Integrand size = 41, antiderivative size = 145 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\frac {3 A b^2 \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3}}-\frac {3 b B \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{d \sqrt {\sin ^2(c+d x)}}+\frac {3 (A-2 C) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{8 d \sqrt {\sin ^2(c+d x)}} \]
3/2*A*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(2/3)-3*b*B*(b*cos(d*x+c))^(1/3)*hyp ergeom([1/6, 1/2],[7/6],cos(d*x+c)^2)*sin(d*x+c)/d/(sin(d*x+c)^2)^(1/2)+3/ 8*(A-2*C)*(b*cos(d*x+c))^(4/3)*hypergeom([1/2, 2/3],[5/3],cos(d*x+c)^2)*si n(d*x+c)/d/(sin(d*x+c)^2)^(1/2)
Time = 0.16 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=-\frac {3 b^2 \csc (c+d x) \left (-2 A \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )+\cos (c+d x) \left (4 B \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )+C \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )\right )\right ) \sqrt {\sin ^2(c+d x)}}{4 d (b \cos (c+d x))^{2/3}} \]
(-3*b^2*Csc[c + d*x]*(-2*A*Hypergeometric2F1[-1/3, 1/2, 2/3, Cos[c + d*x]^ 2] + Cos[c + d*x]*(4*B*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2] + C*Cos[c + d*x]*Hypergeometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]))*Sqrt[Sin [c + d*x]^2])/(4*d*(b*Cos[c + d*x])^(2/3))
Time = 0.56 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.08, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {3042, 2030, 3500, 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.
| \(\displaystyle \int \sec ^3(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
| \(\Big \downarrow \) 2030 |
| \(\displaystyle b^3 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{5/3}}dx\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle b^3 \left (\frac {3 \int \frac {2 b^2 B-b^2 (A-2 C) \cos (c+d x)}{3 (b \cos (c+d x))^{2/3}}dx}{2 b^3}+\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle b^3 \left (\frac {\int \frac {2 b^2 B-b^2 (A-2 C) \cos (c+d x)}{(b \cos (c+d x))^{2/3}}dx}{2 b^3}+\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^3 \left (\frac {\int \frac {2 b^2 B-b^2 (A-2 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx}{2 b^3}+\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle b^3 \left (\frac {2 b^2 B \int \frac {1}{(b \cos (c+d x))^{2/3}}dx-b (A-2 C) \int \sqrt [3]{b \cos (c+d x)}dx}{2 b^3}+\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b^3 \left (\frac {2 b^2 B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx-b (A-2 C) \int \sqrt [3]{b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{2 b^3}+\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}\right )\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle b^3 \left (\frac {\frac {3 (A-2 C) \sin (c+d x) (b \cos (c+d x))^{4/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {2}{3},\frac {5}{3},\cos ^2(c+d x)\right )}{4 d \sqrt {\sin ^2(c+d x)}}-\frac {6 b B \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{2 b^3}+\frac {3 A \sin (c+d x)}{2 b d (b \cos (c+d x))^{2/3}}\right )\) |
b^3*((3*A*Sin[c + d*x])/(2*b*d*(b*Cos[c + d*x])^(2/3)) + ((-6*b*B*(b*Cos[c + d*x])^(1/3)*Hypergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d* x])/(d*Sqrt[Sin[c + d*x]^2]) + (3*(A - 2*C)*(b*Cos[c + d*x])^(4/3)*Hyperge ometric2F1[1/2, 2/3, 5/3, Cos[c + d*x]^2]*Sin[c + d*x])/(4*d*Sqrt[Sin[c + d*x]^2]))/(2*b^3))
3.4.49.3.1 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[(-(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] + Simp[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]
\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{3}\left (d x +c \right )\right )d x\]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 3,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)^3, x)
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\text {Timed out} \]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 3,x, algorithm="maxima")
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{3} \,d x } \]
integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 3,x, algorithm="giac")
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^3} \,d x \]