3.145 \(\int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx\)

Optimal. Leaf size=180 \[ -\frac{(3-n) (8-n) (16-n) (a \sec (c+d x)+a)^{n+4} \text{Hypergeometric2F1}(6,n+4,n+5,\sec (c+d x)+1)}{42 a^4 d (1-n) (n+4)}+\frac{\cos ^7(c+d x) \left (6 (8-n)-\left (n^2-25 n+108\right ) \sec (c+d x)\right ) (a \sec (c+d x)+a)^{n+4}}{42 a^4 d (1-n)}-\frac{\cos ^7(c+d x) (1-\sec (c+d x))^2 (a \sec (c+d x)+a)^{n+4}}{a^4 d (1-n)} \]

[Out]

-((3 - n)*(8 - n)*(16 - n)*Hypergeometric2F1[6, 4 + n, 5 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(4 + n))/
(42*a^4*d*(1 - n)*(4 + n)) - (Cos[c + d*x]^7*(1 - Sec[c + d*x])^2*(a + a*Sec[c + d*x])^(4 + n))/(a^4*d*(1 - n)
) + (Cos[c + d*x]^7*(a + a*Sec[c + d*x])^(4 + n)*(6*(8 - n) - (108 - 25*n + n^2)*Sec[c + d*x]))/(42*a^4*d*(1 -
n))

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Rubi [A]  time = 0.1687, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3873, 100, 145, 65} \[ -\frac{(3-n) (8-n) (16-n) (a \sec (c+d x)+a)^{n+4} \, _2F_1(6,n+4;n+5;\sec (c+d x)+1)}{42 a^4 d (1-n) (n+4)}+\frac{\cos ^7(c+d x) \left (6 (8-n)-\left (n^2-25 n+108\right ) \sec (c+d x)\right ) (a \sec (c+d x)+a)^{n+4}}{42 a^4 d (1-n)}-\frac{\cos ^7(c+d x) (1-\sec (c+d x))^2 (a \sec (c+d x)+a)^{n+4}}{a^4 d (1-n)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^7,x]

[Out]

-((3 - n)*(8 - n)*(16 - n)*Hypergeometric2F1[6, 4 + n, 5 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(4 + n))/
(42*a^4*d*(1 - n)*(4 + n)) - (Cos[c + d*x]^7*(1 - Sec[c + d*x])^2*(a + a*Sec[c + d*x])^(4 + n))/(a^4*d*(1 - n)
) + (Cos[c + d*x]^7*(a + a*Sec[c + d*x])^(4 + n)*(6*(8 - n) - (108 - 25*n + n^2)*Sec[c + d*x]))/(42*a^4*d*(1 -
n))

Rule 3873

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

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

Rule 145

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g +
e*h) + d*e*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(
f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*(b*c - a*d)^2*(m + 1)*(m
+ 2)), x] + Dist[(f*h)/b^2 - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)
) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)), Int[(a + b*x)^(m +
2)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] &&  !L
tQ[n, -2]))

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+a \sec (c+d x))^n \sin ^7(c+d x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(-a-a x)^3 (a-a x)^{3+n}}{x^8} \, dx,x,-\sec (c+d x)\right )}{a^6 d}\\ &=-\frac{\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}-\frac{\operatorname{Subst}\left (\int \frac{(-a-a x) (a-a x)^{3+n} \left (a^3 (8-n)+a^3 (4-n) x\right )}{x^8} \, dx,x,-\sec (c+d x)\right )}{a^7 d (1-n)}\\ &=-\frac{\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}+\frac{\cos ^7(c+d x) (a+a \sec (c+d x))^{4+n} \left (6 (8-n)-\left (108-25 n+n^2\right ) \sec (c+d x)\right )}{42 a^4 d (1-n)}+\frac{((3-n) (8-n) (16-n)) \operatorname{Subst}\left (\int \frac{(a-a x)^{3+n}}{x^6} \, dx,x,-\sec (c+d x)\right )}{42 a^3 d (1-n)}\\ &=-\frac{(3-n) (8-n) (16-n) \, _2F_1(6,4+n;5+n;1+\sec (c+d x)) (a+a \sec (c+d x))^{4+n}}{42 a^4 d (1-n) (4+n)}-\frac{\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}+\frac{\cos ^7(c+d x) (a+a \sec (c+d x))^{4+n} \left (6 (8-n)-\left (108-25 n+n^2\right ) \sec (c+d x)\right )}{42 a^4 d (1-n)}\\ \end{align*}

Mathematica [A]  time = 1.5236, size = 113, normalized size = 0.63 \[ \frac{(\sec (c+d x)+1)^4 (a (\sec (c+d x)+1))^n \left ((n+4) \cos ^5(c+d x) \left (\left (n^2-25 n+24\right ) \cos (c+d x)+6 (n-1) \cos ^2(c+d x)+42\right )-\left (n^3-27 n^2+200 n-384\right ) \text{Hypergeometric2F1}(6,n+4,n+5,\sec (c+d x)+1)\right )}{42 d (n-1) (n+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^7,x]

[Out]

(((4 + n)*Cos[c + d*x]^5*(42 + (24 - 25*n + n^2)*Cos[c + d*x] + 6*(-1 + n)*Cos[c + d*x]^2) - (-384 + 200*n - 2
7*n^2 + n^3)*Hypergeometric2F1[6, 4 + n, 5 + n, 1 + Sec[c + d*x]])*(1 + Sec[c + d*x])^4*(a*(1 + Sec[c + d*x]))
^n)/(42*d*(-1 + n)*(4 + n))

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Maple [F]  time = 0.701, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \sin \left ( dx+c \right ) \right ) ^{7}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^7,x)

[Out]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^7,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^7,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^7, 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 )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1\right )}{\left (a \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+a*sec(d*x+c))^n*sin(d*x+c)^7,x, algorithm="fricas")

[Out]

integral(-(cos(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1)*(a*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+a*sec(d*x+c))**n*sin(d*x+c)**7,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^7,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^7, x)