3.1036 \(\int (g \cos (e+f x))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-p} \, dx\)

Optimal. Leaf size=151 \[ \frac{(A+B) (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1}}{f g (p+1)}-\frac{B 2^{\frac{1}{2}-\frac{p}{2}} (1-\sin (e+f x))^{\frac{p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \, _2F_1\left (\frac{p+1}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{f g (p+1)} \]

[Out]

((A + B)*(g*Cos[e + f*x])^(1 + p)*(c - c*Sin[e + f*x])^(-1 - p))/(f*g*(1 + p)) - (2^(1/2 - p/2)*B*(g*Cos[e + f
*x])^(1 + p)*Hypergeometric2F1[(1 + p)/2, (1 + p)/2, (3 + p)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^((1 +
p)/2)*(c - c*Sin[e + f*x])^(-1 - p))/(f*g*(1 + p))

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Rubi [A]  time = 0.264259, antiderivative size = 151, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2859, 2689, 70, 69} \[ \frac{(A+B) (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1}}{f g (p+1)}-\frac{B 2^{\frac{1}{2}-\frac{p}{2}} (1-\sin (e+f x))^{\frac{p+1}{2}} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^{p+1} \, _2F_1\left (\frac{p+1}{2},\frac{p+1}{2};\frac{p+3}{2};\frac{1}{2} (\sin (e+f x)+1)\right )}{f g (p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(-1 - p),x]

[Out]

((A + B)*(g*Cos[e + f*x])^(1 + p)*(c - c*Sin[e + f*x])^(-1 - p))/(f*g*(1 + p)) - (2^(1/2 - p/2)*B*(g*Cos[e + f
*x])^(1 + p)*Hypergeometric2F1[(1 + p)/2, (1 + p)/2, (3 + p)/2, (1 + Sin[e + f*x])/2]*(1 - Sin[e + f*x])^((1 +
p)/2)*(c - c*Sin[e + f*x])^(-1 - p))/(f*g*(1 + p))

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

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))^p (A+B \sin (e+f x)) (c-c \sin (e+f x))^{-1-p} \, dx &=\frac{(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}-\frac{B \int (g \cos (e+f x))^p (c-c \sin (e+f x))^{-p} \, dx}{c}\\ &=\frac{(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}-\frac{\left (B c (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{\frac{1}{2} (-1-p)} (c+c \sin (e+f x))^{\frac{1}{2} (-1-p)}\right ) \operatorname{Subst}\left (\int (c-c x)^{\frac{1}{2} (-1+p)-p} (c+c x)^{\frac{1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac{(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}-\frac{\left (2^{-\frac{1}{2}-\frac{p}{2}} B c (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-\frac{1}{2}+\frac{1}{2} (-1-p)-\frac{p}{2}} \left (\frac{c-c \sin (e+f x)}{c}\right )^{\frac{1}{2}+\frac{p}{2}} (c+c \sin (e+f x))^{\frac{1}{2} (-1-p)}\right ) \operatorname{Subst}\left (\int \left (\frac{1}{2}-\frac{x}{2}\right )^{\frac{1}{2} (-1+p)-p} (c+c x)^{\frac{1}{2} (-1+p)} \, dx,x,\sin (e+f x)\right )}{f g}\\ &=\frac{(A+B) (g \cos (e+f x))^{1+p} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}-\frac{2^{\frac{1}{2}-\frac{p}{2}} B (g \cos (e+f x))^{1+p} \, _2F_1\left (\frac{1+p}{2},\frac{1+p}{2};\frac{3+p}{2};\frac{1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac{1+p}{2}} (c-c \sin (e+f x))^{-1-p}}{f g (1+p)}\\ \end{align*}

Mathematica [C]  time = 3.44466, size = 300, normalized size = 1.99 \[ -\frac{2^{-p} (c-c \sin (e+f x))^{-p-1} (g \cos (e+f x))^p \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{-2 (-p-1)-2 p} \left (\frac{1-\tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{\frac{1}{\cos (e+f x)+1}}}\right )^{2 p} \left (\frac{1-\tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{\sec ^2\left (\frac{1}{2} (e+f x)\right )}}\right )^{-2 p} \left (p (A+B) \left (\tan \left (\frac{1}{2} (e+f x)\right )+1\right )-i B (p+1) \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right ) \, _2F_1\left (1,-p;1-p;-\frac{i \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right )}{\tan \left (\frac{1}{2} (e+f x)\right )+1}\right )+i B (p+1) \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right ) \, _2F_1\left (1,-p;1-p;\frac{i \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right )}{\tan \left (\frac{1}{2} (e+f x)\right )+1}\right )\right )}{f p (p+1) \left (\tan \left (\frac{1}{2} (e+f x)\right )-1\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(g*Cos[e + f*x])^p*(A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(-1 - p),x]

[Out]

-(((g*Cos[e + f*x])^p*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^(-2*(-1 - p) - 2*p)*(c - c*Sin[e + f*x])^(-1 - p)*
((1 - Tan[(e + f*x)/2])/Sqrt[(1 + Cos[e + f*x])^(-1)])^(2*p)*((-I)*B*(1 + p)*Hypergeometric2F1[1, -p, 1 - p, (
(-I)*(-1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x)/2])]*(-1 + Tan[(e + f*x)/2]) + I*B*(1 + p)*Hypergeometric2F1[
1, -p, 1 - p, (I*(-1 + Tan[(e + f*x)/2]))/(1 + Tan[(e + f*x)/2])]*(-1 + Tan[(e + f*x)/2]) + (A + B)*p*(1 + Tan
[(e + f*x)/2])))/(2^p*f*p*(1 + p)*((1 - Tan[(e + f*x)/2])/Sqrt[Sec[(e + f*x)/2]^2])^(2*p)*(-1 + Tan[(e + f*x)/
2])))

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Maple [F]  time = 0.432, size = 0, normalized size = 0. \begin{align*} \int \left ( g\cos \left ( fx+e \right ) \right ) ^{p} \left ( A+B\sin \left ( fx+e \right ) \right ) \left ( c-c\sin \left ( fx+e \right ) \right ) ^{-1-p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-1-p),x)

[Out]

int((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-1-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 ) + A\right )} \left (g \cos \left (f x + e\right )\right )^{p}{\left (-c \sin \left (f x + e\right ) + c\right )}^{-p - 1}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-1-p),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(-p - 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-1-p),x, algorithm="fricas")

[Out]

integral((B*sin(f*x + e) + A)*(g*cos(f*x + e))^p*(-c*sin(f*x + e) + c)^(-p - 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))**p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(-1-p),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((g*cos(f*x+e))^p*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(-1-p),x, algorithm="giac")

[Out]

Exception raised: AttributeError