\(\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx\) [1509]

Optimal result

Integrand size = 29, antiderivative size = 295 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=-\frac {\left (6 a^2 b^2 \left (1-n^2\right )-a^4 \left (3-4 n+n^2\right )-b^4 \left (3+4 n+n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{8 d (1+n)}-\frac {a b n \left (a^2 (2-n)-b^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{2 d (2+n)}+\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d} \]

[Out]

-1/8*(6*a^2*b^2*(-n^2+1)-a^4*(n^2-4*n+3)-b^4*(n^2+4*n+3))*hypergeom([1, 1/2+1/2*n],[3/2+1/2*n],sin(d*x+c)^2)*s
in(d*x+c)^(1+n)/d/(1+n)-1/2*a*b*n*(a^2*(2-n)-b^2*(2+n))*hypergeom([1, 1+1/2*n],[1/2*n+2],sin(d*x+c)^2)*sin(d*x
+c)^(2+n)/d/(2+n)+1/4*sec(d*x+c)^4*sin(d*x+c)^(1+n)*(a^4+6*a^2*b^2+b^4+4*a*b*(a^2+b^2)*sin(d*x+c))/d+1/8*sec(d
*x+c)^2*sin(d*x+c)^(1+n)*(a^4*(3-n)-6*a^2*b^2*(1+n)-b^4*(5+n)+4*a*b*(a^2*(2-n)-b^2*(2+n))*sin(d*x+c))/d

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2916, 1820, 822, 371} \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=-\frac {a b n \left (a^2 (2-n)-b^2 (n+2)\right ) \sin ^{n+2}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {n+2}{2},\frac {n+4}{2},\sin ^2(c+d x)\right )}{2 d (n+2)}-\frac {\left (-\left (a^4 \left (n^2-4 n+3\right )\right )+6 a^2 b^2 \left (1-n^2\right )-b^4 \left (n^2+4 n+3\right )\right ) \sin ^{n+1}(c+d x) \operatorname {Hypergeometric2F1}\left (1,\frac {n+1}{2},\frac {n+3}{2},\sin ^2(c+d x)\right )}{8 d (n+1)}+\frac {\sec ^4(c+d x) \sin ^{n+1}(c+d x) \left (a^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+b^4\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{n+1}(c+d x) \left (a^4 (3-n)+4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)-b^4 (n+5)\right )}{8 d} \]

[In]

Int[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^4,x]

[Out]

-1/8*((6*a^2*b^2*(1 - n^2) - a^4*(3 - 4*n + n^2) - b^4*(3 + 4*n + n^2))*Hypergeometric2F1[1, (1 + n)/2, (3 + n
)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(1 + n))/(d*(1 + n)) - (a*b*n*(a^2*(2 - n) - b^2*(2 + n))*Hypergeometric2F1[
1, (2 + n)/2, (4 + n)/2, Sin[c + d*x]^2]*Sin[c + d*x]^(2 + n))/(2*d*(2 + n)) + (Sec[c + d*x]^4*Sin[c + d*x]^(1
+ n)*(a^4 + 6*a^2*b^2 + b^4 + 4*a*b*(a^2 + b^2)*Sin[c + d*x]))/(4*d) + (Sec[c + d*x]^2*Sin[c + d*x]^(1 + n)*(
a^4*(3 - n) - 6*a^2*b^2*(1 + n) - b^4*(5 + n) + 4*a*b*(a^2*(2 - n) - b^2*(2 + n))*Sin[c + d*x]))/(8*d)

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 822

Int[((e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[f, Int[(e*x)^m*(a + c*
x^2)^p, x], x] + Dist[g/e, Int[(e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, p}, x] &&  !Ration
alQ[m] &&  !IGtQ[p, 0]

Rule 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x
^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x
], x, 1]}, Simp[(-(c*x)^(m + 1))*(f + g*x)*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[1/(2*a*(p + 1)), I
nt[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(m + 2*p + 3) + g*(m + 2*p + 4)*x, x], x], x]] /;
FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]

