\(\int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx\) [614]

Optimal result

Integrand size = 23, antiderivative size = 157 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=-\frac {a b (3+p) (e \cos (c+d x))^{1+p}}{d e (1+p) (2+p)}-\frac {\left (b^2+a^2 (2+p)\right ) (e \cos (c+d x))^{1+p} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+p}{2},\frac {3+p}{2},\cos ^2(c+d x)\right ) \sin (c+d x)}{d e (1+p) (2+p) \sqrt {\sin ^2(c+d x)}}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p)} \] Output:

-a*b*(3+p)*(e*cos(d*x+c))^(p+1)/d/e/(p+1)/(2+p)-(b^2+a^2*(2+p))*(e*cos(d*x 
+c))^(p+1)*hypergeom([1/2, 1/2*p+1/2],[3/2+1/2*p],cos(d*x+c)^2)*sin(d*x+c) 
/d/e/(p+1)/(2+p)/(sin(d*x+c)^2)^(1/2)-b*(e*cos(d*x+c))^(p+1)*(a+b*sin(d*x+ 
c))/d/e/(2+p)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.

Time = 27.07 (sec) , antiderivative size = 7484, normalized size of antiderivative = 47.67 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3171, 3042, 3148, 3042, 3122}

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.

\(\displaystyle \int (a+b \sin (c+d x))^2 (e \cos (c+d x))^p \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+b \sin (c+d x))^2 (e \cos (c+d x))^pdx\)

\(\Big \downarrow \) 3171

\(\displaystyle \frac {\int (e \cos (c+d x))^p \left ((p+2) a^2+b (p+3) \sin (c+d x) a+b^2\right )dx}{p+2}-\frac {b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (e \cos (c+d x))^p \left ((p+2) a^2+b (p+3) \sin (c+d x) a+b^2\right )dx}{p+2}-\frac {b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)}\)

\(\Big \downarrow \) 3148

\(\displaystyle \frac {\left (a^2 (p+2)+b^2\right ) \int (e \cos (c+d x))^pdx-\frac {a b (p+3) (e \cos (c+d x))^{p+1}}{d e (p+1)}}{p+2}-\frac {b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\left (a^2 (p+2)+b^2\right ) \int \left (e \sin \left (c+d x+\frac {\pi }{2}\right )\right )^pdx-\frac {a b (p+3) (e \cos (c+d x))^{p+1}}{d e (p+1)}}{p+2}-\frac {b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)}\)

\(\Big \downarrow \) 3122

\(\displaystyle \frac {-\frac {\left (a^2 (p+2)+b^2\right ) \sin (c+d x) (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {p+1}{2},\frac {p+3}{2},\cos ^2(c+d x)\right )}{d e (p+1) \sqrt {\sin ^2(c+d x)}}-\frac {a b (p+3) (e \cos (c+d x))^{p+1}}{d e (p+1)}}{p+2}-\frac {b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)}\)

Input:

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

Output:

-((b*(e*Cos[c + d*x])^(1 + p)*(a + b*Sin[c + d*x]))/(d*e*(2 + p))) + (-((a 
*b*(3 + p)*(e*Cos[c + d*x])^(1 + p))/(d*e*(1 + p))) - ((b^2 + a^2*(2 + p)) 
*(e*Cos[c + d*x])^(1 + p)*Hypergeometric2F1[1/2, (1 + p)/2, (3 + p)/2, Cos 
[c + d*x]^2]*Sin[c + d*x])/(d*e*(1 + p)*Sqrt[Sin[c + d*x]^2]))/(2 + p)
 

Defintions of rubi rules used

rule 3042

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

rule 3122

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2 
F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n}, x] 
 &&  !IntegerQ[2*n]
 

rule 3148

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[(-b)*((g*Cos[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + 
 Simp[a   Int[(g*Cos[e + f*x])^p, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && 
 (IntegerQ[2*p] || NeQ[a^2 - b^2, 0])
 

rule 3171

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

Maple [F]

\[\int \left (e \cos \left (d x +c \right )\right )^{p} \left (a +b \sin \left (d x +c \right )\right )^{2}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\int \left (e \cos {\left (c + d x \right )}\right )^{p} \left (a + b \sin {\left (c + d x \right )}\right )^{2}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\int { {\left (b \sin \left (d x + c\right ) + a\right )}^{2} \left (e \cos \left (d x + c\right )\right )^{p} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\int {\left (e\,\cos \left (c+d\,x\right )\right )}^p\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^2 \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\frac {e^{p} \left (-2 \cos \left (d x +c \right )^{p} \cos \left (d x +c \right ) a b +\left (\int \cos \left (d x +c \right )^{p}d x \right ) a^{2} d p +\left (\int \cos \left (d x +c \right )^{p}d x \right ) a^{2} d +\left (\int \cos \left (d x +c \right )^{p} \sin \left (d x +c \right )^{2}d x \right ) b^{2} d p +\left (\int \cos \left (d x +c \right )^{p} \sin \left (d x +c \right )^{2}d x \right ) b^{2} d \right )}{d \left (p +1\right )} \] Input:

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

Output:

(e**p*( - 2*cos(c + d*x)**p*cos(c + d*x)*a*b + int(cos(c + d*x)**p,x)*a**2 
*d*p + int(cos(c + d*x)**p,x)*a**2*d + int(cos(c + d*x)**p*sin(c + d*x)**2 
,x)*b**2*d*p + int(cos(c + d*x)**p*sin(c + d*x)**2,x)*b**2*d))/(d*(p + 1))