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

Optimal. Leaf size=405 \[ \frac{\cos (c+d x) (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{a^2 d e (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{b^2 e (e \sin (c+d x))^{m-1} \left (-\frac{a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} \left (\frac{a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} F_1\left (2-m;\frac{1-m}{2},\frac{1-m}{2};3-m;-\frac{a-b}{b+a \cos (c+d x)},\frac{a+b}{b+a \cos (c+d x)}\right )}{a^3 d (2-m) (a \cos (c+d x)+b)}-\frac{2 b e (e \sin (c+d x))^{m-1} \left (-\frac{a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} \left (\frac{a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} F_1\left (1-m;\frac{1-m}{2},\frac{1-m}{2};2-m;-\frac{a-b}{b+a \cos (c+d x)},\frac{a+b}{b+a \cos (c+d x)}\right )}{a^3 d (1-m)} \]

[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])

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Rubi [A]  time = 0.455521, antiderivative size = 405, normalized size of antiderivative = 1., 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 = {3872, 2912, 2643, 2703} \[ \frac{b^2 e (e \sin (c+d x))^{m-1} \left (-\frac{a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} \left (\frac{a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} F_1\left (2-m;\frac{1-m}{2},\frac{1-m}{2};3-m;-\frac{a-b}{b+a \cos (c+d x)},\frac{a+b}{b+a \cos (c+d x)}\right )}{a^3 d (2-m) (a \cos (c+d x)+b)}-\frac{2 b e (e \sin (c+d x))^{m-1} \left (-\frac{a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} \left (\frac{a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac{1-m}{2}} F_1\left (1-m;\frac{1-m}{2},\frac{1-m}{2};2-m;-\frac{a-b}{b+a \cos (c+d x)},\frac{a+b}{b+a \cos (c+d x)}\right )}{a^3 d (1-m)}+\frac{\cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{a^2 d e (m+1) \sqrt{\cos ^2(c+d x)}} \]

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 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 2912

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 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2703

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

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*}

Mathematica [B]  time = 14.9641, size = 1494, normalized size = 3.69 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(b*(-2*a^2 - a*b + b^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)] + 2*a*b^2*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)] + (a - b)*(a + b)^2*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*(Sec[(c + d*
x)/2]^2)^m*Sin[c + d*x]^m*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])/(a^2*(a - b)*(a + b)^2*d*(1 + m)*(a + b*Sec[c +
d*x])^2*(((b*(-2*a^2 - a*b + b^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)] + 2*a*b^2*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)] + (a - b)*(a + b)^2*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*(Sec
[(c + d*x)/2]^2)^(1 + m)*Sin[c + d*x]^m)/(a^2*(a - b)*(a + b)^2*(1 + m)) + (2*m*Cos[c + d*x]*(b*(-2*a^2 - a*b
+ b^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)] + 2*a*b
^2*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)] + (a - b)*(
a + b)^2*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*(Sec[(c + d*x)/2]^2)^m*Sin[c + d
*x]^(-1 + m)*Tan[(c + d*x)/2])/(a^2*(a - b)*(a + b)^2*(1 + m)) + (2*m*(b*(-2*a^2 - a*b + b^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)] + 2*a*b^2*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)] + (a - b)*(a + b)^2*Hypergeometric
2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*(Sec[(c + d*x)/2]^2)^m*Sin[c + d*x]^m*Tan[(c + d*x)/2]^
2)/(a^2*(a - b)*(a + b)^2*(1 + m)) + (2*(Sec[(c + d*x)/2]^2)^m*Sin[c + d*x]^m*Tan[(c + d*x)/2]*(b*(-2*a^2 - a*
b + b^2)*(((a - b)*(1 + m)*AppellF1[1 + (1 + m)/2, m, 2, 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*Tan[(c + d*x)/2])/((a + b)*(3 + m)) - (m*(1 + m)*AppellF1[1 + (1 + m)/
2, 1 + m, 1, 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*Tan[
(c + d*x)/2])/(3 + m)) + 2*a*b^2*((2*(a - b)*(1 + m)*AppellF1[1 + (1 + m)/2, m, 3, 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*Tan[(c + d*x)/2])/((a + b)*(3 + m)) - (m*(1
+ m)*AppellF1[1 + (1 + m)/2, 1 + m, 2, 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*Tan[(c + d*x)/2])/(3 + m)) + ((a - b)*(a + b)^2*(1 + m)*Csc[(c + d*x)/2]*Sec[(c + d*x)/2
]*(-Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2] + (1 + Tan[(c + d*x)/2]^2)^(-1 - m)))/
2))/(a^2*(a - b)*(a + b)^2*(1 + m))))

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Maple [F]  time = 0.314, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( e\sin \left ( dx+c \right ) \right ) ^{m}}{ \left ( a+b\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+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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \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+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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (e \sin \left (d x + c\right )\right )^{m}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \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+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., size = 0, normalized size = 0. \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (e \sin \left (d x + c\right )\right )^{m}}{{\left (b \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+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)