Integrand size = 21, antiderivative size = 282 \[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=-\frac {a^2 b (1-2 m) \sec ^{-2+m}(c+d x) \sin (c+d x)}{d (1-m) (2-m)}-\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}-\frac {b \left (b^2 (2-m)+3 a^2 (3-m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(c+d x)\right ) \sec ^{-4+m}(c+d x) \sin (c+d x)}{d (2-m) (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (3 b^2 (1-m)+a^2 (2-m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right ) \sec ^{-3+m}(c+d x) \sin (c+d x)}{d (1-m) (3-m) \sqrt {\sin ^2(c+d x)}} \]
-a^2*b*(1-2*m)*sec(d*x+c)^(-2+m)*sin(d*x+c)/d/(m^2-3*m+2)-a^2*sec(d*x+c)^( -2+m)*(b+a*sec(d*x+c))*sin(d*x+c)/d/(1-m)-b*(b^2*(2-m)+3*a^2*(3-m))*hyperg eom([1/2, 2-1/2*m],[3-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-4+m)*sin(d*x+c)/d/ (m^2-6*m+8)/(sin(d*x+c)^2)^(1/2)-a*(3*b^2*(1-m)+a^2*(2-m))*hypergeom([1/2, 3/2-1/2*m],[5/2-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-3+m)*sin(d*x+c)/d/(m^2- 4*m+3)/(sin(d*x+c)^2)^(1/2)
Time = 0.54 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.79 \[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\frac {\csc (c+d x) \sec ^{-4+m}(c+d x) \left (b^3 m \left (2-3 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sec ^2(c+d x)\right )+\frac {1}{2} a (-3+m) \left (6 b^2 (-1+m) m \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+m),\frac {m}{2},\sec ^2(c+d x)\right )+2 a (-2+m) \left (3 b m \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sec ^2(c+d x)\right )+a (-1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right )\right )\right ) \sec ^3(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-3+m) (-2+m) (-1+m) m} \]
(Csc[c + d*x]*Sec[c + d*x]^(-4 + m)*(b^3*m*(2 - 3*m + m^2)*Hypergeometric2 F1[1/2, (-3 + m)/2, (-1 + m)/2, Sec[c + d*x]^2] + (a*(-3 + m)*(6*b^2*(-1 + m)*m*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (-2 + m)/2, m/2, Sec[c + d*x]^ 2] + 2*a*(-2 + m)*(3*b*m*Cos[c + d*x]*Hypergeometric2F1[1/2, (-1 + m)/2, ( 1 + m)/2, Sec[c + d*x]^2] + a*(-1 + m)*Hypergeometric2F1[1/2, m/2, (2 + m) /2, Sec[c + d*x]^2]))*Sec[c + d*x]^3)/2)*Sqrt[-Tan[c + d*x]^2])/(d*(-3 + m )*(-2 + m)*(-1 + m)*m)
Time = 1.51 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 3717, 3042, 4329, 25, 3042, 4535, 3042, 4259, 3042, 3122, 4534, 3042, 4259, 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.
