3.1513 \(\int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=360 \[ \frac {\left (3 a^2-9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^3}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^3}-\frac {b^6 \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a d (n+1) \left (a^2-b^2\right )^3}+\frac {(3 a-5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^2}+\frac {(3 a+5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^2}+\frac {\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;-\sin (c+d x))}{8 d (n+1) (a-b)}+\frac {\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;\sin (c+d x))}{8 d (n+1) (a+b)} \]

[Out]

1/16*(3*a^2-9*a*b+8*b^2)*hypergeom([1, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^(1+n)/(a-b)^3/d/(1+n)+1/16*(3*a^2+9*
a*b+8*b^2)*hypergeom([1, 1+n],[2+n],sin(d*x+c))*sin(d*x+c)^(1+n)/(a+b)^3/d/(1+n)-b^6*hypergeom([1, 1+n],[2+n],
-b*sin(d*x+c)/a)*sin(d*x+c)^(1+n)/a/(a^2-b^2)^3/d/(1+n)+1/16*(3*a-5*b)*hypergeom([2, 1+n],[2+n],-sin(d*x+c))*s
in(d*x+c)^(1+n)/(a-b)^2/d/(1+n)+1/16*(3*a+5*b)*hypergeom([2, 1+n],[2+n],sin(d*x+c))*sin(d*x+c)^(1+n)/(a+b)^2/d
/(1+n)+1/8*hypergeom([3, 1+n],[2+n],-sin(d*x+c))*sin(d*x+c)^(1+n)/(a-b)/d/(1+n)+1/8*hypergeom([3, 1+n],[2+n],s
in(d*x+c))*sin(d*x+c)^(1+n)/(a+b)/d/(1+n)

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Rubi [A]  time = 0.54, antiderivative size = 360, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2837, 961, 64} \[ \frac {\left (3 a^2-9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^3}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \sin ^{n+1}(c+d x) \, _2F_1(1,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^3}-\frac {b^6 \sin ^{n+1}(c+d x) \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a d (n+1) \left (a^2-b^2\right )^3}+\frac {(3 a-5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;-\sin (c+d x))}{16 d (n+1) (a-b)^2}+\frac {(3 a+5 b) \sin ^{n+1}(c+d x) \, _2F_1(2,n+1;n+2;\sin (c+d x))}{16 d (n+1) (a+b)^2}+\frac {\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;-\sin (c+d x))}{8 d (n+1) (a-b)}+\frac {\sin ^{n+1}(c+d x) \, _2F_1(3,n+1;n+2;\sin (c+d x))}{8 d (n+1) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 - 9*a*b + 8*b^2)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(16*(a - b)^3
*d*(1 + n)) + ((3*a^2 + 9*a*b + 8*b^2)*Hypergeometric2F1[1, 1 + n, 2 + n, Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/
(16*(a + b)^3*d*(1 + n)) - (b^6*Hypergeometric2F1[1, 1 + n, 2 + n, -((b*Sin[c + d*x])/a)]*Sin[c + d*x]^(1 + n)
)/(a*(a^2 - b^2)^3*d*(1 + n)) + ((3*a - 5*b)*Hypergeometric2F1[2, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1
+ n))/(16*(a - b)^2*d*(1 + n)) + ((3*a + 5*b)*Hypergeometric2F1[2, 1 + n, 2 + n, Sin[c + d*x]]*Sin[c + d*x]^(
1 + n))/(16*(a + b)^2*d*(1 + n)) + (Hypergeometric2F1[3, 1 + n, 2 + n, -Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(8
*(a - b)*d*(1 + n)) + (Hypergeometric2F1[3, 1 + n, 2 + n, Sin[c + d*x]]*Sin[c + d*x]^(1 + n))/(8*(a + b)*d*(1
+ n))