Rule 2916

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

Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^4}{\left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}-\frac {b^3 \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (-a^4 (3-n)+6 a^2 b^2 (1+n)+b^4 (1+n)-4 a \left (a^2 (2-n)-b^2 (2+n)\right ) x+4 b^2 x^2\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 d} \\ & = \frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d}+\frac {b \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (8 b^4+(1-n) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)\right )-4 a n \left (a^2 (2-n)-b^2 (2+n)\right ) x\right )}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = \frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d}-\frac {\left (a b^2 n \left (a^2 (2-n)-b^2 (2+n)\right )\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^{1+n}}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 d}-\frac {\left (b \left (6 a^2 b^2 \left (1-n^2\right )-a^4 \left (3-4 n+n^2\right )-b^4 \left (3+4 n+n^2\right )\right )\right ) \text {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d} \\ & = -\frac {\left (6 a^2 b^2 \left (1-n^2\right )-a^4 \left (3-4 n+n^2\right )-b^4 \left (3+4 n+n^2\right )\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right ) \sin ^{1+n}(c+d x)}{8 d (1+n)}-\frac {a b n \left (a^2 (2-n)-b^2 (2+n)\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {2+n}{2},\frac {4+n}{2},\sin ^2(c+d x)\right ) \sin ^{2+n}(c+d x)}{2 d (2+n)}+\frac {\sec ^4(c+d x) \sin ^{1+n}(c+d x) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \sin (c+d x)\right )}{4 d}+\frac {\sec ^2(c+d x) \sin ^{1+n}(c+d x) \left (a^4 (3-n)-6 a^2 b^2 (1+n)-b^4 (5+n)+4 a b \left (a^2 (2-n)-b^2 (2+n)\right ) \sin (c+d x)\right )}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.56 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\frac {\left (6 \left (a^2-b^2\right )^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+n}{2},\frac {3+n}{2},\sin ^2(c+d x)\right )+(a-b)^3 (3 a+5 b) \operatorname {Hypergeometric2F1}(2,1+n,2+n,-\sin (c+d x))+(3 a-5 b) (a+b)^3 \operatorname {Hypergeometric2F1}(2,1+n,2+n,\sin (c+d x))+2 (a-b)^4 \operatorname {Hypergeometric2F1}(3,1+n,2+n,-\sin (c+d x))+2 (a+b)^4 \operatorname {Hypergeometric2F1}(3,1+n,2+n,\sin (c+d x))\right ) \sin ^{1+n}(c+d x)}{16 d (1+n)} \]

[In]

Integrate[Sec[c + d*x]^5*Sin[c + d*x]^n*(a + b*Sin[c + d*x])^4,x]

[Out]

((6*(a^2 - b^2)^2*Hypergeometric2F1[1, (1 + n)/2, (3 + n)/2, Sin[c + d*x]^2] + (a - b)^3*(3*a + 5*b)*Hypergeom
etric2F1[2, 1 + n, 2 + n, -Sin[c + d*x]] + (3*a - 5*b)*(a + b)^3*Hypergeometric2F1[2, 1 + n, 2 + n, Sin[c + d*
x]] + 2*(a - b)^4*Hypergeometric2F1[3, 1 + n, 2 + n, -Sin[c + d*x]] + 2*(a + b)^4*Hypergeometric2F1[3, 1 + n,
2 + n, Sin[c + d*x]])*Sin[c + d*x]^(1 + n))/(16*d*(1 + n))

Maple [F]

\[\int \left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{4}d x\]

[In]

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^4,x)

[Out]

int(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^4,x)

Fricas [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^4,x, algorithm="fricas")

[Out]

integral(-(4*(a*b^3*cos(d*x + c)^2 - a^3*b - a*b^3)*sec(d*x + c)^5*sin(d*x + c) - (b^4*cos(d*x + c)^4 + a^4 +
6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)*cos(d*x + c)^2)*sec(d*x + c)^5)*sin(d*x + c)^n, x)

Sympy [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)**5*sin(d*x+c)**n*(a+b*sin(d*x+c))**4,x)

[Out]

Timed out

Maxima [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^4,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^4*sin(d*x + c)^n*sec(d*x + c)^5, x)

Giac [F]

\[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{4} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5} \,d x } \]

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^n*(a+b*sin(d*x+c))^4,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^4*sin(d*x + c)^n*sec(d*x + c)^5, x)

Mupad [F(-1)]

Timed out. \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx=\int \frac {{\sin \left (c+d\,x\right )}^n\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^4}{{\cos \left (c+d\,x\right )}^5} \,d x \]

[In]

int((sin(c + d*x)^n*(a + b*sin(c + d*x))^4)/cos(c + d*x)^5,x)

[Out]

int((sin(c + d*x)^n*(a + b*sin(c + d*x))^4)/cos(c + d*x)^5, x)