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

Optimal. Leaf size=230 \[ \frac{2^{n+\frac{1}{2}} \sin (c+d x) \cos ^n(c+d x) (\cos (c+d x)+1)^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac{1}{2};n-4,\frac{1}{2}-n;\frac{3}{2};1-\cos (c+d x),\frac{1}{2} (1-\cos (c+d x))\right )}{d}-\frac{\cot (c+d x) (n-n \cos (c+d x)) (\cos (c+d x)+1)^{\frac{1}{2}-n} (a \sec (c+d x)+a)^n F_1\left (1-n;-\frac{1}{2},\frac{1}{2}-n;2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sqrt{1-\cos (c+d x)}}-\frac{\sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d} \]

[Out]

-((AppellF1[1 - n, -1/2, 1/2 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 + Cos[c + d*x])^(1/2 - n)*(n - n*Cos[
c + d*x])*Cot[c + d*x]*(a + a*Sec[c + d*x])^n)/(d*(1 - n)*Sqrt[1 - Cos[c + d*x]])) - (Cos[c + d*x]*(a + a*Sec[
c + d*x])^n*Sin[c + d*x])/d + (2^(1/2 + n)*AppellF1[1/2, -4 + n, 1/2 - n, 3/2, 1 - Cos[c + d*x], (1 - Cos[c +
d*x])/2]*Cos[c + d*x]^n*(1 + Cos[c + d*x])^(-1/2 - n)*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/d

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Rubi [A]  time = 0.667858, antiderivative size = 230, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {3876, 2881, 2787, 2786, 2785, 133, 3046, 3008, 135} \[ \frac{2^{n+\frac{1}{2}} \sin (c+d x) \cos ^n(c+d x) (\cos (c+d x)+1)^{-n-\frac{1}{2}} (a \sec (c+d x)+a)^n F_1\left (\frac{1}{2};n-4,\frac{1}{2}-n;\frac{3}{2};1-\cos (c+d x),\frac{1}{2} (1-\cos (c+d x))\right )}{d}-\frac{\cot (c+d x) (n-n \cos (c+d x)) (\cos (c+d x)+1)^{\frac{1}{2}-n} (a \sec (c+d x)+a)^n F_1\left (1-n;-\frac{1}{2},\frac{1}{2}-n;2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n) \sqrt{1-\cos (c+d x)}}-\frac{\sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^n}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((AppellF1[1 - n, -1/2, 1/2 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*(1 + Cos[c + d*x])^(1/2 - n)*(n - n*Cos[
c + d*x])*Cot[c + d*x]*(a + a*Sec[c + d*x])^n)/(d*(1 - n)*Sqrt[1 - Cos[c + d*x]])) - (Cos[c + d*x]*(a + a*Sec[
c + d*x])^n*Sin[c + d*x])/d + (2^(1/2 + n)*AppellF1[1/2, -4 + n, 1/2 - n, 3/2, 1 - Cos[c + d*x], (1 - Cos[c +
d*x])/2]*Cos[c + d*x]^n*(1 + Cos[c + d*x])^(-1/2 - n)*(a + a*Sec[c + d*x])^n*Sin[c + d*x])/d

Rule 3876

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(Sin[
e + f*x]^FracPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(b + a*Sin[e + f*x])^FracPart[m], Int[((g*Cos[e + f*x])
^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])

Rule 2881

Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/d^4, Int[(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^m, x], x] + Int[(d*Sin[e + f*x])^
n*(a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&
!IGtQ[m, 0]

Rule 2787

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Sin[e + f*x])^FracPart[m])/(1 + (b*Sin[e + f*x])/a)^FracPart[m], Int[(1 + (b*Sin[e + f*x])/a)^
m*(d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] &&  !GtQ
[a, 0]

Rule 2786

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[((d/b
)^IntPart[n]*(d*Sin[e + f*x])^FracPart[n])/(b*Sin[e + f*x])^FracPart[n], Int[(a + b*Sin[e + f*x])^m*(b*Sin[e +
f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[m] && GtQ[a, 0] &&  !Gt
Q[d/b, 0]

