3.184 \(\int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n} \, dx\)

Optimal. Leaf size=131 \[ \frac{c 2^{-\frac{m}{2}+\frac{n}{2}+1} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (1-\sin (e+f x))^{\frac{m-n}{2}} (g \cos (e+f x))^{-m-n} \, _2F_1\left (\frac{m-n}{2},\frac{m-n}{2};\frac{1}{2} (m-n+2);\frac{1}{2} (\sin (e+f x)+1)\right )}{f g (m-n)} \]

[Out]

(2^(1 - m/2 + n/2)*c*(g*Cos[e + f*x])^(-m - n)*Hypergeometric2F1[(m - n)/2, (m - n)/2, (2 + m - n)/2, (1 + Sin
[e + f*x])/2]*(1 - Sin[e + f*x])^((m - n)/2)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n)/(f*g*(m - n))

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Rubi [A]  time = 0.36937, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.089, Rules used = {2853, 2689, 70, 69} \[ \frac{c 2^{-\frac{m}{2}+\frac{n}{2}+1} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^n (1-\sin (e+f x))^{\frac{m-n}{2}} (g \cos (e+f x))^{-m-n} \, _2F_1\left (\frac{m-n}{2},\frac{m-n}{2};\frac{1}{2} (m-n+2);\frac{1}{2} (\sin (e+f x)+1)\right )}{f g (m-n)} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(1 + n),x]

[Out]

(2^(1 - m/2 + n/2)*c*(g*Cos[e + f*x])^(-m - n)*Hypergeometric2F1[(m - n)/2, (m - n)/2, (2 + m - n)/2, (1 + Sin
[e + f*x])/2]*(1 - Sin[e + f*x])^((m - n)/2)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n)/(f*g*(m - n))

Rule 2853

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*(c + d*Sin[e
+ f*x])^FracPart[m])/(g^(2*IntPart[m])*(g*Cos[e + f*x])^(2*FracPart[m])), Int[(g*Cos[e + f*x])^(2*m + p)*(c +
d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 -
b^2, 0] && (FractionQ[m] ||  !FractionQ[n])

Rule 2689

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

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rubi steps

\begin{align*} \int (g \cos (e+f x))^{-1-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{1+n} \, dx &=\left ((g \cos (e+f x))^{-2 m} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int (g \cos (e+f x))^{-1+m-n} (c-c \sin (e+f x))^{1-m+n} \, dx\\ &=\frac{\left (c^2 (g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac{1}{2} (-m+n)} (c+c \sin (e+f x))^{\frac{1}{2} (-m+n)}\right ) \operatorname{Subst}\left (\int (c-c x)^{1-m+\frac{1}{2} (-2+m-n)+n} (c+c x)^{\frac{1}{2} (-2+m-n)} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac{\left (2^{-\frac{m}{2}+\frac{n}{2}} c^2 (g \cos (e+f x))^{-m-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{\frac{m}{2}+\frac{n}{2}+\frac{1}{2} (-m+n)} \left (\frac{c-c \sin (e+f x)}{c}\right )^{\frac{m}{2}-\frac{n}{2}} (c+c \sin (e+f x))^{\frac{1}{2} (-m+n)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{1-m+\frac{1}{2} (-2+m-n)+n} (c+c x)^{\frac{1}{2} (-2+m-n)} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac{2^{1-\frac{m}{2}+\frac{n}{2}} c (g \cos (e+f x))^{-m-n} \, _2F_1\left (\frac{m-n}{2},\frac{m-n}{2};\frac{1}{2} (2+m-n);\frac{1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac{m-n}{2}} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n}{f g (m-n)}\\ \end{align*}

Mathematica [C]  time = 25.637, size = 207, normalized size = 1.58 \[ \frac{i c (\sin (e+f x)-1) (a (\sin (e+f x)+1))^m (c-c \sin (e+f x))^n (g \cos (e+f x))^{-m-n} \left (\, _2F_1\left (1,-m+n+1;-m+n+2;-\frac{i \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right )}{\tan \left (\frac{1}{2} (e+f x)\right )+1}\right )-\, _2F_1\left (1,-m+n+1;-m+n+2;\frac{i \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right )}{\tan \left (\frac{1}{2} (e+f x)\right )+1}\right )\right )}{f g (m-n-1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^(-1 - m - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(1 + n),x]

[Out]

(I*c*(g*Cos[e + f*x])^(-m - n)*(Hypergeometric2F1[1, 1 - m + n, 2 - m + n, ((-I)*(-1 + Tan[(e + f*x)/2]))/(1 +
Tan[(e + f*x)/2])] - Hypergeometric2F1[1, 1 - m + n, 2 - m + n, (I*(-1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x
)/2])])*(-1 + Sin[e + f*x])*(a*(1 + Sin[e + f*x]))^m*(c - c*Sin[e + f*x])^n)/(f*g*(-1 + m - n)*(Cos[(e + f*x)/
2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [F]  time = 29.86, size = 0, normalized size = 0. \begin{align*} \int \left ( g\cos \left ( fx+e \right ) \right ) ^{-1-m-n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sin \left ( fx+e \right ) \right ) ^{1+n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1+n),x)

[Out]

int((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1+n),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (g \cos \left (f x + e\right )\right )^{-m - n - 1}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1+n),x, algorithm="maxima")

[Out]

integrate((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n + 1), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (g \cos \left (f x + e\right )\right )^{-m - n - 1}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1+n),x, algorithm="fricas")

[Out]

integral((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n + 1), 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((g*cos(f*x+e))**(-1-m-n)*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**(1+n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (g \cos \left (f x + e\right )\right )^{-m - n - 1}{\left (a \sin \left (f x + e\right ) + a\right )}^{m}{\left (-c \sin \left (f x + e\right ) + c\right )}^{n + 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^(-1-m-n)*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(1+n),x, algorithm="giac")

[Out]

integrate((g*cos(f*x + e))^(-m - n - 1)*(a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^(n + 1), x)