\(\int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx\) [89]

Optimal result

Integrand size = 23, antiderivative size = 159 \[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=-\frac {g \operatorname {AppellF1}\left (1+m,\frac {1-p}{2},\frac {1-p}{2},2+m,\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right ) (a+b \cos (c+d x))^{1+m} \left (1-\frac {a+b \cos (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \cos (c+d x)}{a+b}\right )^{\frac {1-p}{2}} (g \sin (c+d x))^{-1+p}}{b d (1+m)} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(676\) vs. \(2(159)=318\).

Time = 4.11 (sec) , antiderivative size = 676, normalized size of antiderivative = 4.25 \[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {1+p}{2},1+m+p,-m,\frac {3+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) (a+b \cos (c+d x))^m (g \sin (c+d x))^p \tan \left (\frac {1}{2} (c+d x)\right )}{d \left (\operatorname {AppellF1}\left (\frac {1+p}{2},1+m+p,-m,\frac {3+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )-\frac {4 b m \operatorname {AppellF1}\left (\frac {1+p}{2},1+m+p,-m,\frac {3+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sin ^2\left (\frac {1}{2} (c+d x)\right )}{a+b \cos (c+d x)}+2 p \operatorname {AppellF1}\left (\frac {1+p}{2},1+m+p,-m,\frac {3+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \cot (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+2 (m+p) \operatorname {AppellF1}\left (\frac {1+p}{2},1+m+p,-m,\frac {3+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {2 (-a+b) m \operatorname {AppellF1}\left (\frac {1+p}{2},1+m+p,-m,\frac {3+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b \cos (c+d x)}+\frac {2 (1+p) \left (\frac {(a-b) m \operatorname {AppellF1}\left (\frac {3+p}{2},1+m+p,1-m,\frac {5+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )}{a+b}-(1+m+p) \operatorname {AppellF1}\left (\frac {3+p}{2},2+m+p,-m,\frac {5+p}{2},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{3+p}\right )} \] Input:

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

Output:

(2*AppellF1[(1 + p)/2, 1 + m + p, -m, (3 + p)/2, -Tan[(c + d*x)/2]^2, ((-a 
 + b)*Tan[(c + d*x)/2]^2)/(a + b)]*(a + b*Cos[c + d*x])^m*(g*Sin[c + d*x]) 
^p*Tan[(c + d*x)/2])/(d*(AppellF1[(1 + p)/2, 1 + m + p, -m, (3 + p)/2, -Ta 
n[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sec[(c + d*x)/2]^ 
2 - (4*b*m*AppellF1[(1 + p)/2, 1 + m + p, -m, (3 + p)/2, -Tan[(c + d*x)/2] 
^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sin[(c + d*x)/2]^2)/(a + b*Cos[ 
c + d*x]) + 2*p*AppellF1[(1 + p)/2, 1 + m + p, -m, (3 + p)/2, -Tan[(c + d* 
x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Cot[c + d*x]*Tan[(c + d*x) 
/2] + 2*(m + p)*AppellF1[(1 + p)/2, 1 + m + p, -m, (3 + p)/2, -Tan[(c + d* 
x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Tan[(c + d*x)/2]^2 + (2*(- 
a + b)*m*AppellF1[(1 + p)/2, 1 + m + p, -m, (3 + p)/2, -Tan[(c + d*x)/2]^2 
, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)]*Tan[(c + d*x)/2]^2)/(a + b*Cos[c 
+ d*x]) + (2*(1 + p)*(((a - b)*m*AppellF1[(3 + p)/2, 1 + m + p, 1 - m, (5 
+ p)/2, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)])/(a + 
b) - (1 + m + p)*AppellF1[(3 + p)/2, 2 + m + p, -m, (5 + p)/2, -Tan[(c + d 
*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2*Tan[( 
c + d*x)/2]^2)/(3 + p)))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3042, 3183, 155}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int (g \sin (c+d x))^p (a+b \cos (c+d x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \left (g \cos \left (c+d x-\frac {\pi }{2}\right )\right )^p \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )^mdx\)

\(\Big \downarrow \) 3183

\(\displaystyle \frac {g (g \sin (c+d x))^{p-1} \left (1-\frac {a+b \cos (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \cos (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \int (a+b \cos (c+d x))^m \left (-\frac {\cos (c+d x) b}{a-b}-\frac {b}{a-b}\right )^{\frac {p-1}{2}} \left (\frac {b}{a+b}-\frac {b \cos (c+d x)}{a+b}\right )^{\frac {p-1}{2}}d(-\cos (c+d x))}{d}\)

\(\Big \downarrow \) 155

\(\displaystyle -\frac {g (g \sin (c+d x))^{p-1} (a+b \cos (c+d x))^{m+1} \left (1-\frac {a+b \cos (c+d x)}{a-b}\right )^{\frac {1-p}{2}} \left (1-\frac {a+b \cos (c+d x)}{a+b}\right )^{\frac {1-p}{2}} \operatorname {AppellF1}\left (m+1,\frac {1-p}{2},\frac {1-p}{2},m+2,\frac {a+b \cos (c+d x)}{a-b},\frac {a+b \cos (c+d x)}{a+b}\right )}{b d (m+1)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 155

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_)) 
^(p_), x_] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*Simplify[b/(b*c - a*d)]^n* 
Simplify[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] &&  !IntegerQ[n] &&  !IntegerQ[p] && GtQ[Sim 
plify[b/(b*c - a*d)], 0] && GtQ[Simplify[b/(b*e - a*f)], 0] &&  !(GtQ[Simpl 
ify[d/(d*a - c*b)], 0] && GtQ[Simplify[d/(d*e - c*f)], 0] && SimplerQ[c + d 
*x, a + b*x]) &&  !(GtQ[Simplify[f/(f*a - e*b)], 0] && GtQ[Simplify[f/(f*c 
- e*d)], 0] && SimplerQ[e + f*x, a + b*x])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3183

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_), x_Symbol] :> Simp[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]
 

Maple [F]

\[\int \left (a +\cos \left (d x +c \right ) b \right )^{m} \left (g \sin \left (d x +c \right )\right )^{p}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{m} \left (g \sin \left (d x + c\right )\right )^{p} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=\int \left (g \sin {\left (c + d x \right )}\right )^{p} \left (a + b \cos {\left (c + d x \right )}\right )^{m}\, dx \] Input:

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

Output:

Integral((g*sin(c + d*x))**p*(a + b*cos(c + d*x))**m, x)
 

Maxima [F]

\[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{m} \left (g \sin \left (d x + c\right )\right )^{p} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=\int { {\left (b \cos \left (d x + c\right ) + a\right )}^{m} \left (g \sin \left (d x + c\right )\right )^{p} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=\int {\left (g\,\sin \left (c+d\,x\right )\right )}^p\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^m \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (a+b \cos (c+d x))^m (g \sin (c+d x))^p \, dx=g^{p} \left (\int \sin \left (d x +c \right )^{p} \left (\cos \left (d x +c \right ) b +a \right )^{m}d x \right ) \] Input:

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

Output:

g**p*int(sin(c + d*x)**p*(cos(c + d*x)*b + a)**m,x)