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

Optimal. Leaf size=95 \[ \frac {\sqrt {2} 7^n \cos (c+d x) (4-3 \sin (c+d x))^{-n} (3 \sin (c+d x)-4)^n F_1\left (\frac {1}{2};-n,\frac {1}{2};\frac {3}{2};\frac {3}{7} (\sin (c+d x)+1),\frac {1}{2} (\sin (c+d x)+1)\right )}{d \sqrt {1-\sin (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 95, 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} \[ \frac {\sqrt {2} 7^n \cos (c+d x) (4-3 \sin (c+d x))^{-n} (3 \sin (c+d x)-4)^n F_1\left (\frac {1}{2};-n,\frac {1}{2};\frac {3}{2};\frac {3}{7} (\sin (c+d x)+1),\frac {1}{2} (\sin (c+d x)+1)\right )}{d \sqrt {1-\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(-4 + 3*Sin[c + d*x])^n,x]

[Out]

(Sqrt[2]*7^n*AppellF1[1/2, -n, 1/2, 3/2, (3*(1 + Sin[c + d*x]))/7, (1 + Sin[c + d*x])/2]*Cos[c + d*x]*(-4 + 3*
Sin[c + d*x])^n)/(d*(4 - 3*Sin[c + d*x])^n*Sqrt[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 (\cos (c+d x) (4-3 \sin (c+d x))^{-n} (-4+3 \sin (c+d x))^n\right ) \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 {\sqrt {2} 7^n F_1\left (\frac {1}{2};-n,\frac {1}{2};\frac {3}{2};\frac {3}{7} (1+\sin (c+d x)),\frac {1}{2} (1+\sin (c+d x))\right ) \cos (c+d x) (4-3 \sin (c+d x))^{-n} (-4+3 \sin (c+d x))^n}{d \sqrt {1-\sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 95, normalized size = 1.00 \[ \frac {\sqrt {\sin (c+d x)-1} \sqrt {\sin (c+d x)+1} \sec (c+d x) (3 \sin (c+d x)-4)^{n+1} F_1\left (n+1;\frac {1}{2},\frac {1}{2};n+2;\frac {1}{7} (4-3 \sin (c+d x)),4-3 \sin (c+d x)\right )}{\sqrt {7} d (n+1)} \]

Warning: Unable to verify antiderivative.

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

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fricas [F]  time = 0.87, 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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