3.268 \(\int (a+b \sec (c+d x))^n \sin ^5(c+d x) \, dx\)

Optimal. Leaf size=150 \[ -\frac{2 b^3 (a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (4,n+1,n+2,\frac{b \sec (c+d x)}{a}+1\right )}{a^4 d (n+1)}+\frac{b^5 (a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (6,n+1,n+2,\frac{b \sec (c+d x)}{a}+1\right )}{a^6 d (n+1)}+\frac{b (a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (2,n+1,n+2,\frac{b \sec (c+d x)}{a}+1\right )}{a^2 d (n+1)} \]

[Out]

(b*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a^2*d*(1 + n)) -
(2*b^3*Hypergeometric2F1[4, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a^4*d*(1 + n)
) + (b^5*Hypergeometric2F1[6, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a^6*d*(1 +
n))

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Rubi [A]  time = 0.125513, antiderivative size = 150, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3874, 180, 65} \[ -\frac{2 b^3 (a+b \sec (c+d x))^{n+1} \, _2F_1\left (4,n+1;n+2;\frac{b \sec (c+d x)}{a}+1\right )}{a^4 d (n+1)}+\frac{b^5 (a+b \sec (c+d x))^{n+1} \, _2F_1\left (6,n+1;n+2;\frac{b \sec (c+d x)}{a}+1\right )}{a^6 d (n+1)}+\frac{b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac{b \sec (c+d x)}{a}+1\right )}{a^2 d (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sec[c + d*x])^n*Sin[c + d*x]^5,x]

[Out]

(b*Hypergeometric2F1[2, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a^2*d*(1 + n)) -
(2*b^3*Hypergeometric2F1[4, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a^4*d*(1 + n)
) + (b^5*Hypergeometric2F1[6, 1 + n, 2 + n, 1 + (b*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a^6*d*(1 +
n))

Rule 3874

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[f^(-1), Subs
t[Int[((-1 + x)^((p - 1)/2)*(1 + x)^((p - 1)/2)*(a + b*x)^m)/x^(p + 1), x], x, Csc[e + f*x]], x] /; FreeQ[{a,
b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rubi steps

\begin{align*} \int (a+b \sec (c+d x))^n \sin ^5(c+d x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(-1+x)^2 (1+x)^2 (a-b x)^n}{x^6} \, dx,x,-\sec (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{(a-b x)^n}{x^6}-\frac{2 (a-b x)^n}{x^4}+\frac{(a-b x)^n}{x^2}\right ) \, dx,x,-\sec (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(a-b x)^n}{x^6} \, dx,x,-\sec (c+d x)\right )}{d}-\frac{\operatorname{Subst}\left (\int \frac{(a-b x)^n}{x^2} \, dx,x,-\sec (c+d x)\right )}{d}+\frac{2 \operatorname{Subst}\left (\int \frac{(a-b x)^n}{x^4} \, dx,x,-\sec (c+d x)\right )}{d}\\ &=\frac{b \, _2F_1\left (2,1+n;2+n;1+\frac{b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a^2 d (1+n)}-\frac{2 b^3 \, _2F_1\left (4,1+n;2+n;1+\frac{b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a^4 d (1+n)}+\frac{b^5 \, _2F_1\left (6,1+n;2+n;1+\frac{b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a^6 d (1+n)}\\ \end{align*}

