3.2.45 \(\int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx\) [145]

Optimal. Leaf size=180 \[ -\frac {(3-n) (8-n) (16-n) \, _2F_1(6,4+n;5+n;1+\sec (c+d x)) (a+a \sec (c+d x))^{4+n}}{42 a^4 d (1-n) (4+n)}-\frac {\cos ^7(c+d x) (1-\sec (c+d x))^2 (a+a \sec (c+d x))^{4+n}}{a^4 d (1-n)}+\frac {\cos ^7(c+d x) (a+a \sec (c+d x))^{4+n} \left (6 (8-n)-\left (108-25 n+n^2\right ) \sec (c+d x)\right )}{42 a^4 d (1-n)} \]

[Out]

-1/42*(3-n)*(8-n)*(16-n)*hypergeom([6, 4+n],[5+n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(4+n)/a^4/d/(-n^2-3*n+4)-cos(
d*x+c)^7*(1-sec(d*x+c))^2*(a+a*sec(d*x+c))^(4+n)/a^4/d/(1-n)+1/42*cos(d*x+c)^7*(a+a*sec(d*x+c))^(4+n)*(48-6*n-
(n^2-25*n+108)*sec(d*x+c))/a^4/d/(1-n)

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Rubi [A]
time = 0.12, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3958, 102, 150, 67} \begin {gather*} -\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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/42*((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))/(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 67

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

Rule 102

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 150

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)/(b^2*(b*c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(
n + 1), 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 3958

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]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^n \sin ^7(c+d x) \, dx &=-\frac {\text {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 {\text {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)) \text {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*}

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Mathematica [A]
time = 0.99, size = 113, normalized size = 0.63 \begin {gather*} \frac {\left ((4+n) \cos ^5(c+d x) \left (42+\left (24-25 n+n^2\right ) \cos (c+d x)+6 (-1+n) \cos ^2(c+d x)\right )-\left (-384+200 n-27 n^2+n^3\right ) \, _2F_1(6,4+n;5+n;1+\sec (c+d x))\right ) (1+\sec (c+d x))^4 (a (1+\sec (c+d x)))^n}{42 d (-1+n) (4+n)} \end {gather*}

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 = 0.32, size = 0, normalized size = 0.00 \[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \left (\sin ^{7}\left (d x +c \right )\right )\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 8010 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (c+d\,x\right )}^7\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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