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

Optimal. Leaf size=232 \[ \frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{a d e (m+1) \sqrt {\cos ^2(c+d x)}}-\frac {b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^2 d (1-m)} \]

[Out]

-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^2/d/(1-m)
+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/d/e/(1+m)/(cos(d*x+c)^
2)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 232, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3872, 2867, 2643, 2703} \[ \frac {\cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac {1}{2},\frac {m+1}{2};\frac {m+3}{2};\sin ^2(c+d x)\right )}{a d e (m+1) \sqrt {\cos ^2(c+d x)}}-\frac {b e (e \sin (c+d x))^{m-1} \left (-\frac {a (1-\cos (c+d x))}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} \left (\frac {a (\cos (c+d x)+1)}{a \cos (c+d x)+b}\right )^{\frac {1-m}{2}} F_1\left (1-m;\frac {1-m}{2},\frac {1-m}{2};2-m;-\frac {a-b}{b+a \cos (c+d x)},\frac {a+b}{b+a \cos (c+d x)}\right )}{a^2 d (1-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2867

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

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]

Rubi steps

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

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Mathematica [B]  time = 5.97, size = 687, normalized size = 2.96 \[ \frac {2 \tan \left (\frac {1}{2} (c+d x)\right ) (e \sin (c+d x))^m \left ((a+b) \, _2F_1\left (\frac {m+1}{2},m+1;\frac {m+3}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b F_1\left (\frac {m+1}{2};m,1;\frac {m+3}{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 )}{d (a+b \sec (c+d x)) \left (2 m \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left ((a+b) \, _2F_1\left (\frac {m+1}{2},m+1;\frac {m+3}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b F_1\left (\frac {m+1}{2};m,1;\frac {m+3}{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 )+2 m \tan \left (\frac {1}{2} (c+d x)\right ) \cot (c+d x) \left ((a+b) \, _2F_1\left (\frac {m+1}{2},m+1;\frac {m+3}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b F_1\left (\frac {m+1}{2};m,1;\frac {m+3}{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 ) \left ((a+b) \, _2F_1\left (\frac {m+1}{2},m+1;\frac {m+3}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b F_1\left (\frac {m+1}{2};m,1;\frac {m+3}{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 )+\frac {(m+1) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (\frac {2 b \tan ^2\left (\frac {1}{2} (c+d x)\right ) \left ((b-a) F_1\left (\frac {m+3}{2};m,2;\frac {m+5}{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 )+m (a+b) F_1\left (\frac {m+3}{2};m+1,1;\frac {m+5}{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 )}{m+3}-(a+b)^2 \left (\, _2F_1\left (\frac {m+1}{2},m+1;\frac {m+3}{2};-\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )^{-m-1}\right )\right )}{a+b}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(-(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)]) + (a
+ b)*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*(e*Sin[c + d*x])^m*Tan[(c + d*x)/2])
/(d*(a + b*Sec[c + d*x])*((-(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)]) + (a + b)*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*Sec[(c + d*x)
/2]^2 + 2*m*Cot[c + d*x]*(-(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)]) + (a + b)*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/
2] + 2*m*(-(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)])
+ (a + b)*Hypergeometric2F1[(1 + m)/2, 1 + m, (3 + m)/2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2 + ((1 + m)*
Sec[(c + d*x)/2]^2*(-((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))) + (2*b*((-a + b)*AppellF1[(3 + m)/2, m, 2, (5 + m)/2, -Tan[(c + d*x)/2]^2, ((a - b)*T
an[(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)])*Tan[(c + d*x)/2]^2)/(3 + m)))/(a + b)))

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fricas [F]  time = 0.50, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e \sin \left (d x + c\right )\right )^{m}}{b \sec \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.98, size = 0, normalized size = 0.00 \[ \int \frac {\left (e \sin \left (d x +c \right )\right )^{m}}{a +b \sec \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\sin \left (c+d\,x\right )\right )}^m}{b+a\,\cos \left (c+d\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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