Integrand size = 41, antiderivative size = 232 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 b C \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \sin (c+d x)}{d (10+3 m)}-\frac {3 b (C (7+3 m)+A (10+3 m)) \cos ^{2+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+3 m) (10+3 m) \sqrt {\sin ^2(c+d x)}}-\frac {3 b B \cos ^{3+m}(c+d x) \sqrt [3]{b \cos (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (10+3 m),\frac {1}{6} (16+3 m),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (10+3 m) \sqrt {\sin ^2(c+d x)}} \]
3*b*C*cos(d*x+c)^(2+m)*(b*cos(d*x+c))^(1/3)*sin(d*x+c)/d/(10+3*m)-3*b*(C*( 7+3*m)+A*(10+3*m))*cos(d*x+c)^(2+m)*(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 7 /6+1/2*m],[13/6+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(9*m^2+51*m+70)/(sin(d*x +c)^2)^(1/2)-3*b*B*cos(d*x+c)^(3+m)*(b*cos(d*x+c))^(1/3)*hypergeom([1/2, 5 /3+1/2*m],[8/3+1/2*m],cos(d*x+c)^2)*sin(d*x+c)/d/(10+3*m)/(sin(d*x+c)^2)^( 1/2)
Time = 0.50 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.77 \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {3 \cos ^{1+m}(c+d x) (b \cos (c+d x))^{4/3} \csc (c+d x) \left (C (7+3 m) \sin ^2(c+d x)-B (7+3 m) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{3}+\frac {m}{2},\frac {8}{3}+\frac {m}{2},\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}-(C (7+3 m)+A (10+3 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (7+3 m),\frac {1}{6} (13+3 m),\cos ^2(c+d x)\right ) \sqrt {\sin ^2(c+d x)}\right )}{d (7+3 m) (10+3 m)} \]
(3*Cos[c + d*x]^(1 + m)*(b*Cos[c + d*x])^(4/3)*Csc[c + d*x]*(C*(7 + 3*m)*S in[c + d*x]^2 - B*(7 + 3*m)*Cos[c + d*x]*Hypergeometric2F1[1/2, 5/3 + m/2, 8/3 + m/2, Cos[c + d*x]^2]*Sqrt[Sin[c + d*x]^2] - (C*(7 + 3*m) + A*(10 + 3*m))*Hypergeometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d*x]^2]*Sq rt[Sin[c + d*x]^2]))/(d*(7 + 3*m)*(10 + 3*m))
Time = 0.64 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {2034, 3042, 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.
| \(\displaystyle \int (b \cos (c+d x))^{4/3} \cos ^m(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\) |
| \(\Big \downarrow \) 2034 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \int \cos ^{m+\frac {4}{3}}(c+d x) \left (C \cos ^2(c+d x)+B \cos (c+d x)+A\right )dx}{\sqrt [3]{\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (C \sin \left (c+d x+\frac {\pi }{2}\right )^2+B \sin \left (c+d x+\frac {\pi }{2}\right )+A\right )dx}{\sqrt [3]{\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \left (\frac {3 \int \frac {1}{3} \cos ^{m+\frac {4}{3}}(c+d x) \left (3 C \left (m+\frac {7}{3}\right )+3 A \left (m+\frac {10}{3}\right )+B (3 m+10) \cos (c+d x)\right )dx}{3 m+10}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x)}{d (3 m+10)}\right )}{\sqrt [3]{\cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \left (\frac {\int \cos ^{m+\frac {4}{3}}(c+d x) (C (3 m+7)+A (3 m+10)+B (3 m+10) \cos (c+d x))dx}{3 m+10}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x)}{d (3 m+10)}\right )}{\sqrt [3]{\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \left (\frac {\int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}} \left (C (3 m+7)+A (3 m+10)+B (3 m+10) \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx}{3 m+10}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x)}{d (3 m+10)}\right )}{\sqrt [3]{\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \left (\frac {(A (3 m+10)+C (3 m+7)) \int \cos ^{m+\frac {4}{3}}(c+d x)dx+B (3 m+10) \int \cos ^{m+\frac {7}{3}}(c+d x)dx}{3 m+10}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x)}{d (3 m+10)}\right )}{\sqrt [3]{\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \left (\frac {(A (3 m+10)+C (3 m+7)) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {4}{3}}dx+B (3 m+10) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{m+\frac {7}{3}}dx}{3 m+10}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x)}{d (3 m+10)}\right )}{\sqrt [3]{\cos (c+d x)}}\) |
| \(\Big \downarrow \) 3122 |
| \(\displaystyle \frac {b \sqrt [3]{b \cos (c+d x)} \left (\frac {-\frac {3 (A (3 m+10)+C (3 m+7)) \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+7),\frac {1}{6} (3 m+13),\cos ^2(c+d x)\right )}{d (3 m+7) \sqrt {\sin ^2(c+d x)}}-\frac {3 B \sin (c+d x) \cos ^{m+\frac {10}{3}}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{6} (3 m+10),\frac {1}{6} (3 m+16),\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}}}{3 m+10}+\frac {3 C \sin (c+d x) \cos ^{m+\frac {7}{3}}(c+d x)}{d (3 m+10)}\right )}{\sqrt [3]{\cos (c+d x)}}\) |
(b*(b*Cos[c + d*x])^(1/3)*((3*C*Cos[c + d*x]^(7/3 + m)*Sin[c + d*x])/(d*(1 0 + 3*m)) + ((-3*(C*(7 + 3*m) + A*(10 + 3*m))*Cos[c + d*x]^(7/3 + m)*Hyper geometric2F1[1/2, (7 + 3*m)/6, (13 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x]) /(d*(7 + 3*m)*Sqrt[Sin[c + d*x]^2]) - (3*B*Cos[c + d*x]^(10/3 + m)*Hyperge ometric2F1[1/2, (10 + 3*m)/6, (16 + 3*m)/6, Cos[c + d*x]^2]*Sin[c + d*x])/ (d*Sqrt[Sin[c + d*x]^2]))/(10 + 3*m)))/Cos[c + d*x]^(1/3)
3.4.63.3.1 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_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Simp[b^IntPart [n]*((b*v)^FracPart[n]/(a^IntPart[n]*(a*v)^FracPart[n])) Int[(a*v)^(m + n )*Fx, x], x] /; FreeQ[{a, b, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && !IntegerQ[m + n]
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]
\[\int \left (\cos ^{m}\left (d x +c \right )\right ) \left (\cos \left (d x +c \right ) b \right )^{\frac {4}{3}} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )d x\]
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2 ),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)*cos(d*x + c)^m, x)
Timed out. \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\text {Timed out} \]
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2 ),x, algorithm="maxima")
\[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {4}{3}} \cos \left (d x + c\right )^{m} \,d x } \]
integrate(cos(d*x+c)^m*(b*cos(d*x+c))^(4/3)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2 ),x, algorithm="giac")
Timed out. \[ \int \cos ^m(c+d x) (b \cos (c+d x))^{4/3} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\int {\cos \left (c+d\,x\right )}^m\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{4/3}\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right ) \,d x \]