Integrand size = 39, antiderivative size = 148 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {3 C (b \cos (c+d x))^{4/3} \sin (c+d x)}{7 d}-\frac {3 (7 A+4 C) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{28 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{7 b d \sqrt {\sin ^2(c+d x)}} \]
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)*si n(d*x+c)/b/d/(sin(d*x+c)^2)^(1/2)
Time = 0.42 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {3 b \sqrt [3]{b \cos (c+d x)} \left ((7 A+4 C) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}+4 B \cos (c+d x) \cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}-2 C \sin (2 (c+d x))\right )}{28 d} \]
(-3*b*(b*Cos[c + d*x])^(1/3)*((7*A + 4*C)*Cot[c + d*x]*Hypergeometric2F1[1 /2, 2/3, 5/3, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] + 4*B*Cos[c + d*x]*Cot[ c + d*x]*Hypergeometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sqrt[Sin[c + d* x]^2] - 2*C*Sin[2*(c + d*x)]))/(28*d)
Time = 0.54 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.205, Rules used = {3042, 2030, 3502, 27, 3042, 3227, 3042, 3122}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sec (c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b \int \sqrt [3]{b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )} \left (C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+A\right )dx\) |
\(\Big \downarrow \) 3502 |
\(\displaystyle b \left (\frac {3 \int \frac {1}{3} \sqrt [3]{b \cos (c+d x)} (b (7 A+4 C)+7 b B \cos (c+d x))dx}{7 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 b d}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle b \left (\frac {\int \sqrt [3]{b \cos (c+d x)} (b (7 A+4 C)+7 b B \cos (c+d x))dx}{7 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 b d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {\int \sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )} \left (b (7 A+4 C)+7 b B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{7 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 b d}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle b \left (\frac {b (7 A+4 C) \int \sqrt [3]{b \cos (c+d x)}dx+7 B \int (b \cos (c+d x))^{4/3}dx}{7 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 b d}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \left (\frac {b (7 A+4 C) \int \sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}dx+7 B \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3}dx}{7 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 b d}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b \left (\frac {-\frac {3 (7 A+4 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) (b \cos (c+d x))^{7/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{6},\frac {13}{6},\cos ^2(c+d x)\right )}{b d \sqrt {\sin ^2(c+d x)}}}{7 b}+\frac {3 C \sin (c+d x) (b \cos (c+d x))^{4/3}}{7 b d}\right )\) |
b*((3*C*(b*Cos[c + d*x])^(4/3)*Sin[c + d*x])/(7*b*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])/(4*d*Sqrt[Sin[c + d*x]^2]) - (3*B*(b*Cos[c + d*x])^(7/3)*Hyperge ometric2F1[1/2, 7/6, 13/6, Cos[c + d*x]^2]*Sin[c + d*x])/(b*d*Sqrt[Sin[c + d*x]^2]))/(7*b))
3.4.47.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m Int[(b*v) ^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
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]))*Hypergeometric2 F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[2*n]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[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]
\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sec \left (d x +c \right )d x\]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
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")
integral((C*b*cos(d*x + c)^3 + B*b*cos(d*x + c)^2 + A*b*cos(d*x + c))*(b*c os(d*x + c))^(1/3)*sec(d*x + c), x)
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\text {Timed out} \]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
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")
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right ) \,d x } \]
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")
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{\cos \left (c+d\,x\right )} \,d x \]