3.7.53 \(\int (e \cos (c+d x))^{2-m} (a+b \sin (c+d x))^m \, dx\) [653]

Optimal. Leaf size=152 \[ \frac {e F_1\left (1+m;\frac {1}{2} (-1+m),\frac {1}{2} (-1+m);2+m;\frac {a+b \sin (c+d x)}{a-b},\frac {a+b \sin (c+d x)}{a+b}\right ) (e \cos (c+d x))^{1-m} (a+b \sin (c+d x))^{1+m} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {1}{2} (-1+m)} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {1}{2} (-1+m)}}{b d (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2783, 143} \begin {gather*} \frac {e (e \cos (c+d x))^{1-m} (a+b \sin (c+d x))^{m+1} \left (1-\frac {a+b \sin (c+d x)}{a-b}\right )^{\frac {m-1}{2}} \left (1-\frac {a+b \sin (c+d x)}{a+b}\right )^{\frac {m-1}{2}} F_1\left (m+1;\frac {m-1}{2},\frac {m-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)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [F]
time = 2.85, size = 0, normalized size = 0.00 \begin {gather*} \int (e \cos (c+d x))^{2-m} (a+b \sin (c+d x))^m \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((e*cos(c + d*x))**(2 - m)*(a + b*sin(c + d*x))**m, x)

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

[Out]

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

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

[Out]

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

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