\(\int (c \cos (e+f x))^m (d \sin (e+f x))^n (a+b \sin ^2(e+f x))^p \, dx\) [592]

Optimal result

Integrand size = 35, antiderivative size = 137 \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {c \operatorname {AppellF1}\left (\frac {1+n}{2},\frac {1-m}{2},-p,\frac {3+n}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{1+n} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{d f (1+n)} \]

[Out]

c*AppellF1(1/2+1/2*n,1/2-1/2*m,-p,3/2+1/2*n,sin(f*x+e)^2,-b*sin(f*x+e)^2/a)*(c*cos(f*x+e))^(-1+m)*(cos(f*x+e)^
2)^(1/2-1/2*m)*(d*sin(f*x+e))^(1+n)*(a+b*sin(f*x+e)^2)^p/d/f/(1+n)/((1+b*sin(f*x+e)^2/a)^p)

Rubi [A] (verified)

Time = 0.23 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3281, 525, 524} \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {c \cos ^2(e+f x)^{\frac {1-m}{2}} (c \cos (e+f x))^{m-1} (d \sin (e+f x))^{n+1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac {b \sin ^2(e+f x)}{a}+1\right )^{-p} \operatorname {AppellF1}\left (\frac {n+1}{2},\frac {1-m}{2},-p,\frac {n+3}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right )}{d f (n+1)} \]

[In]

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

[Out]

(c*AppellF1[(1 + n)/2, (1 - m)/2, -p, (3 + n)/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*(c*Cos[e + f*x])^(-1
+ m)*(Cos[e + f*x]^2)^((1 - m)/2)*(d*Sin[e + f*x])^(1 + n)*(a + b*Sin[e + f*x]^2)^p)/(d*f*(1 + n)*(1 + (b*Sin
[e + f*x]^2)/a)^p)

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 3281

Int[(cos[(e_.) + (f_.)*(x_)]*(c_.))^(m_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff*c^(2*IntPart[(m - 1)/2] + 1)*(
(c*Cos[e + f*x])^(2*FracPart[(m - 1)/2])/(f*(Cos[e + f*x]^2)^FracPart[(m - 1)/2])), Subst[Int[(d*ff*x)^n*(1 -
ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x
] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (c (c \cos (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \cos ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}}\right ) \text {Subst}\left (\int (d x)^n \left (1-x^2\right )^{\frac {1}{2} (-1+m)} \left (a+b x^2\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (c (c \cos (e+f x))^{2 \left (-\frac {1}{2}+\frac {m}{2}\right )} \cos ^2(e+f x)^{\frac {1}{2}-\frac {m}{2}} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int (d x)^n \left (1-x^2\right )^{\frac {1}{2} (-1+m)} \left (1+\frac {b x^2}{a}\right )^p \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {c \operatorname {AppellF1}\left (\frac {1+n}{2},\frac {1-m}{2},-p,\frac {3+n}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (c \cos (e+f x))^{-1+m} \cos ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^{1+n} \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p}}{d f (1+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99 \[ \int (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \, dx=\frac {\operatorname {AppellF1}\left (\frac {1+n}{2},\frac {1-m}{2},-p,\frac {3+n}{2},\sin ^2(e+f x),-\frac {b \sin ^2(e+f x)}{a}\right ) (c \cos (e+f x))^m \cos ^2(e+f x)^{\frac {1-m}{2}} (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \left (1+\frac {b \sin ^2(e+f x)}{a}\right )^{-p} \tan (e+f x)}{f (1+n)} \]

[In]

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

[Out]

(AppellF1[(1 + n)/2, (1 - m)/2, -p, (3 + n)/2, Sin[e + f*x]^2, -((b*Sin[e + f*x]^2)/a)]*(c*Cos[e + f*x])^m*(Co
s[e + f*x]^2)^((1 - m)/2)*(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x]^2)^p*Tan[e + f*x])/(f*(1 + n)*(1 + (b*Sin[e +
f*x]^2)/a)^p)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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