3.2.39 \(\int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx\) [139]

Optimal. Leaf size=236 \[ -\frac {4 e^5 (e \sin (c+d x))^{-5+m}}{a^3 d (5-m)}+\frac {e^5 \cos (c+d x) \, _2F_1\left (-\frac {5}{2},\frac {1}{2} (-5+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {3 e^5 \cos (c+d x) \, _2F_1\left (-\frac {3}{2},\frac {1}{2} (-5+m);\frac {1}{2} (-3+m);\sin ^2(c+d x)\right ) (e \sin (c+d x))^{-5+m}}{a^3 d (5-m) \sqrt {\cos ^2(c+d x)}}+\frac {7 e^3 (e \sin (c+d x))^{-3+m}}{a^3 d (3-m)}-\frac {3 e (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)} \]

[Out]

-4*e^5*(e*sin(d*x+c))^(-5+m)/a^3/d/(5-m)+7*e^3*(e*sin(d*x+c))^(-3+m)/a^3/d/(3-m)-3*e*(e*sin(d*x+c))^(-1+m)/a^3
/d/(1-m)+e^5*cos(d*x+c)*hypergeom([-5/2, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^(-5+m)/a^3/d/(5
-m)/(cos(d*x+c)^2)^(1/2)+3*e^5*cos(d*x+c)*hypergeom([-3/2, -5/2+1/2*m],[-3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c
))^(-5+m)/a^3/d/(5-m)/(cos(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.45, antiderivative size = 236, normalized size of antiderivative = 1.00, 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 = {3957, 2954, 2952, 2644, 14, 2657, 276} \begin {gather*} \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)} \end {gather*}

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 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 276

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]

Rule 2644

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 2657

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)/(a*f*(m + 1)*(Cos[e + f*x]^
2)^FracPart[(n - 1)/2]))*Hypergeometric2F1[(1 + m)/2, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2], x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rule 2952

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 2954

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 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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]

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 \text {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 ) \text {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 \text {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 ) \text {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*}

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Mathematica [F]
time = 6.41, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(e \sin (c+d x))^m}{(a+a \sec (c+d x))^3} \, dx \end {gather*}

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 = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +a \sec \left (d x +c \right )\right )^{3}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (e \sin {\left (c + d x \right )}\right )^{m}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \end {gather*}

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]

Integral((e*sin(c + d*x))**m/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\cos \left (c+d\,x\right )}^3\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{a^3\,{\left (\cos \left (c+d\,x\right )+1\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^m/(a + a/cos(c + d*x))^3,x)

[Out]

int((cos(c + d*x)^3*(e*sin(c + d*x))^m)/(a^3*(cos(c + d*x) + 1)^3), x)

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