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

Optimal. Leaf size=137 \[ \frac{c \cos ^2(e+f x)^{\frac{1-m}{2}} (c \cos (e+f x))^{m-1} (d \sin (e+f x))^{n+1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{n+1}{2};\frac{1-m}{2},-p;\frac{n+3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{d f (n+1)} \]

[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)

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Rubi [A]  time = 0.210649, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.086, Rules used = {3202, 511, 510} \[ \frac{c \cos ^2(e+f x)^{\frac{1-m}{2}} (c \cos (e+f x))^{m-1} (d \sin (e+f x))^{n+1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{n+1}{2};\frac{1-m}{2},-p;\frac{n+3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{d f (n+1)} \]

Antiderivative was successfully verified.

[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 3202

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]

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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])

Rubi steps

\begin{align*} \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{\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 ) \operatorname{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 ) \operatorname{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 F_1\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]  time = 0.779783, size = 135, normalized size = 0.99 \[ \frac{\tan (e+f x) \cos ^2(e+f x)^{\frac{1-m}{2}} (c \cos (e+f x))^m (d \sin (e+f x))^n \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{n+1}{2};\frac{1-m}{2},-p;\frac{n+3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f (n+1)} \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 3.26, size = 0, normalized size = 0. \begin{align*} \int \left ( c\cos \left ( fx+e \right ) \right ) ^{m} \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \left (c \cos \left (f x + e\right )\right )^{m} \left (d \sin \left (f x + e\right )\right )^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)