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

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

[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])

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Rubi [A]  time = 0.148493, antiderivative size = 148, normalized size of antiderivative = 1., 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, 3023, 2748, 2643} \[ -\frac{3 (7 A+4 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )}{28 d \sqrt{\sin ^2(c+d x)}}-\frac{3 B \sin (c+d x) (b \cos (c+d x))^{7/3} \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right )}{7 b d \sqrt{\sin ^2(c+d x)}}+\frac{3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 d} \]

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 3023

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]

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 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 (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*}

Mathematica [A]  time = 0.224451, size = 109, normalized size = 0.74 \[ -\frac{3 b \sin (2 (c+d x)) \sqrt [3]{b \cos (c+d x)} \left ((7 A+4 C) \, _2F_1\left (\frac{1}{2},\frac{2}{3};\frac{5}{3};\cos ^2(c+d x)\right )+4 B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{7}{6};\frac{13}{6};\cos ^2(c+d x)\right )-4 C \sqrt{\sin ^2(c+d x)}\right )}{56 d \sqrt{\sin ^2(c+d x)}} \]

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 = 0.362, size = 0, normalized size = 0. \begin{align*} \int \left ( b\cos \left ( dx+c \right ) \right ) ^{{\frac{4}{3}}} \left ( A+B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sec \left ( dx+c \right ) \, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\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{4}{3}} \sec \left (d x + c\right )\,{d x} \end{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b \cos \left (d x + c\right )^{3} + B b \cos \left (d x + c\right )^{2} + A b \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{\frac{1}{3}} \sec \left (d x + c\right ), x\right ) \end{align*}

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)]  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((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., size = 0, normalized size = 0. \begin{align*} \int{\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{4}{3}} \sec \left (d x + c\right )\,{d x} \end{align*}

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)