Rule 2785

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Dist[(b*(d
/b)^n*Cos[e + f*x])/(f*Sqrt[a + b*Sin[e + f*x]]*Sqrt[a - b*Sin[e + f*x]]), Subst[Int[((a - x)^n*(2*a - x)^(m -
1/2))/Sqrt[x], x], x, a - b*Sin[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !In
tegerQ[m] && GtQ[a, 0] && GtQ[d/b, 0]

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 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp
[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-
1)] && NeQ[m + n + 2, 0]

Rule 3008

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

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^n \sin ^4(c+d x) \, dx &=\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n \sin ^4(c+d x) \, dx\\ &=\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{4-n} (-a-a \cos (c+d x))^n \, dx+\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n \left (1-2 \cos ^2(c+d x)\right ) \, dx\\ &=-\frac{\cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d}+\left ((-\cos (c+d x))^n (1+\cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{4-n} (1+\cos (c+d x))^n \, dx+\frac{\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n (2 a n-2 a n \cos (c+d x)) \, dx}{2 a}\\ &=-\frac{\cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d}+\left (\cos ^n(c+d x) (1+\cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int \cos ^{4-n}(c+d x) (1+\cos (c+d x))^n \, dx-\frac{\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{\frac{1}{2}-n} \sqrt{2 a n-2 a n \cos (c+d x)} \csc (c+d x) (a+a \sec (c+d x))^n\right ) \operatorname{Subst}\left (\int (-x)^{-n} (-a-a x)^{-\frac{1}{2}+n} \sqrt{2 a n-2 a n x} \, dx,x,\cos (c+d x)\right )}{2 a d}\\ &=-\frac{\cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d}-\frac{\left ((-\cos (c+d x))^n (1+\cos (c+d x))^{\frac{1}{2}-n} \sqrt{2 a n-2 a n \cos (c+d x)} \csc (c+d x) (a+a \sec (c+d x))^n\right ) \operatorname{Subst}\left (\int (-x)^{-n} (1+x)^{-\frac{1}{2}+n} \sqrt{2 a n-2 a n x} \, dx,x,\cos (c+d x)\right )}{2 a d}+\frac{\left (\cos ^n(c+d x) (1+\cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)\right ) \operatorname{Subst}\left (\int \frac{(1-x)^{4-n} (2-x)^{-\frac{1}{2}+n}}{\sqrt{x}} \, dx,x,1-\cos (c+d x)\right )}{d \sqrt{1-\cos (c+d x)}}\\ &=-\frac{\cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d}+\frac{2^{\frac{1}{2}+n} F_1\left (\frac{1}{2};-4+n,\frac{1}{2}-n;\frac{3}{2};1-\cos (c+d x),\frac{1}{2} (1-\cos (c+d x))\right ) \cos ^n(c+d x) (1+\cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d}-\frac{\left ((-\cos (c+d x))^n (1+\cos (c+d x))^{\frac{1}{2}-n} (2 a n-2 a n \cos (c+d x)) \csc (c+d x) (a+a \sec (c+d x))^n\right ) \operatorname{Subst}\left (\int \sqrt{1-x} (-x)^{-n} (1+x)^{-\frac{1}{2}+n} \, dx,x,\cos (c+d x)\right )}{2 a d \sqrt{1-\cos (c+d x)}}\\ &=-\frac{F_1\left (1-n;-\frac{1}{2},\frac{1}{2}-n;2-n;\cos (c+d x),-\cos (c+d x)\right ) (1+\cos (c+d x))^{\frac{1}{2}-n} (n-n \cos (c+d x)) \cot (c+d x) (a+a \sec (c+d x))^n}{d (1-n) \sqrt{1-\cos (c+d x)}}-\frac{\cos (c+d x) (a+a \sec (c+d x))^n \sin (c+d x)}{d}+\frac{2^{\frac{1}{2}+n} F_1\left (\frac{1}{2};-4+n,\frac{1}{2}-n;\frac{3}{2};1-\cos (c+d x),\frac{1}{2} (1-\cos (c+d x))\right ) \cos ^n(c+d x) (1+\cos (c+d x))^{-\frac{1}{2}-n} (a+a \sec (c+d x))^n \sin (c+d x)}{d}\\ \end{align*}

Mathematica [C]  time = 23.103, size = 7069, normalized size = 30.73 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.66, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \sin \left ( dx+c \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1)*(a*sec(d*x + c) + a)^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((a+a*sec(d*x+c))**n*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 (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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