3.138 \(\int \frac{(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=207 \[ -\frac{e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \text{Hypergeometric2F1}\left (-\frac{3}{2},\frac{m-3}{2},\frac{m-1}{2},\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{m-3}{2},\frac{m-1}{2},\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}+\frac{2 e^3 (e \sin (c+d x))^{m-3}}{a^2 d (3-m)}-\frac{2 e (e \sin (c+d x))^{m-1}}{a^2 d (1-m)} \]

[Out]

(2*e^3*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)) - (e^3*Cos[c + d*x]*Hypergeometric2F1[-3/2, (-3 + m)/2, (-1
+ m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)*Sqrt[Cos[c + d*x]^2]) - (e^3*Cos[c + d*x]*Hy
pergeometric2F1[-1/2, (-3 + m)/2, (-1 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)*Sqrt[C
os[c + d*x]^2]) - (2*e*(e*Sin[c + d*x])^(-1 + m))/(a^2*d*(1 - m))

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Rubi [A]  time = 0.525747, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3872, 2875, 2873, 2577, 2564, 14} \[ -\frac{e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \, _2F_1\left (-\frac{3}{2},\frac{m-3}{2};\frac{m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{e^3 \cos (c+d x) (e \sin (c+d x))^{m-3} \, _2F_1\left (-\frac{1}{2},\frac{m-3}{2};\frac{m-1}{2};\sin ^2(c+d x)\right )}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}+\frac{2 e^3 (e \sin (c+d x))^{m-3}}{a^2 d (3-m)}-\frac{2 e (e \sin (c+d x))^{m-1}}{a^2 d (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^2,x]

[Out]

(2*e^3*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)) - (e^3*Cos[c + d*x]*Hypergeometric2F1[-3/2, (-3 + m)/2, (-1
+ m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)*Sqrt[Cos[c + d*x]^2]) - (e^3*Cos[c + d*x]*Hy
pergeometric2F1[-1/2, (-3 + m)/2, (-1 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-3 + m))/(a^2*d*(3 - m)*Sqrt[C
os[c + d*x]^2]) - (2*e*(e*Sin[c + d*x])^(-1 + m))/(a^2*d*(1 - m))

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 2875

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

Rule 2873

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

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]

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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

\begin{align*} \int \frac{(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx &=\int \frac{\cos ^2(c+d x) (e \sin (c+d x))^m}{(-a-a \cos (c+d x))^2} \, dx\\ &=\frac{e^4 \int \cos ^2(c+d x) (-a+a \cos (c+d x))^2 (e \sin (c+d x))^{-4+m} \, dx}{a^4}\\ &=\frac{e^4 \int \left (a^2 \cos ^2(c+d x) (e \sin (c+d x))^{-4+m}-2 a^2 \cos ^3(c+d x) (e \sin (c+d x))^{-4+m}+a^2 \cos ^4(c+d x) (e \sin (c+d x))^{-4+m}\right ) \, dx}{a^4}\\ &=\frac{e^4 \int \cos ^2(c+d x) (e \sin (c+d x))^{-4+m} \, dx}{a^2}+\frac{e^4 \int \cos ^4(c+d x) (e \sin (c+d x))^{-4+m} \, dx}{a^2}-\frac{\left (2 e^4\right ) \int \cos ^3(c+d x) (e \sin (c+d x))^{-4+m} \, dx}{a^2}\\ &=-\frac{e^3 \cos (c+d x) \, _2F_1\left (-\frac{3}{2},\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{e^3 \cos (c+d x) \, _2F_1\left (-\frac{1}{2},\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int x^{-4+m} \left (1-\frac{x^2}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=-\frac{e^3 \cos (c+d x) \, _2F_1\left (-\frac{3}{2},\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{e^3 \cos (c+d x) \, _2F_1\left (-\frac{1}{2},\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{\left (2 e^3\right ) \operatorname{Subst}\left (\int \left (x^{-4+m}-\frac{x^{-2+m}}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^2 d}\\ &=\frac{2 e^3 (e \sin (c+d x))^{-3+m}}{a^2 d (3-m)}-\frac{e^3 \cos (c+d x) \, _2F_1\left (-\frac{3}{2},\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{e^3 \cos (c+d x) \, _2F_1\left (-\frac{1}{2},\frac{1}{2} (-3+m);\frac{1}{2} (-1+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-3+m}}{a^2 d (3-m) \sqrt{\cos ^2(c+d x)}}-\frac{2 e (e \sin (c+d x))^{-1+m}}{a^2 d (1-m)}\\ \end{align*}

Mathematica [F]  time = 0.673564, size = 0, normalized size = 0. \[ \int \frac{(e \sin (c+d x))^m}{(a+a \sec (c+d x))^2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^2,x]

[Out]

Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^2, x]

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Maple [F]  time = 0.364, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\sin \left ( dx+c \right ) \right ) ^{m}}{ \left ( a+a\sec \left ( dx+c \right ) \right ) ^{2}}}\, 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))^2,x)

[Out]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^2,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}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a)^2, 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^{2} \sec \left (d x + c\right )^{2} + 2 \, a^{2} \sec \left (d x + c\right ) + a^{2}}, 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))^2,x, algorithm="fricas")

[Out]

integral((e*sin(d*x + c))^m/(a^2*sec(d*x + c)^2 + 2*a^2*sec(d*x + c) + a^2), 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 ^{2}{\left (c + d x \right )} + 2 \sec{\left (c + d x \right )} + 1}\, dx}{a^{2}} \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))**2,x)

[Out]

Integral((e*sin(c + d*x))**m/(sec(c + d*x)**2 + 2*sec(c + d*x) + 1), x)/a**2

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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}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}\,{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))^2,x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a)^2, x)