Mathematica [B]  time = 8.16185, size = 562, normalized size = 3.75 \[ -\frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \cos (c+d x) (a+b \sec (c+d x))^n \left (-10 a \sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (b \left (12 a^2 b (n-1)+24 a^3-4 a b^2 \left (n^2-3 n+2\right )-b^3 \left (n^3-6 n^2+11 n-6\right )\right ) \text{Hypergeometric2F1}\left (2,1-n,2-n,\frac{a \cos (c+d x)}{a \cos (c+d x)+b}\right )+(n-1) \left (-14 a^2+2 a b (n-1)+b^2 \left (n^2-5 n+6\right )\right ) (a \cos (c+d x)+b)^2\right )+\sec ^6\left (\frac{1}{2} (c+d x)\right ) \left (b \left (120 a^3 b (n-1)+120 a^4-10 a b^3 \left (n^3-6 n^2+11 n-6\right )-b^4 \left (n^4-10 n^3+35 n^2-50 n+24\right )\right ) \text{Hypergeometric2F1}\left (2,1-n,2-n,\frac{a \cos (c+d x)}{a \cos (c+d x)+b}\right )+(n-1) \left (2 a^2 b (18-7 n)-84 a^3+4 a b^2 \left (2 n^2-9 n+9\right )+b^3 \left (n^3-9 n^2+26 n-24\right )\right ) (a \cos (c+d x)+b)^2\right )+a (1-n) \left (96 a^2+4 a b (6-4 n)-4 b^2 \left (n^2-7 n+12\right )\right ) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2+192 a^3 (n-1) (a \cos (c+d x)+b)^2+40 a^2 (n-1) (2 a-b (n-3)) \sec ^4\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2-240 a^3 (n-1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2-24 a^2 (n-1) (2 a-b (n-4)) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2\right )}{120 a^4 d (n-1) (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sec[c + d*x])^n*Sin[c + d*x]^5,x]

[Out]

-(Cos[(c + d*x)/2]^6*Cos[c + d*x]*(192*a^3*(-1 + n)*(b + a*Cos[c + d*x])^2 - 240*a^3*(-1 + n)*(b + a*Cos[c + d
*x])^2*Sec[(c + d*x)/2]^2 - 24*a^2*(2*a - b*(-4 + n))*(-1 + n)*(b + a*Cos[c + d*x])^2*Sec[(c + d*x)/2]^2 + 40*
a^2*(2*a - b*(-3 + n))*(-1 + n)*(b + a*Cos[c + d*x])^2*Sec[(c + d*x)/2]^4 + a*(1 - n)*(96*a^2 + 4*a*b*(6 - 4*n
) - 4*b^2*(12 - 7*n + n^2))*(b + a*Cos[c + d*x])^2*Sec[(c + d*x)/2]^4 - 10*a*((-1 + n)*(-14*a^2 + 2*a*b*(-1 +
n) + b^2*(6 - 5*n + n^2))*(b + a*Cos[c + d*x])^2 + b*(24*a^3 + 12*a^2*b*(-1 + n) - 4*a*b^2*(2 - 3*n + n^2) - b
^3*(-6 + 11*n - 6*n^2 + n^3))*Hypergeometric2F1[2, 1 - n, 2 - n, (a*Cos[c + d*x])/(b + a*Cos[c + d*x])])*Sec[(
c + d*x)/2]^6 + ((-1 + n)*(-84*a^3 + 2*a^2*b*(18 - 7*n) + 4*a*b^2*(9 - 9*n + 2*n^2) + b^3*(-24 + 26*n - 9*n^2
+ n^3))*(b + a*Cos[c + d*x])^2 + b*(120*a^4 + 120*a^3*b*(-1 + n) - 10*a*b^3*(-6 + 11*n - 6*n^2 + n^3) - b^4*(2
4 - 50*n + 35*n^2 - 10*n^3 + n^4))*Hypergeometric2F1[2, 1 - n, 2 - n, (a*Cos[c + d*x])/(b + a*Cos[c + d*x])])*
Sec[(c + d*x)/2]^6)*(a + b*Sec[c + d*x])^n)/(120*a^4*d*(-1 + n)*(b + a*Cos[c + d*x]))

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Maple [F]  time = 0.753, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( dx+c \right ) \right ) ^{n} \left ( \sin \left ( dx+c \right ) \right ) ^{5}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^n*sin(d*x+c)^5,x)

[Out]

int((a+b*sec(d*x+c))^n*sin(d*x+c)^5,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^n*sin(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^n*sin(d*x + c)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^n*sin(d*x+c)^5,x, algorithm="fricas")

[Out]

integral((cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*(b*sec(d*x + c) + a)^n*sin(d*x + c), 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*sec(d*x+c))**n*sin(d*x+c)**5,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^n*sin(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^n*sin(d*x + c)^5, x)