\(\int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx\) [617]

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2771, 2748, 2722} \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=-\frac {\left (a^2 (p+2)+b^2\right ) \sin (c+d x) (e \cos (c+d x))^{p+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {p+1}{2},\frac {p+3}{2},\cos ^2(c+d x)\right )}{d e (p+1) (p+2) \sqrt {\sin ^2(c+d x)}}-\frac {a b (p+3) (e \cos (c+d x))^{p+1}}{d e (p+1) (p+2)}-\frac {b (a+b \sin (c+d x)) (e \cos (c+d x))^{p+1}}{d e (p+2)} \]

[In]

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

[Out]

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

Rule 2722

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]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2748

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-b)*((g*Co
s[e + f*x])^(p + 1)/(f*g*(p + 1))), x] + Dist[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 2771

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] + Dist[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])

Rubi steps \begin{align*} \text {integral}& = -\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p)}+\frac {\int (e \cos (c+d x))^p \left (b^2+a^2 (2+p)+a b (3+p) \sin (c+d x)\right ) \, dx}{2+p} \\ & = -\frac {a b (3+p) (e \cos (c+d x))^{1+p}}{d e (1+p) (2+p)}-\frac {b (e \cos (c+d x))^{1+p} (a+b \sin (c+d x))}{d e (2+p)}+\frac {\left (b^2+a^2 (2+p)\right ) \int (e \cos (c+d x))^p \, dx}{2+p} \\ & = -\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)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 25.89 (sec) , antiderivative size = 7484, normalized size of antiderivative = 47.67 \[ \int (e \cos (c+d x))^p (a+b \sin (c+d x))^2 \, dx=\text {Result too large to show} \]

[In]

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

[Out]

Result too large to show

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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