 C) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{28 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}} \]

[Out]

3/7*C*(b*cos(d*x+c))^(4/3)*sin(d*x+c)/d-3/28*(7*A+4*C)*(b*cos(d*x+c))^(4/3)*hypergeom([1/2, 2/3],[5/3],cos(d*x
+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)-3/7*B*(b*cos(d*x+c))^(7/3)*hypergeom([1/2, 7/6],[13/6],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 = 0.11, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 3102, 2827, 2722} \begin {gather*} -\frac {3 (7 A+4 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )}{28 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt {\sin ^2(c+d x)}}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*C*(b*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*d) - (3*(7*A + 4*C)*(b*Cos[c + d*x])^(4/3)*Hypergeometric2F1[1/2,
2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(28*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(7/3)*Hypergeomet
ric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(7*b*d*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 3102

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

Rubi steps

\begin {align*} \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=b \int \sqrt [3]{b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}+\frac {3}{7} \int \sqrt [3]{b \cos (c+d x)} \left (\frac {1}{3} b (7 A+4 C)+\frac {7}{3} b B \cos (c+d x)\right ) \, dx\\ &=\frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}+B \int (b \cos (c+d x))^{4/3} \, dx+\frac {1}{7} (b (7 A+4 C)) \int \sqrt [3]{b \cos (c+d x)} \, dx\\ &=\frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}-\frac {3 (7 A+4 C) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{28 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.26, size = 109, normalized size = 0.74 \begin {gather*} -\frac {3 b \sqrt [3]{b \cos (c+d x)} \left ((7 A+4 C) \, _2F_1\left (\frac {1}{2},\frac {2}{3};\frac {5}{3};\cos ^2(c+d x)\right )+4 B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {7}{6};\frac {13}{6};\cos ^2(c+d x)\right )-4 C \sqrt {\sin ^2(c+d x)}\right ) \sin (2 (c+d x))}{56 d \sqrt {\sin ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*(b*Cos[c + d*x])^(1/3)*((7*A + 4*C)*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2] + 4*B*Cos[c + d*x]*
Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2] - 4*C*Sqrt[Sin[c + d*x]^2])*Sin[2*(c + d*x)])/(56*d*Sqrt[Sin
[c + d*x]^2])

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Maple [F]
time = 0.26, size = 0, normalized size = 0.00 \[\int \left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((C*b*cos(d*x + c)^3 + B*b*cos(d*x + c)^2 + A*b*cos(d*x + c))*(b*cos(d*x + c))^(1/3)*sec(d*x + c), 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((b*cos(d*x+c))**(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c),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((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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