3.875 \(\int (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx\)

Optimal. Leaf size=266 \[ -\frac{b \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \cos (e+f x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n}{2},\frac{n+2}{2},\cos ^2(e+f x)\right )}{f (2-n) n \sqrt{\sin ^2(e+f x)}}-\frac{a \left (a^2 (1-n)-3 b^2 n\right ) \sin (e+f x) \cos (e+f x) (d \cos (e+f x))^n \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+1}{2},\frac{n+3}{2},\cos ^2(e+f x)\right )}{f (1-n) (n+1) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5-2 n) \tan (e+f x) (d \cos (e+f x))^n}{f (1-n) (2-n)}+\frac{b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \cos (e+f x))^n}{f (2-n)} \]

[Out]

-((b*(b^2*(1 - n) + 3*a^2*(2 - n))*(d*Cos[e + f*x])^n*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[e + f*x]^2]*S
in[e + f*x])/(f*(2 - n)*n*Sqrt[Sin[e + f*x]^2])) - (a*(a^2*(1 - n) - 3*b^2*n)*Cos[e + f*x]*(d*Cos[e + f*x])^n*
Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*(1 - n)*(1 + n)*Sqrt[Sin[e + f*x
]^2]) + (a*b^2*(5 - 2*n)*(d*Cos[e + f*x])^n*Tan[e + f*x])/(f*(1 - n)*(2 - n)) + (b^2*(d*Cos[e + f*x])^n*(a + b
*Sec[e + f*x])*Tan[e + f*x])/(f*(2 - n))

________________________________________________________________________________________

Rubi [A]  time = 0.456245, antiderivative size = 266, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {4264, 3842, 4047, 3772, 2643, 4046} \[ -\frac{b \left (3 a^2 (2-n)+b^2 (1-n)\right ) \sin (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{n}{2};\frac{n+2}{2};\cos ^2(e+f x)\right )}{f (2-n) n \sqrt{\sin ^2(e+f x)}}-\frac{a \left (a^2 (1-n)-3 b^2 n\right ) \sin (e+f x) \cos (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\cos ^2(e+f x)\right )}{f (1-n) (n+1) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5-2 n) \tan (e+f x) (d \cos (e+f x))^n}{f (1-n) (2-n)}+\frac{b^2 \tan (e+f x) (a+b \sec (e+f x)) (d \cos (e+f x))^n}{f (2-n)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[e + f*x])^n*(a + b*Sec[e + f*x])^3,x]

[Out]

-((b*(b^2*(1 - n) + 3*a^2*(2 - n))*(d*Cos[e + f*x])^n*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[e + f*x]^2]*S
in[e + f*x])/(f*(2 - n)*n*Sqrt[Sin[e + f*x]^2])) - (a*(a^2*(1 - n) - 3*b^2*n)*Cos[e + f*x]*(d*Cos[e + f*x])^n*
Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[e + f*x]^2]*Sin[e + f*x])/(f*(1 - n)*(1 + n)*Sqrt[Sin[e + f*x
]^2]) + (a*b^2*(5 - 2*n)*(d*Cos[e + f*x])^n*Tan[e + f*x])/(f*(1 - n)*(2 - n)) + (b^2*(d*Cos[e + f*x])^n*(a + b
*Sec[e + f*x])*Tan[e + f*x])/(f*(2 - n))

