3.1509 \(\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx\)

Optimal. Leaf size=295 \[ -\frac{\left (6 a^2 b^2 \left (1-n^2\right )+a^4 \left (-\left (n^2-4 n+3\right )\right )-b^4 \left (n^2+4 n+3\right )\right ) \sin ^{n+1}(c+d x) \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )}{8 d (n+1)}-\frac{a b n \left (a^2 (2-n)-b^2 (n+2)\right ) \sin ^{n+2}(c+d x) \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};\sin ^2(c+d x)\right )}{2 d (n+2)}+\frac{\sec ^4(c+d x) \sin ^{n+1}(c+d x) \left (4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+a^4+b^4\right )}{4 d}+\frac{\sec ^2(c+d x) \sin ^{n+1}(c+d x) \left (4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)+a^4 (3-n)-b^4 (n+5)\right )}{8 d} \]

[Out]

-((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))/(8*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)

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Rubi [A]  time = 0.530405, antiderivative size = 295, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2837, 1806, 808, 364} \[ -\frac{\left (6 a^2 b^2 \left (1-n^2\right )+a^4 \left (-\left (n^2-4 n+3\right )\right )-b^4 \left (n^2+4 n+3\right )\right ) \sin ^{n+1}(c+d x) \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )}{8 d (n+1)}-\frac{a b n \left (a^2 (2-n)-b^2 (n+2)\right ) \sin ^{n+2}(c+d x) \, _2F_1\left (1,\frac{n+2}{2};\frac{n+4}{2};\sin ^2(c+d x)\right )}{2 d (n+2)}+\frac{\sec ^4(c+d x) \sin ^{n+1}(c+d x) \left (4 a b \left (a^2+b^2\right ) \sin (c+d x)+6 a^2 b^2+a^4+b^4\right )}{4 d}+\frac{\sec ^2(c+d x) \sin ^{n+1}(c+d x) \left (4 a b \left (a^2 (2-n)-b^2 (n+2)\right ) \sin (c+d x)-6 a^2 b^2 (n+1)+a^4 (3-n)-b^4 (n+5)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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))/(8*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 2837

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*x)/b)^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]

Rule 1806

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)), Int
[(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]] /; F
reeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && LtQ[p, -1] &&  !GtQ[m, 0]

Rule 808

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 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^4 \, dx &=\frac{b^5 \operatorname{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 \operatorname{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 \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \, _2F_1\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 ) \, _2F_1\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]  time = 0.196805, size = 164, normalized size = 0.56 \[ \frac{\sin ^{n+1}(c+d x) \left (6 \left (a^2-b^2\right )^2 \, _2F_1\left (1,\frac{n+1}{2};\frac{n+3}{2};\sin ^2(c+d x)\right )+2 (a-b)^4 \, _2F_1(3,n+1;n+2;-\sin (c+d x))+(3 a+5 b) (a-b)^3 \, _2F_1(2,n+1;n+2;-\sin (c+d x))+(3 a-5 b) (a+b)^3 \, _2F_1(2,n+1;n+2;\sin (c+d x))+2 (a+b)^4 \, _2F_1(3,n+1;n+2;\sin (c+d x))\right )}{16 d (n+1)} \]

Antiderivative was successfully verified.

[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))

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Maple [F]  time = 1.954, size = 0, normalized size = 0. \begin{align*} \int \left ( \sec \left ( dx+c \right ) \right ) ^{5} \left ( \sin \left ( dx+c \right ) \right ) ^{n} \left ( a+b\sin \left ( dx+c \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (4 \,{\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a b^{3}\right )} \sec \left (d x + c\right )^{5} \sin \left (d x + c\right ) -{\left (b^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sec \left (d x + c\right )^{5}\right )} \sin \left (d x + c\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)