3.139 \(\int \frac{(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=236 \[ \frac{e^5 \cos (c+d x) (e \sin (c+d x))^{m-5} \text{Hypergeometric2F1}\left (-\frac{5}{2},\frac{m-5}{2},\frac{m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{3 e^5 \cos (c+d x) (e \sin (c+d x))^{m-5} \text{Hypergeometric2F1}\left (-\frac{3}{2},\frac{m-5}{2},\frac{m-3}{2},\sin ^2(c+d x)\right )}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}-\frac{4 e^5 (e \sin (c+d x))^{m-5}}{a^3 d (5-m)}+\frac{7 e^3 (e \sin (c+d x))^{m-3}}{a^3 d (3-m)}-\frac{3 e (e \sin (c+d x))^{m-1}}{a^3 d (1-m)} \]

[Out]

(-4*e^5*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)) + (e^5*Cos[c + d*x]*Hypergeometric2F1[-5/2, (-5 + m)/2, (-3
+ m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)*Sqrt[Cos[c + d*x]^2]) + (3*e^5*Cos[c + d*x]
*Hypergeometric2F1[-3/2, (-5 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)*Sqr
t[Cos[c + d*x]^2]) + (7*e^3*(e*Sin[c + d*x])^(-3 + m))/(a^3*d*(3 - m)) - (3*e*(e*Sin[c + d*x])^(-1 + m))/(a^3*
d*(1 - m))

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Rubi [A]  time = 0.635327, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {3872, 2875, 2873, 2564, 14, 2577, 270} \[ \frac{e^5 \cos (c+d x) (e \sin (c+d x))^{m-5} \, _2F_1\left (-\frac{5}{2},\frac{m-5}{2};\frac{m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{3 e^5 \cos (c+d x) (e \sin (c+d x))^{m-5} \, _2F_1\left (-\frac{3}{2},\frac{m-5}{2};\frac{m-3}{2};\sin ^2(c+d x)\right )}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}-\frac{4 e^5 (e \sin (c+d x))^{m-5}}{a^3 d (5-m)}+\frac{7 e^3 (e \sin (c+d x))^{m-3}}{a^3 d (3-m)}-\frac{3 e (e \sin (c+d x))^{m-1}}{a^3 d (1-m)} \]

Antiderivative was successfully verified.

[In]

Int[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^3,x]

[Out]

(-4*e^5*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)) + (e^5*Cos[c + d*x]*Hypergeometric2F1[-5/2, (-5 + m)/2, (-3
+ m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)*Sqrt[Cos[c + d*x]^2]) + (3*e^5*Cos[c + d*x]
*Hypergeometric2F1[-3/2, (-5 + m)/2, (-3 + m)/2, Sin[c + d*x]^2]*(e*Sin[c + d*x])^(-5 + m))/(a^3*d*(5 - m)*Sqr
t[Cos[c + d*x]^2]) + (7*e^3*(e*Sin[c + d*x])^(-3 + m))/(a^3*d*(3 - m)) - (3*e*(e*Sin[c + d*x])^(-1 + m))/(a^3*
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 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]]

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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx &=-\int \frac{\cos ^3(c+d x) (e \sin (c+d x))^m}{(-a-a \cos (c+d x))^3} \, dx\\ &=-\frac{e^6 \int \cos ^3(c+d x) (-a+a \cos (c+d x))^3 (e \sin (c+d x))^{-6+m} \, dx}{a^6}\\ &=-\frac{e^6 \int \left (-a^3 \cos ^3(c+d x) (e \sin (c+d x))^{-6+m}+3 a^3 \cos ^4(c+d x) (e \sin (c+d x))^{-6+m}-3 a^3 \cos ^5(c+d x) (e \sin (c+d x))^{-6+m}+a^3 \cos ^6(c+d x) (e \sin (c+d x))^{-6+m}\right ) \, dx}{a^6}\\ &=\frac{e^6 \int \cos ^3(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3}-\frac{e^6 \int \cos ^6(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3}-\frac{\left (3 e^6\right ) \int \cos ^4(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3}+\frac{\left (3 e^6\right ) \int \cos ^5(c+d x) (e \sin (c+d x))^{-6+m} \, dx}{a^3}\\ &=\frac{e^5 \cos (c+d x) \, _2F_1\left (-\frac{5}{2},\frac{1}{2} (-5+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{3 e^5 \cos (c+d x) \, _2F_1\left (-\frac{3}{2},\frac{1}{2} (-5+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{e^5 \operatorname{Subst}\left (\int x^{-6+m} \left (1-\frac{x^2}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^3 d}+\frac{\left (3 e^5\right ) \operatorname{Subst}\left (\int x^{-6+m} \left (1-\frac{x^2}{e^2}\right )^2 \, dx,x,e \sin (c+d x)\right )}{a^3 d}\\ &=\frac{e^5 \cos (c+d x) \, _2F_1\left (-\frac{5}{2},\frac{1}{2} (-5+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{3 e^5 \cos (c+d x) \, _2F_1\left (-\frac{3}{2},\frac{1}{2} (-5+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{e^5 \operatorname{Subst}\left (\int \left (x^{-6+m}-\frac{x^{-4+m}}{e^2}\right ) \, dx,x,e \sin (c+d x)\right )}{a^3 d}+\frac{\left (3 e^5\right ) \operatorname{Subst}\left (\int \left (x^{-6+m}-\frac{2 x^{-4+m}}{e^2}+\frac{x^{-2+m}}{e^4}\right ) \, dx,x,e \sin (c+d x)\right )}{a^3 d}\\ &=-\frac{4 e^5 (e \sin (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac{e^5 \cos (c+d x) \, _2F_1\left (-\frac{5}{2},\frac{1}{2} (-5+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{3 e^5 \cos (c+d x) \, _2F_1\left (-\frac{3}{2},\frac{1}{2} (-5+m);\frac{1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt{\cos ^2(c+d x)}}+\frac{7 e^3 (e \sin (c+d x))^{-3+m}}{a^3 d (3-m)}-\frac{3 e (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)}\\ \end{align*}

Mathematica [F]  time = 1.21067, size = 0, normalized size = 0. \[ \int \frac{(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^3,x]

[Out]

Integrate[(e*Sin[c + d*x])^m/(a + a*Sec[c + d*x])^3, x]

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Maple [F]  time = 0.409, 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 ) ^{3}}}\, 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))^3,x)

[Out]

int((e*sin(d*x+c))^m/(a+a*sec(d*x+c))^3,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 )}^{3}}\,{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))^3,x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a)^3, 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^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}, 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))^3,x, algorithm="fricas")

[Out]

integral((e*sin(d*x + c))^m/(a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3*a^3*sec(d*x + c) + a^3), 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((e*sin(d*x+c))**m/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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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 )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^m/(a*sec(d*x + c) + a)^3, x)