3.3.61 \(\int \frac {(e \sin (c+d x))^m}{(a+b \sec (c+d x))^2} \, dx\) [261]

Optimal. Leaf size=405 \[ -\frac {2 b e F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)}+\frac {b^2 e F_1\left (2-m;\frac {1-m}{2},\frac {1-m}{2};3-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (2-m) (b+a \cos (c+d x))}+\frac {\cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a^2 d e (1+m) \sqrt {\cos ^2(c+d x)}} \]

[Out]

-2*b*e*AppellF1(1-m,1/2-1/2*m,1/2-1/2*m,2-m,(-a+b)/(b+a*cos(d*x+c)),(a+b)/(b+a*cos(d*x+c)))*(-a*(1-cos(d*x+c))
/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(a*(1+cos(d*x+c))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(e*sin(d*x+c))^(-1+m)/a^3/d/(1-
m)+b^2*e*AppellF1(2-m,1/2-1/2*m,1/2-1/2*m,3-m,(-a+b)/(b+a*cos(d*x+c)),(a+b)/(b+a*cos(d*x+c)))*(-a*(1-cos(d*x+c
))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(a*(1+cos(d*x+c))/(b+a*cos(d*x+c)))^(1/2-1/2*m)*(e*sin(d*x+c))^(-1+m)/a^3/d/(
2-m)/(b+a*cos(d*x+c))+cos(d*x+c)*hypergeom([1/2, 1/2+1/2*m],[3/2+1/2*m],sin(d*x+c)^2)*(e*sin(d*x+c))^(1+m)/a^2
/d/e/(1+m)/(cos(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.31, antiderivative size = 405, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3957, 2991, 2722, 2782} \begin {gather*} \frac {b^2 e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (2-m;\frac {1-m}{2},\frac {1-m}{2};3-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^3 d (2-m) (a \cos (c+d x)+b)}-\frac {2 b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^3 d (1-m)}+\frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{a^2 d e (m+1) \sqrt {\cos ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*e*AppellF1[1 - m, (1 - m)/2, (1 - m)/2, 2 - m, -((a - b)/(b + a*Cos[c + d*x])), (a + b)/(b + a*Cos[c + d
*x])]*(-((a*(1 - Cos[c + d*x]))/(b + a*Cos[c + d*x])))^((1 - m)/2)*((a*(1 + Cos[c + d*x]))/(b + a*Cos[c + d*x]
))^((1 - m)/2)*(e*Sin[c + d*x])^(-1 + m))/(a^3*d*(1 - m)) + (b^2*e*AppellF1[2 - m, (1 - m)/2, (1 - m)/2, 3 - m
, -((a - b)/(b + a*Cos[c + d*x])), (a + b)/(b + a*Cos[c + d*x])]*(-((a*(1 - Cos[c + d*x]))/(b + a*Cos[c + d*x]
)))^((1 - m)/2)*((a*(1 + Cos[c + d*x]))/(b + a*Cos[c + d*x]))^((1 - m)/2)*(e*Sin[c + d*x])^(-1 + m))/(a^3*d*(2
- m)*(b + a*Cos[c + d*x])) + (Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + m)/2, (3 + m)/2, Sin[c + d*x]^2]*(e*Si
n[c + d*x])^(1 + m))/(a^2*d*e*(1 + m)*Sqrt[Cos[c + d*x]^2])

Rule 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2782

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p)*((-b)*((1 - Sin[e + f*x])/(a + b*Sin[e + f*x])
))^((p - 1)/2)*(b*((1 + Sin[e + f*x])/(a + b*Sin[e + f*x])))^((p - 1)/2)))*AppellF1[-p - m, (1 - p)/2, (1 - p)
/2, 1 - p - m, (a + b)/(a + b*Sin[e + f*x]), (a - b)/(a + b*Sin[e + f*x])], x] /; FreeQ[{a, b, e, f, g, p}, x]
&& NeQ[a^2 - b^2, 0] && ILtQ[m, 0] &&  !IGtQ[m + p + 1, 0]

Rule 2991

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] && NeQ[a^2 - b^2, 0] && IntegerQ[m] && (GtQ[m, 0] || IntegerQ[n])

