3.72 \(\int (-4-3 \sin (c+d x))^n \, dx\)

Optimal. Leaf size=110 \[ -\frac {\sqrt {-\sin (c+d x)-1} \cos (c+d x) (-3 \sin (c+d x)-4)^{n+1} F_1\left (n+1;\frac {1}{2},\frac {1}{2};n+2;3 \sin (c+d x)+4,\frac {1}{7} (3 \sin (c+d x)+4)\right )}{\sqrt {7} d (n+1) \sqrt {1-\sin (c+d x)} (\sin (c+d x)+1)} \]

[Out]

-1/7*AppellF1(1+n,1/2,1/2,2+n,4+3*sin(d*x+c),4/7+3/7*sin(d*x+c))*cos(d*x+c)*(-4-3*sin(d*x+c))^(1+n)*(-1-sin(d*
x+c))^(1/2)/d/(1+n)/(1+sin(d*x+c))*7^(1/2)/(1-sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2665, 139, 138} \[ -\frac {\sqrt {-\sin (c+d x)-1} \cos (c+d x) (-3 \sin (c+d x)-4)^{n+1} F_1\left (n+1;\frac {1}{2},\frac {1}{2};n+2;3 \sin (c+d x)+4,\frac {1}{7} (3 \sin (c+d x)+4)\right )}{\sqrt {7} d (n+1) \sqrt {1-\sin (c+d x)} (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

Int[(-4 - 3*Sin[c + d*x])^n,x]

[Out]

-((AppellF1[1 + n, 1/2, 1/2, 2 + n, 4 + 3*Sin[c + d*x], (4 + 3*Sin[c + d*x])/7]*Cos[c + d*x]*(-4 - 3*Sin[c + d
*x])^(1 + n)*Sqrt[-1 - Sin[c + d*x]])/(Sqrt[7]*d*(1 + n)*Sqrt[1 - Sin[c + d*x]]*(1 + Sin[c + d*x])))

Rule 138

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

Rule 139

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

Rule 2665

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[Cos[c + d*x]/(d*Sqrt[1 + Sin[c + d*x]]*Sqrt
[1 - Sin[c + d*x]]), Subst[Int[(a + b*x)^n/(Sqrt[1 + x]*Sqrt[1 - x]), x], x, Sin[c + d*x]], x] /; FreeQ[{a, b,
c, d, n}, x] && NeQ[a^2 - b^2, 0] &&  !IntegerQ[2*n]

Rubi steps

\begin {align*} \int (-4-3 \sin (c+d x))^n \, dx &=\frac {\cos (c+d x) \operatorname {Subst}\left (\int \frac {(-4-3 x)^n}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (c+d x)\right )}{d \sqrt {1-\sin (c+d x)} \sqrt {1+\sin (c+d x)}}\\ &=\frac {\left (\sqrt {3} \cos (c+d x) \sqrt {-1-\sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {(-4-3 x)^n}{\sqrt {-3-3 x} \sqrt {1-x}} \, dx,x,\sin (c+d x)\right )}{d \sqrt {1-\sin (c+d x)} (1+\sin (c+d x))}\\ &=-\frac {F_1\left (1+n;\frac {1}{2},\frac {1}{2};2+n;4+3 \sin (c+d x),\frac {1}{7} (4+3 \sin (c+d x))\right ) \cos (c+d x) (-4-3 \sin (c+d x))^{1+n} \sqrt {-1-\sin (c+d x)}}{\sqrt {7} d (1+n) \sqrt {1-\sin (c+d x)} (1+\sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 100, normalized size = 0.91 \[ -\frac {\sqrt {-\sin (c+d x)-1} \sqrt {1-\sin (c+d x)} \sec (c+d x) (-3 \sin (c+d x)-4)^{n+1} F_1\left (n+1;\frac {1}{2},\frac {1}{2};n+2;3 \sin (c+d x)+4,\frac {1}{7} (3 \sin (c+d x)+4)\right )}{\sqrt {7} d (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 3*Sin[c + d*x])^n,x]

[Out]

-((AppellF1[1 + n, 1/2, 1/2, 2 + n, 4 + 3*Sin[c + d*x], (4 + 3*Sin[c + d*x])/7]*Sec[c + d*x]*(-4 - 3*Sin[c + d
*x])^(1 + n)*Sqrt[-1 - Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]])/(Sqrt[7]*d*(1 + n)))

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-3 \, \sin \left (d x + c\right ) - 4\right )}^{n}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4-3*sin(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((-3*sin(d*x + c) - 4)^n, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-3 \, \sin \left (d x + c\right ) - 4\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4-3*sin(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((-3*sin(d*x + c) - 4)^n, x)

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maple [F]  time = 0.44, size = 0, normalized size = 0.00 \[ \int \left (-4-3 \sin \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4-3*sin(d*x+c))^n,x)

[Out]

int((-4-3*sin(d*x+c))^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-3 \, \sin \left (d x + c\right ) - 4\right )}^{n}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4-3*sin(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((-3*sin(d*x + c) - 4)^n, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (-3\,\sin \left (c+d\,x\right )-4\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((- 3*sin(c + d*x) - 4)^n,x)

[Out]

int((- 3*sin(c + d*x) - 4)^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (- 3 \sin {\left (c + d x \right )} - 4\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4-3*sin(d*x+c))**n,x)

[Out]

Integral((-3*sin(c + d*x) - 4)**n, x)

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