3.1514 \(\int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx\)

Optimal. Leaf size=487 \[ \frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{8 d (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{8 d (n+1)} \]

[Out]

3/16*AppellF1(1+n,-p,1,2+n,-b*sin(d*x+c)/a,-sin(d*x+c))*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n)/((1+b*sin(
d*x+c)/a)^p)+3/16*AppellF1(1+n,-p,1,2+n,-b*sin(d*x+c)/a,sin(d*x+c))*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n
)/((1+b*sin(d*x+c)/a)^p)+3/16*AppellF1(1+n,-p,2,2+n,-b*sin(d*x+c)/a,-sin(d*x+c))*sin(d*x+c)^(1+n)*(a+b*sin(d*x
+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)+3/16*AppellF1(1+n,-p,2,2+n,-b*sin(d*x+c)/a,sin(d*x+c))*sin(d*x+c)^(1+n)*
(a+b*sin(d*x+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)+1/8*AppellF1(1+n,-p,3,2+n,-b*sin(d*x+c)/a,-sin(d*x+c))*sin(d
*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)+1/8*AppellF1(1+n,-p,3,2+n,-b*sin(d*x+c)/a,sin(d*
x+c))*sin(d*x+c)^(1+n)*(a+b*sin(d*x+c))^p/d/(1+n)/((1+b*sin(d*x+c)/a)^p)

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Rubi [A]  time = 0.57, antiderivative size = 487, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2837, 961, 135, 133, 912} \[ \frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,1;n+2;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{16 d (n+1)}+\frac {3 \sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,2;n+2;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{16 d (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right )}{8 d (n+1)}+\frac {\sin ^{n+1}(c+d x) (a+b \sin (c+d x))^p \left (\frac {b \sin (c+d x)}{a}+1\right )^{-p} F_1\left (n+1;-p,3;n+2;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right )}{8 d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*AppellF1[1 + n, -p, 1, 2 + n, -((b*Sin[c + d*x])/a), -Sin[c + d*x]]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x
])^p)/(16*d*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (3*AppellF1[1 + n, -p, 1, 2 + n, -((b*Sin[c + d*x])/a), Sin[
c + d*x]]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^p)/(16*d*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (3*AppellF1
[1 + n, -p, 2, 2 + n, -((b*Sin[c + d*x])/a), -Sin[c + d*x]]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^p)/(16*d
*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (3*AppellF1[1 + n, -p, 2, 2 + n, -((b*Sin[c + d*x])/a), Sin[c + d*x]]*S
in[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^p)/(16*d*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p) + (AppellF1[1 + n, -p, 3
, 2 + n, -((b*Sin[c + d*x])/a), -Sin[c + d*x]]*Sin[c + d*x]^(1 + n)*(a + b*Sin[c + d*x])^p)/(8*d*(1 + n)*(1 +
(b*Sin[c + d*x])/a)^p) + (AppellF1[1 + n, -p, 3, 2 + n, -((b*Sin[c + d*x])/a), Sin[c + d*x]]*Sin[c + d*x]^(1 +
n)*(a + b*Sin[c + d*x])^p)/(8*d*(1 + n)*(1 + (b*Sin[c + d*x])/a)^p)

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 135

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 912

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] &&  !IntegerQ[m] &&  !IntegerQ[n]

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 \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{\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 (a+x)^p}{8 b^3 (b-x)^3}+\frac {3 \left (\frac {x}{b}\right )^n (a+x)^p}{16 b^4 (b-x)^2}+\frac {\left (\frac {x}{b}\right )^n (a+x)^p}{8 b^3 (b+x)^3}+\frac {3 \left (\frac {x}{b}\right )^n (a+x)^p}{16 b^4 (b+x)^2}+\frac {3 \left (\frac {x}{b}\right )^n (a+x)^p}{8 b^4 \left (b^2-x^2\right )}\right ) \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{(b-x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{(b+x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {(3 b) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{(b-x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac {b^2 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{(b+x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac {(3 b) \operatorname {Subst}\left (\int \left (\frac {\left (\frac {x}{b}\right )^n (a+x)^p}{2 b (b-x)}+\frac {\left (\frac {x}{b}\right )^n (a+x)^p}{2 b (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac {\left (3 b (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (1+\frac {x}{a}\right )^p}{(b-x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (3 b (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (1+\frac {x}{a}\right )^p}{(b+x)^2} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (b^2 (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (1+\frac {x}{a}\right )^p}{(b-x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}+\frac {\left (b^2 (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (1+\frac {x}{a}\right )^p}{(b+x)^3} \, dx,x,b \sin (c+d x)\right )}{8 d}\\ &=\frac {3 F_1\left (1+n;-p,2;2+n;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 F_1\left (1+n;-p,2;2+n;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {F_1\left (1+n;-p,3;2+n;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac {F_1\left (1+n;-p,3;2+n;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac {3 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {3 \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n (a+x)^p}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac {3 F_1\left (1+n;-p,2;2+n;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 F_1\left (1+n;-p,2;2+n;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {F_1\left (1+n;-p,3;2+n;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac {F_1\left (1+n;-p,3;2+n;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac {\left (3 (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (1+\frac {x}{a}\right )^p}{b-x} \, dx,x,b \sin (c+d x)\right )}{16 d}+\frac {\left (3 (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {x}{b}\right )^n \left (1+\frac {x}{a}\right )^p}{b+x} \, dx,x,b \sin (c+d x)\right )}{16 d}\\ &=\frac {3 F_1\left (1+n;-p,1;2+n;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 F_1\left (1+n;-p,1;2+n;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 F_1\left (1+n;-p,2;2+n;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {3 F_1\left (1+n;-p,2;2+n;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{16 d (1+n)}+\frac {F_1\left (1+n;-p,3;2+n;-\frac {b \sin (c+d x)}{a},-\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}+\frac {F_1\left (1+n;-p,3;2+n;-\frac {b \sin (c+d x)}{a},\sin (c+d x)\right ) \sin ^{1+n}(c+d x) (a+b \sin (c+d x))^p \left (1+\frac {b \sin (c+d x)}{a}\right )^{-p}}{8 d (1+n)}\\ \end {align*}

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Mathematica [F]  time = 15.01, size = 0, normalized size = 0.00 \[ \int \sec ^5(c+d x) \sin ^n(c+d x) (a+b \sin (c+d x))^p \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 1.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sin \left (d x + c\right ) + a\right )}^{p} \sin \left (d x + c\right )^{n} \sec \left (d x + c\right )^{5}, x\right ) \]

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))^p,x, algorithm="fricas")

[Out]

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

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

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))^p,x, algorithm="giac")

[Out]

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

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maple [F]  time = 2.26, size = 0, normalized size = 0.00 \[ \int \left (\sec ^{5}\left (d x +c \right )\right ) \left (\sin ^{n}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{p}\, dx \]

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))^p,x)

[Out]

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

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maxima [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))^p,x, algorithm="maxima")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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