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+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) (e \sin (c+d x))^m}{(-b-a \cos (c+d x))^2} \, dx\\ &=\int \left (\frac {(e \sin (c+d x))^m}{a^2}+\frac {b^2 (e \sin (c+d x))^m}{a^2 (b+a \cos (c+d x))^2}-\frac {2 b (e \sin (c+d x))^m}{a^2 (b+a \cos (c+d x))}\right ) \, dx\\ &=\frac {\int (e \sin (c+d x))^m \, dx}{a^2}-\frac {(2 b) \int \frac {(e \sin (c+d x))^m}{b+a \cos (c+d x)} \, dx}{a^2}+\frac {b^2 \int \frac {(e \sin (c+d x))^m}{(b+a \cos (c+d x))^2} \, dx}{a^2}\\ &=-\frac {2 b e F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (1-m)}+\frac {b^2 e F_1\left (2-m;\frac {1-m}{2},\frac {1-m}{2};3-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right ) \left (-\frac {a (1-\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} \left (\frac {a (1+\cos (c+d x))}{b+a \cos (c+d x)}\right )^{\frac {1-m}{2}} (e \sin (c+d x))^{-1+m}}{a^3 d (2-m) (b+a \cos (c+d x))}+\frac {\cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1+m}{2};\frac {3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{a^2 d e (1+m) \sqrt {\cos ^2(c+d x)}}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(1433\) vs. \(2(405)=810\).
time = 9.85, size = 1433, normalized size = 3.54 \begin {gather*} -\frac {4 b F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) (b+a \cos (c+d x)) \sec ^2(c+d x) (e \sin (c+d x))^m \tan \left (\frac {1}{2} (c+d x)\right )}{a^2 d (a+b \sec (c+d x))^2 \left (F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+2 m F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \cot (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+2 m F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-\frac {2 (1+m) \left ((-a+b) F_1\left (\frac {3+m}{2};m,2;\frac {5+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) m F_1\left (\frac {3+m}{2};1+m,1;\frac {5+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{(a+b) (3+m)}\right )}+\frac {2 b^2 \left ((a+b) F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a F_1\left (\frac {1+m}{2};m,2;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2(c+d x) (e \sin (c+d x))^m \tan \left (\frac {1}{2} (c+d x)\right )}{a^2 d (a+b \sec (c+d x))^2 \left (\left ((a+b) F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a F_1\left (\frac {1+m}{2};m,2;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )+2 m \left ((a+b) F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a F_1\left (\frac {1+m}{2};m,2;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \cot (c+d x) \tan \left (\frac {1}{2} (c+d x)\right )+2 m \left ((a+b) F_1\left (\frac {1+m}{2};m,1;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a F_1\left (\frac {1+m}{2};m,2;\frac {3+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )-\frac {2 (1+m) \left (\left (-a^2+b^2\right ) F_1\left (\frac {3+m}{2};m,2;\frac {5+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+4 a (a-b) F_1\left (\frac {3+m}{2};m,3;\frac {5+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) m \left ((a+b) F_1\left (\frac {3+m}{2};1+m,1;\frac {5+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )-2 a F_1\left (\frac {3+m}{2};1+m,2;\frac {5+m}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(a-b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right )\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{(a+b) (3+m)}\right )}-\frac {(b+a \cos (c+d x))^2 \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3}{2};\cos ^2(c+d x)\right ) (e \sin (c+d x))^m \sin ^2(c+d x)^{\frac {1}{2} (-1-m)} \tan (c+d x)}{a^2 d (a+b \sec (c+d x))^2} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-4*b*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*(b + a*C
os[c + d*x])*Sec[c + d*x]^2*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^2*d*(a + b*Sec[c + d*x])^2*(AppellF1[(1 +
m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*Sec[(c + d*x)/2]^2 + 2*m*App
ellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)]*Cot[c + d*x]*Tan[
(c + d*x)/2] + 2*m*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a +
b)]*Tan[(c + d*x)/2]^2 - (2*(1 + m)*((-a + b)*AppellF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a -
b)*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*m*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a
- b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2)/((a + b)*(3 + m)))) + (2*b^2*((a +
b)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*Appel
lF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[c + d*x]^2*(e*
Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^2*d*(a + b*Sec[c + d*x])^2*(((a + b)*AppellF1[(1 + m)/2, m, 1, (3 + m)/2,
-Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*AppellF1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c
+ d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Sec[(c + d*x)/2]^2 + 2*m*((a + b)*AppellF1[(1 + m)/2, m,
1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*AppellF1[(1 + m)/2, m, 2, (3 +
m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Cot[c + d*x]*Tan[(c + d*x)/2] + 2*m*((a + b)
*AppellF1[(1 + m)/2, m, 1, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)] - 2*a*AppellF
1[(1 + m)/2, m, 2, (3 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])*Tan[(c + d*x)/2]^2 -
(2*(1 + m)*((-a^2 + b^2)*AppellF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^
2)/(a + b)] + 4*a*(a - b)*AppellF1[(3 + m)/2, m, 3, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d*x)/2]^
2)/(a + b)] + (a + b)*m*((a + b)*AppellF1[(3 + m)/2, 1 + m, 1, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c
+ d*x)/2]^2)/(a + b)] - 2*a*AppellF1[(3 + m)/2, 1 + m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*Tan[(c + d
*x)/2]^2)/(a + b)]))*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]^2)/((a + b)*(3 + m)))) - ((b + a*Cos[c + d*x])^2*Hype
rgeometric2F1[1/2, (1 - m)/2, 3/2, Cos[c + d*x]^2]*(e*Sin[c + d*x])^m*(Sin[c + d*x]^2)^((-1 - m)/2)*Tan[c + d*
x])/(a^2*d*(a + b*Sec[c + d*x])^2)

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{\left (a +b \sec \left (d x +c \right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((e*sin(d*x+c))^m/(a+b*sec(d*x+c))^2,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+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate((e*sin(d*x + c))^m/(b*sec(d*x + c) + a)^2, 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+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((e*sin(d*x + c))^m/(b*sec(d*x + c) + a)^2, 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 )}^2\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{{\left (b+a\,\cos \left (c+d\,x\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^m/(a + b/cos(c + d*x))^2,x)

[Out]

int((cos(c + d*x)^2*(e*sin(c + d*x))^m)/(b + a*cos(c + d*x))^2, x)

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