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

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 (d \cos (e+f x))^n (a+b \sec (e+f x))^3 \, dx &=\left ((d \cos (e+f x))^n (d \sec (e+f x))^n\right ) \int (d \sec (e+f x))^{-n} (a+b \sec (e+f x))^3 \, dx\\ &=\frac{b^2 (d \cos (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2-n)}+\frac{\left ((d \cos (e+f x))^n (d \sec (e+f x))^n\right ) \int (d \sec (e+f x))^{-n} \left (a d \left (a^2 (2-n)-b^2 n\right )+b d \left (b^2 (1-n)+3 a^2 (2-n)\right ) \sec (e+f x)+a b^2 d (5-2 n) \sec ^2(e+f x)\right ) \, dx}{d (2-n)}\\ &=\frac{b^2 (d \cos (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2-n)}+\frac{\left ((d \cos (e+f x))^n (d \sec (e+f x))^n\right ) \int (d \sec (e+f x))^{-n} \left (a d \left (a^2 (2-n)-b^2 n\right )+a b^2 d (5-2 n) \sec ^2(e+f x)\right ) \, dx}{d (2-n)}+\frac{\left (b \left (b^2 (1-n)+3 a^2 (2-n)\right ) (d \cos (e+f x))^n (d \sec (e+f x))^n\right ) \int (d \sec (e+f x))^{1-n} \, dx}{d (2-n)}\\ &=\frac{a b^2 (5-2 n) (d \cos (e+f x))^n \tan (e+f x)}{f (1-n) (2-n)}+\frac{b^2 (d \cos (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2-n)}+\frac{\left (b \left (b^2 (1-n)+3 a^2 (2-n)\right ) \left (\frac{\cos (e+f x)}{d}\right )^{-n} (d \cos (e+f x))^n\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^{-1+n} \, dx}{d (2-n)}+\frac{\left (a \left (a^2 (1-n)-3 b^2 n\right ) (d \cos (e+f x))^n (d \sec (e+f x))^n\right ) \int (d \sec (e+f x))^{-n} \, dx}{1-n}\\ &=-\frac{b \left (b^2 (1-n)+3 a^2 (2-n)\right ) (d \cos (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{n}{2};\frac{2+n}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{f (2-n) n \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5-2 n) (d \cos (e+f x))^n \tan (e+f x)}{f (1-n) (2-n)}+\frac{b^2 (d \cos (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2-n)}+\frac{\left (a \left (a^2 (1-n)-3 b^2 n\right ) \left (\frac{\cos (e+f x)}{d}\right )^{-n} (d \cos (e+f x))^n\right ) \int \left (\frac{\cos (e+f x)}{d}\right )^n \, dx}{1-n}\\ &=-\frac{b \left (b^2 (1-n)+3 a^2 (2-n)\right ) (d \cos (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{n}{2};\frac{2+n}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{f (2-n) n \sqrt{\sin ^2(e+f x)}}-\frac{a \left (a^2 (1-n)-3 b^2 n\right ) \cos (e+f x) (d \cos (e+f x))^n \, _2F_1\left (\frac{1}{2},\frac{1+n}{2};\frac{3+n}{2};\cos ^2(e+f x)\right ) \sin (e+f x)}{f (1-n) (1+n) \sqrt{\sin ^2(e+f x)}}+\frac{a b^2 (5-2 n) (d \cos (e+f x))^n \tan (e+f x)}{f (1-n) (2-n)}+\frac{b^2 (d \cos (e+f x))^n (a+b \sec (e+f x)) \tan (e+f x)}{f (2-n)}\\ \end{align*}

Mathematica [A]  time = 0.689, size = 222, normalized size = 0.83 \[ -\frac{\sqrt{\sin ^2(e+f x)} \csc (e+f x) \sec ^2(e+f x) (d \cos (e+f x))^n \left (\frac{1}{2} a (n-2) \cos (e+f x) \left (2 a (n-1) \cos (e+f x) \left (a n \cos (e+f x) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n+1}{2},\frac{n+3}{2},\cos ^2(e+f x)\right )+3 b (n+1) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n}{2},\frac{n+2}{2},\cos ^2(e+f x)\right )\right )+6 b^2 n (n+1) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n-1}{2},\frac{n+1}{2},\cos ^2(e+f x)\right )\right )+b^3 n \left (n^2-1\right ) \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{n-2}{2},\frac{n}{2},\cos ^2(e+f x)\right )\right )}{f (n-2) (n-1) n (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d*Cos[e + f*x])^n*(a + b*Sec[e + f*x])^3,x]

[Out]

-(((d*Cos[e + f*x])^n*Csc[e + f*x]*(b^3*n*(-1 + n^2)*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Cos[e + f*x]^2] +
(a*(-2 + n)*Cos[e + f*x]*(6*b^2*n*(1 + n)*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[e + f*x]^2] + 2*a
*(-1 + n)*Cos[e + f*x]*(3*b*(1 + n)*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[e + f*x]^2] + a*n*Cos[e + f*x]*
Hypergeometric2F1[1/2, (1 + n)/2, (3 + n)/2, Cos[e + f*x]^2])))/2)*Sec[e + f*x]^2*Sqrt[Sin[e + f*x]^2])/(f*(-2
+ n)*(-1 + n)*n*(1 + n)))

________________________________________________________________________________________

Maple [F]  time = 3.103, size = 0, normalized size = 0. \begin{align*} \int \left ( d\cos \left ( fx+e \right ) \right ) ^{n} \left ( a+b\sec \left ( fx+e \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(f*x+e))^n*(a+b*sec(f*x+e))^3,x)

[Out]

int((d*cos(f*x+e))^n*(a+b*sec(f*x+e))^3,x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n*(a+b*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e) + a)^3*(d*cos(f*x + e))^n, x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n*(a+b*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

integral((b^3*sec(f*x + e)^3 + 3*a*b^2*sec(f*x + e)^2 + 3*a^2*b*sec(f*x + e) + a^3)*(d*cos(f*x + e))^n, x)

________________________________________________________________________________________

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((d*cos(f*x+e))**n*(a+b*sec(f*x+e))**3,x)

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n*(a+b*sec(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e) + a)^3*(d*cos(f*x + e))^n, x)