3.137 \(\int \frac{(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx\)

Optimal. Leaf size=100 \[ \frac{e \cos (c+d x) (e \sin (c+d x))^{m-1} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m-1}{2},\frac{m+1}{2},\sin ^2(c+d x)\right )}{a d (1-m) \sqrt{\cos ^2(c+d x)}}-\frac{e (e \sin (c+d x))^{m-1}}{a d (1-m)} \]

[Out]

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

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Rubi [A]  time = 0.198809, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3872, 2839, 2564, 30, 2577} \[ \frac{e \cos (c+d x) (e \sin (c+d x))^{m-1} \, _2F_1\left (-\frac{1}{2},\frac{m-1}{2};\frac{m+1}{2};\sin ^2(c+d x)\right )}{a d (1-m) \sqrt{\cos ^2(c+d x)}}-\frac{e (e \sin (c+d x))^{m-1}}{a d (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x]),x]

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2839

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

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

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

Mathematica [F]  time = 29.8222, size = 0, normalized size = 0. \[ \int \frac{(e \sin (c+d x))^m}{a+a \sec (c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x]),x]

[Out]

Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x]), x]

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Maple [F]  time = 0.619, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\sin \left ( dx+c \right ) \right ) ^{m}}{a+a\sec \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c)),x)

[Out]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (e \sin \left (d x + c\right )\right )^{m}}{a \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

integral((e*sin(d*x + c))^m/(a*sec(d*x + c) + a), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (e \sin{\left (c + d x \right )}\right )^{m}}{\sec{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))**m/(a+a*sec(d*x+c)),x)

[Out]

Integral((e*sin(c + d*x))**m/(sec(c + d*x) + 1), x)/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^m/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a), x)