Rule 64

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

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

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]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x) \sin ^n(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {b^5 \operatorname {Subst}\left (\int \left (\frac {\left (\frac {x}{b}\right )^n}{8 b^3 (a+b) (b-x)^3}+\frac {(3 a+5 b) \left (\frac {x}{b}\right )^n}{16 b^4 (a+b)^2 (b-x)^2}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \left (\frac {x}{b}\right )^n}{16 b^5 (a+b)^3 (b-x)}-\frac {\left (\frac {x}{b}\right )^n}{(a-b)^3 (a+b)^3 (a+x)}-\frac {\left (\frac {x}{b}\right )^n}{8 b^3 (-a+b) (b+x)^3}+\frac {(3 a-5 b) \left (\frac {x}{b}\right )^n}{16 (a-b)^2 b^4 (b+x)^2}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \left (\frac {x}{b}\right )^n}{16 (a-b)^3 b^5 (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {((3 a-5 b) b) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b+x)^2} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^2 d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b+x)^3} \, dx,x,b \sin (c+d x)\right )}{8 (a-b) d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b-x)^3} \, dx,x,b \sin (c+d x)\right )}{8 (a+b) d}+\frac {(b (3 a+5 b)) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{(b-x)^2} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^2 d}-\frac {b^5 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{a+x} \, dx,x,b \sin (c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac {\left (3 a^2-9 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 (a-b)^3 d}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 (a+b)^3 d}\\ &=\frac {\left (3 a^2-9 a b+8 b^2\right ) \, _2F_1(1,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^3 d (1+n)}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \, _2F_1(1,1+n;2+n;\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^3 d (1+n)}-\frac {b^6 \, _2F_1\left (1,1+n;2+n;-\frac {b \sin (c+d x)}{a}\right ) \sin ^{1+n}(c+d x)}{a \left (a^2-b^2\right )^3 d (1+n)}+\frac {(3 a-5 b) \, _2F_1(2,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a-b)^2 d (1+n)}+\frac {(3 a+5 b) \, _2F_1(2,1+n;2+n;\sin (c+d x)) \sin ^{1+n}(c+d x)}{16 (a+b)^2 d (1+n)}+\frac {\, _2F_1(3,1+n;2+n;-\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a-b) d (1+n)}+\frac {\, _2F_1(3,1+n;2+n;\sin (c+d x)) \sin ^{1+n}(c+d x)}{8 (a+b) d (1+n)}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 241, normalized size = 0.67 \[ \frac {\sin ^{n+1}(c+d x) \left (\frac {\left (3 a^2-9 a b+8 b^2\right ) \, _2F_1(1,n+1;n+2;-\sin (c+d x))}{(a-b)^3}+\frac {\left (3 a^2+9 a b+8 b^2\right ) \, _2F_1(1,n+1;n+2;\sin (c+d x))}{(a+b)^3}-\frac {16 b^6 \, _2F_1\left (1,n+1;n+2;-\frac {b \sin (c+d x)}{a}\right )}{a (a-b)^3 (a+b)^3}+\frac {(3 a-5 b) \, _2F_1(2,n+1;n+2;-\sin (c+d x))}{(a-b)^2}+\frac {(3 a+5 b) \, _2F_1(2,n+1;n+2;\sin (c+d x))}{(a+b)^2}+\frac {2 \, _2F_1(3,n+1;n+2;-\sin (c+d x))}{a-b}+\frac {2 \, _2F_1(3,n+1;n+2;\sin (c+d x))}{a+b}\right )}{16 d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((((3*a^2 - 9*a*b + 8*b^2)*Hypergeometric2F1[1, 1 + n, 2 + n, -Sin[c + d*x]])/(a - b)^3 + ((3*a^2 + 9*a*b + 8*
b^2)*Hypergeometric2F1[1, 1 + n, 2 + n, Sin[c + d*x]])/(a + b)^3 - (16*b^6*Hypergeometric2F1[1, 1 + n, 2 + n,
-((b*Sin[c + d*x])/a)])/(a*(a - b)^3*(a + b)^3) + ((3*a - 5*b)*Hypergeometric2F1[2, 1 + n, 2 + n, -Sin[c + d*x
]])/(a - b)^2 + ((3*a + 5*b)*Hypergeometric2F1[2, 1 + n, 2 + n, Sin[c + d*x]])/(a + b)^2 + (2*Hypergeometric2F
1[3, 1 + n, 2 + n, -Sin[c + d*x]])/(a - b) + (2*Hypergeometric2F1[3, 1 + n, 2 + n, Sin[c + d*x]])/(a + b))*Sin
[c + d*x]^(1 + n))/(16*d*(1 + n))

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}}{b \sin \left (d x + c\right ) + a}, x\right ) \]

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)),x, algorithm="fricas")

[Out]

integral(sin(d*x + c)^n*sec(d*x + c)^5/(b*sin(d*x + c) + a), x)

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

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)),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.91, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

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)),x)

[Out]

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

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

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)),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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