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(1/2+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)

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Rubi [A]  time = 0.21, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.84, 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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fricas [F]  time = 1.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \left (c \cos \left (f x + e\right )\right )^{m} \left (d \sin \left (f x + e\right )\right )^{n}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (c \cos \left (f x + e\right )\right )^{m} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \]

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)

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maple [F]  time = 8.13, size = 0, normalized size = 0.00 \[ \int \left (c \cos \left (f x +e \right )\right )^{m} \left (d \sin \left (f x +e \right )\right )^{n} \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{p} \left (c \cos \left (f x + e\right )\right )^{m} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,\cos \left (e+f\,x\right )\right )}^m\,{\left (d\,\sin \left (e+f\,x\right )\right )}^n\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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