\(\int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx\) [775]

Optimal result

Integrand size = 21, antiderivative size = 282 \[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=-\frac {a^2 b (1-2 m) \sec ^{-2+m}(c+d x) \sin (c+d x)}{d (1-m) (2-m)}-\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}-\frac {b \left (b^2 (2-m)+3 a^2 (3-m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(c+d x)\right ) \sec ^{-4+m}(c+d x) \sin (c+d x)}{d (2-m) (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (3 b^2 (1-m)+a^2 (2-m)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right ) \sec ^{-3+m}(c+d x) \sin (c+d x)}{d (1-m) (3-m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-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))*hypergeom([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)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3317, 3927, 4132, 3857, 2722, 4131} \[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=-\frac {b \left (3 a^2 (3-m)+b^2 (2-m)\right ) \sin (c+d x) \sec ^{m-4}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {4-m}{2},\frac {6-m}{2},\cos ^2(c+d x)\right )}{d (2-m) (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3-m}{2},\frac {5-m}{2},\cos ^2(c+d x)\right )}{d (1-m) (3-m) \sqrt {\sin ^2(c+d x)}}-\frac {a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}-\frac {a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (1-m) (2-m)} \]

[In]

Int[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^m,x]

[Out]

-((a^2*b*(1 - 2*m)*Sec[c + d*x]^(-2 + m)*Sin[c + d*x])/(d*(1 - m)*(2 - m))) - (a^2*Sec[c + d*x]^(-2 + m)*(b +
a*Sec[c + d*x])*Sin[c + d*x])/(d*(1 - m)) - (b*(b^2*(2 - m) + 3*a^2*(3 - 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))*Hypergeometric2F1[1/2, (3 - m)/2, (5 - m)/2, Cos[c + d*x]^2]*Sec[c + d*x]^(-3 +
m)*Sin[c + d*x])/(d*(1 - m)*(3 - m)*Sqrt[Sin[c + d*x]^2])

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3317

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[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]

Rule 3857

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]

Rule 3927

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] + Dist[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] &&  !Int
egerQ[m])

Rule 4131

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] + Dist[(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]

Rule 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[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]

Rubi steps \begin{align*} \text {integral}& = \int \sec ^{-3+m}(c+d x) (b+a \sec (c+d x))^3 \, dx \\ & = -\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}+\frac {\int \sec ^{-3+m}(c+d x) \left (-b \left (b^2 (1-m)+a^2 (3-m)\right )-a \left (3 b^2 (1-m)+a^2 (2-m)\right ) \sec (c+d x)-a^2 b (1-2 m) \sec ^2(c+d x)\right ) \, dx}{-1+m} \\ & = -\frac {a^2 \sec ^{-2+m}(c+d x) (b+a \sec (c+d x)) \sin (c+d x)}{d (1-m)}+\left (a \left (3 b^2+\frac {a^2 (2-m)}{1-m}\right )\right ) \int \sec ^{-2+m}(c+d x) \, dx+\frac {\int \sec ^{-3+m}(c+d x) \left (-b \left (b^2 (1-m)+a^2 (3-m)\right )-a^2 b (1-2 m) \sec ^2(c+d x)\right ) \, dx}{-1+m} \\ & = -\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)}+\left (b \left (b^2+\frac {3 a^2 (3-m)}{2-m}\right )\right ) \int \sec ^{-3+m}(c+d x) \, dx+\left (a \left (3 b^2+\frac {a^2 (2-m)}{1-m}\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{2-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 {a \left (3 b^2+\frac {a^2 (2-m)}{1-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 (3-m) \sqrt {\sin ^2(c+d x)}}+\left (b \left (b^2+\frac {3 a^2 (3-m)}{2-m}\right ) \cos ^m(c+d x) \sec ^m(c+d x)\right ) \int \cos ^{3-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+\frac {3 a^2 (3-m)}{2-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 (4-m) \sqrt {\sin ^2(c+d x)}}-\frac {a \left (3 b^2+\frac {a^2 (2-m)}{1-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 (3-m) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.79 \[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\frac {\csc (c+d x) \sec ^{-4+m}(c+d x) \left (b^3 m \left (2-3 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-3+m),\frac {1}{2} (-1+m),\sec ^2(c+d x)\right )+\frac {1}{2} a (-3+m) \left (6 b^2 (-1+m) m \cos ^2(c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-2+m),\frac {m}{2},\sec ^2(c+d x)\right )+2 a (-2+m) \left (3 b m \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{2} (-1+m),\frac {1+m}{2},\sec ^2(c+d x)\right )+a (-1+m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right )\right )\right ) \sec ^3(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}}{d (-3+m) (-2+m) (-1+m) m} \]

[In]

Integrate[(a + b*Cos[c + d*x])^3*Sec[c + d*x]^m,x]

[Out]

(Csc[c + d*x]*Sec[c + d*x]^(-4 + m)*(b^3*m*(2 - 3*m + m^2)*Hypergeometric2F1[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)

Maple [F]

\[\int \left (a +\cos \left (d x +c \right ) b \right )^{3} \left (\sec ^{m}\left (d x +c \right )\right )d x\]

[In]

int((a+cos(d*x+c)*b)^3*sec(d*x+c)^m,x)

[Out]

int((a+cos(d*x+c)*b)^3*sec(d*x+c)^m,x)

Fricas [F]

\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^m,x, algorithm="fricas")

[Out]

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)

Sympy [F]

\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec ^{m}{\left (c + d x \right )}\, dx \]

[In]

integrate((a+b*cos(d*x+c))**3*sec(d*x+c)**m,x)

[Out]

Integral((a + b*cos(c + d*x))**3*sec(c + d*x)**m, x)

Maxima [F]

\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^m,x, algorithm="maxima")

[Out]

integrate((b*cos(d*x + c) + a)^3*sec(d*x + c)^m, x)

Giac [F]

\[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m} \,d x } \]

[In]

integrate((a+b*cos(d*x+c))^3*sec(d*x+c)^m,x, algorithm="giac")

[Out]

integrate((b*cos(d*x + c) + a)^3*sec(d*x + c)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx=\int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^3 \,d x \]

[In]

int((1/cos(c + d*x))^m*(a + b*cos(c + d*x))^3,x)

[Out]

int((1/cos(c + d*x))^m*(a + b*cos(c + d*x))^3, x)