\(\displaystyle \int \sec ^m(c+d x) (a+b \cos (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^m \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^3dx\) |
\(\Big \downarrow \) 3717 |
\(\displaystyle \int \sec ^{m-3}(c+d x) (a \sec (c+d x)+b)^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-3} \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+b\right )^3dx\) |
\(\Big \downarrow \) 4329 |
\(\displaystyle -\frac {\int -\sec ^{m-3}(c+d x) \left (a^2 b (1-2 m) \sec ^2(c+d x)+a \left ((2-m) a^2+3 b^2 (1-m)\right ) \sec (c+d x)+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \sec ^{m-3}(c+d x) \left (a^2 b (1-2 m) \sec ^2(c+d x)+a \left ((2-m) a^2+3 b^2 (1-m)\right ) \sec (c+d x)+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-3} \left (a^2 b (1-2 m) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a \left ((2-m) a^2+3 b^2 (1-m)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 4535 |
\(\displaystyle \frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \int \sec ^{m-2}(c+d x)dx+\int \sec ^{m-3}(c+d x) \left (a^2 b (1-2 m) \sec ^2(c+d x)+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-2}dx+\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-3} \left (a^2 b (1-2 m) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-3} \left (a^2 b (1-2 m) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx+a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \cos ^{2-m}(c+d x)dx}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-3} \left (a^2 b (1-2 m) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx+a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{2-m}dx}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-3} \left (a^2 b (1-2 m) \csc \left (c+d x+\frac {\pi }{2}\right )^2+b \left ((3-m) a^2+b^2 (1-m)\right )\right )dx-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (3-m) \sqrt {\sin ^2(c+d x)}}}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 4534 |
\(\displaystyle \frac {\frac {b (1-m) \left (3 a^2 (3-m)+b^2 (2-m)\right ) \int \sec ^{m-3}(c+d x)dx}{2-m}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (2-m)}}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (1-m) \left (3 a^2 (3-m)+b^2 (2-m)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^{m-3}dx}{2-m}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (2-m)}}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 4259 |
\(\displaystyle \frac {\frac {b (1-m) \left (3 a^2 (3-m)+b^2 (2-m)\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \cos ^{3-m}(c+d x)dx}{2-m}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (2-m)}}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {b (1-m) \left (3 a^2 (3-m)+b^2 (2-m)\right ) \cos ^m(c+d x) \sec ^m(c+d x) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3-m}dx}{2-m}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (2-m)}}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
\(\Big \downarrow \) 3122 |
\(\displaystyle \frac {-\frac {b (1-m) \left (3 a^2 (3-m)+b^2 (2-m)\right ) \sin (c+d x) \sec ^{m-4}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(c+d x)\right )}{d (2-m) (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (2-m)}}{1-m}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}\) |
-((a^2*Sec[c + d*x]^(-2 + m)*(b + a*Sec[c + d*x])*Sin[c + d*x])/(d*(1 - m) )) + (-((a^2*b*(1 - 2*m)*Sec[c + d*x]^(-2 + m)*Sin[c + d*x])/(d*(2 - m))) - (b*(b^2*(2 - m) + 3*a^2*(3 - m))*(1 - m)*Hypergeometric2F1[1/2, (4 - m)/ 2, (6 - m)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-4 + m)*Sin[c + d*x])/(d*(2 - m)*(4 - m)*Sqrt[Sin[c + d*x]^2]) - (a*(3*b^2*(1 - m) + a^2*(2 - m))*Hyperg eometric2F1[1/2, (3 - m)/2, (5 - m)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-3 + m)*Sin[c + d*x])/(d*(3 - m)*Sqrt[Sin[c + d*x]^2]))/(1 - m)
3.8.75.3.1 Defintions of rubi rules used
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[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p) Int[(d*Csc[e + f*x])^(m - n*p )*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && !IntegerQ[m] && IntegersQ[n, p]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^(n - 1)*((Sin[c + d*x]/b)^(n - 1) Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] && !IntegerQ[n]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-b^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*((d*Csc[e + f*x])^n/(f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) Int[ (a + b*Csc[e + f*x])^(m - 3)*(d*Csc[e + f*x])^n*Simp[a^3*d*(m + n - 1) + a* b^2*d*n + b*(b^2*d*(m + n - 2) + 3*a^2*d*(m + n - 1))*Csc[e + f*x] + a*b^2* d*(3*m + 2*n - 4)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, n}, x ] && NeQ[a^2 - b^2, 0] && GtQ[m, 2] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) && !(IGtQ[n, 2] && !IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) )), x] + Simp[(C*m + A*(m + 1))/(m + 1) Int[(b*Csc[e + f*x])^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] && !LeQ[m, -1]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]* (B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.)), x_Symbol] :> Simp[B/b Int[(b*Cs c[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x]^2) , x] /; FreeQ[{b, e, f, A, B, C, m}, x]
\[\int \left (a +\cos \left (d x +c \right ) b \right )^{3} \left (\sec ^{m}\left (d x +c \right )\right )d x\]
\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m} \,d x } \]
integral((b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x + c)^2 + 3*a^2*b*cos(d*x + c) + a^3)*sec(d*x + c)^m, x)
\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec ^{m}{\left (c + d x \right )}\, dx \]
\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m} \,d x } \]
\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m} \,d x } \]
Timed out. \[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]