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

Optimal. Leaf size=282 \[ -\frac{b \left (3 a^2 (3-m)+b^2 (2-m)\right ) \sin (c+d x) \sec ^{m-4}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{4-m}{2};\frac{6-m}{2};\cos ^2(c+d x)\right )}{d (2-m) (4-m) \sqrt{\sin ^2(c+d x)}}-\frac{a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{3-m}{2};\frac{5-m}{2};\cos ^2(c+d x)\right )}{d (1-m) (3-m) \sqrt{\sin ^2(c+d x)}}-\frac{a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}-\frac{a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (1-m) (2-m)} \]

[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])

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Rubi [A]  time = 0.424042, antiderivative size = 282, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3238, 3842, 4047, 3772, 2643, 4046} \[ -\frac{b \left (3 a^2 (3-m)+b^2 (2-m)\right ) \sin (c+d x) \sec ^{m-4}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{4-m}{2};\frac{6-m}{2};\cos ^2(c+d x)\right )}{d (2-m) (4-m) \sqrt{\sin ^2(c+d x)}}-\frac{a \left (a^2 (2-m)+3 b^2 (1-m)\right ) \sin (c+d x) \sec ^{m-3}(c+d x) \, _2F_1\left (\frac{1}{2},\frac{3-m}{2};\frac{5-m}{2};\cos ^2(c+d x)\right )}{d (1-m) (3-m) \sqrt{\sin ^2(c+d x)}}-\frac{a^2 \sin (c+d x) \sec ^{m-2}(c+d x) (a \sec (c+d x)+b)}{d (1-m)}-\frac{a^2 b (1-2 m) \sin (c+d x) \sec ^{m-2}(c+d x)}{d (1-m) (2-m)} \]

Antiderivative was successfully verified.

[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 3238

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 3842

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)), In
t[(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] &&  !Integ
erQ[m])

Rule 4047

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]

Rule 3772

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 4046

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]

Rubi steps

\begin{align*} \int (a+b \cos (c+d x))^3 \sec ^m(c+d x) \, dx &=\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 ) \, _2F_1\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 ) \, _2F_1\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 ) \, _2F_1\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]  time = 0.792781, size = 222, normalized size = 0.79 \[ \frac{\sqrt{-\tan ^2(c+d x)} \csc (c+d x) \sec ^{m-4}(c+d x) \left (\frac{1}{2} a (m-3) \sec ^3(c+d x) \left (2 a (m-2) \left (a (m-1) \, _2F_1\left (\frac{1}{2},\frac{m}{2};\frac{m+2}{2};\sec ^2(c+d x)\right )+3 b m \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{m-1}{2};\frac{m+1}{2};\sec ^2(c+d x)\right )\right )+6 b^2 (m-1) m \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{m-2}{2};\frac{m}{2};\sec ^2(c+d x)\right )\right )+b^3 m \left (m^2-3 m+2\right ) \, _2F_1\left (\frac{1}{2},\frac{m-3}{2};\frac{m-1}{2};\sec ^2(c+d x)\right )\right )}{d (m-3) (m-2) (m-1) m} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 1.984, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\cos \left ( dx+c \right ) \right ) ^{3} \left ( \sec \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}\right )} \sec \left (d x + c\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sec \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)