3.7.38 \(\int \cos ^2(c+d x) (a+b \sin (c+d x))^m \, dx\) [638]

Optimal. Leaf size=127 \[ \frac {F_1\left (1+m;-\frac {1}{2},-\frac {1}{2};2+m;\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^{1+m}}{b d (1+m) \sqrt {1-\frac {a+b \sin (c+d x)}{a-b}} \sqrt {1-\frac {a+b \sin (c+d x)}{a+b}}} \]

[Out]

AppellF1(1+m,-1/2,-1/2,2+m,(a+b*sin(d*x+c))/(a-b),(a+b*sin(d*x+c))/(a+b))*cos(d*x+c)*(a+b*sin(d*x+c))^(1+m)/b/
d/(1+m)/(1+(-a-b*sin(d*x+c))/(a-b))^(1/2)/(1+(-a-b*sin(d*x+c))/(a+b))^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2783, 143} \begin {gather*} \frac {\cos (c+d x) (a+b \sin (c+d x))^{m+1} F_1\left (m+1;-\frac {1}{2},-\frac {1}{2};m+2;\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right )}{b d (m+1) \sqrt {1-\frac {a+b \sin (c+d x)}{a-b}} \sqrt {1-\frac {a+b \sin (c+d x)}{a+b}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*(a + b*Sin[c + d*x])^m,x]

[Out]

(AppellF1[1 + m, -1/2, -1/2, 2 + m, (a + b*Sin[c + d*x])/(a - b), (a + b*Sin[c + d*x])/(a + b)]*Cos[c + d*x]*(
a + b*Sin[c + d*x])^(1 + m))/(b*d*(1 + m)*Sqrt[1 - (a + b*Sin[c + d*x])/(a - b)]*Sqrt[1 - (a + b*Sin[c + d*x])
/(a + b)])

Rule 143

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((a + b*x)
^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n*(b/(b*e - a*f))^p))*AppellF1[m + 1, -n, -p, m + 2, (-d)*((a + b*x)/(b*c
- a*d)), (-f)*((a + b*x)/(b*e - a*f))], 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 2783

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[g*((g*
Cos[e + f*x])^(p - 1)/(f*(1 - (a + b*Sin[e + f*x])/(a - b))^((p - 1)/2)*(1 - (a + b*Sin[e + f*x])/(a + b))^((p
- 1)/2))), Subst[Int[(-b/(a - b) - b*(x/(a - b)))^((p - 1)/2)*(b/(a + b) - b*(x/(a + b)))^((p - 1)/2)*(a + b*
x)^m, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && NeQ[a^2 - b^2, 0] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \sin (c+d x))^m \, dx &=\frac {\cos (c+d x) \text {Subst}\left (\int (a+b x)^m \sqrt {-\frac {b}{a-b}-\frac {b x}{a-b}} \sqrt {\frac {b}{a+b}-\frac {b x}{a+b}} \, dx,x,\sin (c+d x)\right )}{d \sqrt {1-\frac {a+b \sin (c+d x)}{a-b}} \sqrt {1-\frac {a+b \sin (c+d x)}{a+b}}}\\ &=\frac {F_1\left (1+m;-\frac {1}{2},-\frac {1}{2};2+m;\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) \cos (c+d x) (a+b \sin (c+d x))^{1+m}}{b d (1+m) \sqrt {1-\frac {a+b \sin (c+d x)}{a-b}} \sqrt {1-\frac {a+b \sin (c+d x)}{a+b}}}\\ \end {align*}

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Mathematica [F]
time = 2.76, size = 0, normalized size = 0.00 \begin {gather*} \int \cos ^2(c+d x) (a+b \sin (c+d x))^m \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[Cos[c + d*x]^2*(a + b*Sin[c + d*x])^m,x]

[Out]

Integrate[Cos[c + d*x]^2*(a + b*Sin[c + d*x])^m, x]

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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2}\left (d x +c \right )\right ) \left (a +b \sin \left (d x +c \right )\right )^{m}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*sin(d*x+c))^m,x)

[Out]

int(cos(d*x+c)^2*(a+b*sin(d*x+c))^m,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^m*cos(d*x + c)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

integral((b*sin(d*x + c) + a)^m*cos(d*x + c)^2, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+b*sin(d*x+c))**m,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^m*cos(d*x + c)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^2\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + b*sin(c + d*x))^m,x)

[Out]

int(cos(c + d*x)^2*(a + b*sin(c + d*x))^m, x)

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