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

Optimal. Leaf size=150 \[ \frac {b^5 (a+b \sec (c+d x))^{n+1} \, _2F_1\left (6,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a^6 d (n+1)}-\frac {2 b^3 (a+b \sec (c+d x))^{n+1} \, _2F_1\left (4,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a^4 d (n+1)}+\frac {b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a^2 d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 150, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [B]  time = 8.93, size = 562, normalized size = 3.75 \[ -\frac {\cos ^6\left (\frac {1}{2} (c+d x)\right ) \cos (c+d x) (a+b \sec (c+d x))^n \left (192 a^3 (n-1) (a \cos (c+d x)+b)^2-240 a^3 (n-1) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2+a (1-n) \left (96 a^2+4 a b (6-4 n)-4 b^2 \left (n^2-7 n+12\right )\right ) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2+40 a^2 (n-1) (2 a-b (n-3)) \sec ^4\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2-24 a^2 (n-1) (2 a-b (n-4)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)^2-10 a \sec ^6\left (\frac {1}{2} (c+d x)\right ) \left ((n-1) \left (-14 a^2+2 a b (n-1)+b^2 \left (n^2-5 n+6\right )\right ) (a \cos (c+d x)+b)^2+b \left (24 a^3+12 a^2 b (n-1)-4 a b^2 \left (n^2-3 n+2\right )-b^3 \left (n^3-6 n^2+11 n-6\right )\right ) \, _2F_1\left (2,1-n;2-n;\frac {a \cos (c+d x)}{b+a \cos (c+d x)}\right )\right )+\sec ^6\left (\frac {1}{2} (c+d x)\right ) \left (b \left (120 a^4+120 a^3 b (n-1)-10 a b^3 \left (n^3-6 n^2+11 n-6\right )-b^4 \left (n^4-10 n^3+35 n^2-50 n+24\right )\right ) \, _2F_1\left (2,1-n;2-n;\frac {a \cos (c+d x)}{b+a \cos (c+d x)}\right )+(n-1) \left (-84 a^3+2 a^2 b (18-7 n)+4 a b^2 \left (2 n^2-9 n+9\right )+b^3 \left (n^3-9 n^2+26 n-24\right )\right ) (a \cos (c+d x)+b)^2\right )\right )}{120 a^4 d (n-1) (a \cos (c+d x)+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/120*(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*(24 - 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)/(a^4*d*(-1 + n)*(b + a*Cos[c + d*x]))

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fricas [F]  time = 0.69, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right ), x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5}\,{d x} \]

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)

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maple [F]  time = 2.72, size = 0, normalized size = 0.00 \[ \int \left (a +b \sec \left (d x +c \right )\right )^{n} \left (\sin ^{5}\left (d x +c \right )\right )\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{5}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (c+d\,x\right )}^5\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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