Integrand size = 41, antiderivative size = 152 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 A b^3 \sin (c+d x)}{5 d (b \cos (c+d x))^{5/3}}+\frac {3 b^2 B \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right ) \sin (c+d x)}{2 d (b \cos (c+d x))^{2/3} \sqrt {\sin ^2(c+d x)}}-\frac {3 b (2 A+5 C) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right ) \sin (c+d x)}{5 d \sqrt {\sin ^2(c+d x)}} \]
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)
Time = 0.14 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.82 \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {3 (b \cos (c+d x))^{4/3} \csc (c+d x) \left (-2 A \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\cos ^2(c+d x)\right )+5 \cos (c+d x) \left (-B \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )+2 C \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )\right )\right ) \sec ^3(c+d x) \sqrt {\sin ^2(c+d x)}}{10 d} \]
(-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, C os[c + d*x]^2]))*Sec[c + d*x]^3*Sqrt[Sin[c + d*x]^2])/(10*d)
Time = 0.58 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.05, 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.
\(\displaystyle \int \sec ^4(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{4/3} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 2030 |
\(\displaystyle b^4 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+B \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{8/3}}dx\) |
\(\Big \downarrow \) 3500 |
\(\displaystyle b^4 \left (\frac {3 \int \frac {5 B b^2+(2 A+5 C) \cos (c+d x) b^2}{3 (b \cos (c+d x))^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle b^4 \left (\frac {\int \frac {5 B b^2+(2 A+5 C) \cos (c+d x) b^2}{(b \cos (c+d x))^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^4 \left (\frac {\int \frac {5 B b^2+(2 A+5 C) \sin \left (c+d x+\frac {\pi }{2}\right ) b^2}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3227 |
\(\displaystyle b^4 \left (\frac {b (2 A+5 C) \int \frac {1}{(b \cos (c+d x))^{2/3}}dx+5 b^2 B \int \frac {1}{(b \cos (c+d x))^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b^4 \left (\frac {b (2 A+5 C) \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{2/3}}dx+5 b^2 B \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{5/3}}dx}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle b^4 \left (\frac {\frac {15 b B \sin (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},\frac {1}{2},\frac {2}{3},\cos ^2(c+d x)\right )}{2 d \sqrt {\sin ^2(c+d x)} (b \cos (c+d x))^{2/3}}-\frac {3 (2 A+5 C) \sin (c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{2},\frac {7}{6},\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{5 b^3}+\frac {3 A \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/3}}\right )\) |
b^4*((3*A*Sin[c + d*x])/(5*b*d*(b*Cos[c + d*x])^(5/3)) + ((15*b*B*Hypergeo metric2F1[-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*(2*A + 5*C)*(b*Cos[c + d*x])^(1/3)*Hy pergeometric2F1[1/6, 1/2, 7/6, Cos[c + d*x]^2]*Sin[c + d*x])/(d*Sqrt[Sin[c + d*x]^2]))/(5*b^3))
3.4.50.3.1 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]
\[\int \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{4}\left (d x +c \right )\right )d x\]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{4} \,d x } \]
integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 4,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)^4, x)
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \]
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{4} \,d x } \]
integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 4,x, algorithm="maxima")
\[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \sec \left (d x + c\right )^{4} \,d x } \]
integrate((b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^ 4,x, algorithm="giac")
Timed out. \[ \int (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^4} \,d x \]