3.4.50 \(\int (b \cos (c+d x))^{4/3} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [350]

Optimal. Leaf size=152 \[ \frac {3 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}+\frac {3 b^2 B \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b (2 A+5 C) \sqrt [3]{b \cos (c+d x)} \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 152, 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 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}-\frac {3 b (2 A+5 C) \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 )}{5 d \sqrt {\sin ^2(c+d x)}}+\frac {3 b^2 B \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}} \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]^4,x]

[Out]

(3*A*b^3*Sin[c + d*x])/(5*d*(b*Cos[c + d*x])^(5/3)) + (3*b^2*B*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*(2*A + 5*C)*(b*Cos[c + d*x])^(1/3)*H
ypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(5*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 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 (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx &=b^4 \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(b \cos (c+d x))^{8/3}} \, dx\\ &=\frac {3 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}+\frac {1}{5} (3 b) \int \frac {\frac {5 b^2 B}{3}+\frac {1}{3} b^2 (2 A+5 C) \cos (c+d x)}{(b \cos (c+d x))^{5/3}} \, dx\\ &=\frac {3 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}+\left (b^3 B\right ) \int \frac {1}{(b \cos (c+d x))^{5/3}} \, dx+\frac {1}{5} \left (b^2 (2 A+5 C)\right ) \int \frac {1}{(b \cos (c+d x))^{2/3}} \, dx\\ &=\frac {3 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}+\frac {3 b^2 B \, _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 (2 A+5 C) \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)}{5 d \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 124, normalized size = 0.82 \begin {gather*} -\frac {3 (b \cos (c+d x))^{4/3} \csc (c+d x) \left (-2 A \, _2F_1\left (-\frac {5}{6},\frac {1}{2};\frac {1}{6};\cos ^2(c+d x)\right )+5 \cos (c+d x) \left (-B \, _2F_1\left (-\frac {1}{3},\frac {1}{2};\frac {2}{3};\cos ^2(c+d x)\right )+2 C \cos (c+d x) \, _2F_1\left (\frac {1}{6},\frac {1}{2};\frac {7}{6};\cos ^2(c+d x)\right )\right )\right ) \sec ^3(c+d x) \sqrt {\sin ^2(c+d x)}}{10 d} \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]^4,x]

[Out]

(-3*(b*Cos[c + d*x])^(4/3)*Csc[c + d*x]*(-2*A*Hypergeometric2F1[-5/6, 1/2, 1/6, Cos[c + d*x]^2] + 5*Cos[c + d*
x]*(-(B*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^2]) + 2*C*Cos[c + d*x]*Hypergeometric2F1[1/6, 1/2, 7/6,
Cos[c + d*x]^2]))*Sec[c + d*x]^3*Sqrt[Sin[c + d*x]^2])/(10*d)

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Maple [F]
time = 0.30, size = 0, normalized size = 0.00 \[\int \left (b \cos \left (d x +c \right )\right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{4}\left (d x +c \right )\right )\, dx\]

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)^4,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)^4,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)^4,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)^4, 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)^4,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)^4, 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)**4,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)^4,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)^4, 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 )}^4} \,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)^4,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)^4, x)

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