3.2.28 \(\int \csc ^3(a+b x) \sin ^m(2 a+2 b x) \, dx\) [128]
Optimal. Leaf size=85 \[ -\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \csc ^2(a+b x) \, _2F_1\left (\frac {1-m}{2},\frac {1}{2} (-2+m);\frac {m}{2};\sin ^2(a+b x)\right ) \sec (a+b x) \sin ^m(2 a+2 b x)}{b (2-m)} \]
[Out]
-(cos(b*x+a)^2)^(1/2-1/2*m)*csc(b*x+a)^2*hypergeom([-1+1/2*m, 1/2-1/2*m],[1/2*m],sin(b*x+a)^2)*sec(b*x+a)*sin(
2*b*x+2*a)^m/b/(2-m)
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Rubi [A]
time = 0.06, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps
used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {4395, 2657}
\begin {gather*} -\frac {\csc ^2(a+b x) \sec (a+b x) \sin ^m(2 a+2 b x) \cos ^2(a+b x)^{\frac {1-m}{2}} \, _2F_1\left (\frac {1-m}{2},\frac {m-2}{2};\frac {m}{2};\sin ^2(a+b x)\right )}{b (2-m)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^m,x]
[Out]
-(((Cos[a + b*x]^2)^((1 - m)/2)*Csc[a + b*x]^2*Hypergeometric2F1[(1 - m)/2, (-2 + m)/2, m/2, Sin[a + b*x]^2]*S
ec[a + b*x]*Sin[2*a + 2*b*x]^m)/(b*(2 - m)))
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 4395
Int[((f_.)*sin[(a_.) + (b_.)*(x_)])^(n_.)*((g_.)*sin[(c_.) + (d_.)*(x_)])^(p_), x_Symbol] :> Dist[(g*Sin[c + d
*x])^p/(Cos[a + b*x]^p*(f*Sin[a + b*x])^p), Int[Cos[a + b*x]^p*(f*Sin[a + b*x])^(n + p), x], x] /; FreeQ[{a, b
, c, d, f, g, n, p}, x] && EqQ[b*c - a*d, 0] && EqQ[d/b, 2] && !IntegerQ[p]
Rubi steps
\begin {align*} \int \csc ^3(a+b x) \sin ^m(2 a+2 b x) \, dx &=\left (\cos ^{-m}(a+b x) \sin ^{-m}(a+b x) \sin ^m(2 a+2 b x)\right ) \int \cos ^m(a+b x) \sin ^{-3+m}(a+b x) \, dx\\ &=-\frac {\cos ^2(a+b x)^{\frac {1-m}{2}} \csc ^2(a+b x) \, _2F_1\left (\frac {1-m}{2},\frac {1}{2} (-2+m);\frac {m}{2};\sin ^2(a+b x)\right ) \sec (a+b x) \sin ^m(2 a+2 b x)}{b (2-m)}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in
optimal.
time = 18.96, size = 2308, normalized size = 27.15 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Csc[a + b*x]^3*Sin[2*a + 2*b*x]^m,x]
[Out]
(AppellF1[-1 + m/2, -m, 2*m, m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Cos[(a + b*x)/2]^2*Cot[(a + b*x)/2]
^2*Sin[2*(a + b*x)]^m)/(2*b*(-2 + m)*(2*(AppellF1[m/2, 1 - m, 2*m, 1 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)
/2]^2] + 2*AppellF1[m/2, -m, 1 + 2*m, 1 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*(-1 + Cos[a + b*x]) +
AppellF1[-1 + m/2, -m, 2*m, m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*(1 + Cos[a + b*x]))) + ((4 + m)*App
ellF1[1 + m/2, -m, 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Sec[a + b*x]*Sin[(a + b*x)/2]^2*Sin[
2*(a + b*x)]^m)/(2*b*(2 + m)*((4 + m)*AppellF1[1 + m/2, -m, 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2
]^2]*(1 + Sec[a + b*x]) - 4*m*(AppellF1[2 + m/2, 1 - m, 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]
+ 2*AppellF1[2 + m/2, -m, 1 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2])*Sec[a + b*x]*Sin[(a + b
*x)/2]^2)) + ((4 + m)*AppellF1[1 + m/2, -m, 1 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Sin[a +
b*x]^2*Sin[2*(a + b*x)]^m)/(4*b*(2 + m)*(2*(m*AppellF1[2 + m/2, 1 - m, 1 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2,
-Tan[(a + b*x)/2]^2] + (1 + 2*m)*AppellF1[2 + m/2, -m, 2 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]
^2])*(-1 + Cos[a + b*x]) + (4 + m)*AppellF1[1 + m/2, -m, 1 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/
2]^2]*(1 + Cos[a + b*x]))) + (2^(-3 + m)*Cot[(a + b*x)/2]*(Sec[(a + b*x)/2]^2)^(2*m)*(Cos[(a + b*x)/2]*(-Sin[(
a + b*x)/2] + Sin[(3*(a + b*x))/2]))^m*Sin[2*(a + b*x)]^m*((2 + m)*AppellF1[m/2, -m, 2*m, 1 + m/2, Tan[(a + b*
x)/2]^2, -Tan[(a + b*x)/2]^2] - m*AppellF1[1 + m/2, -m, 1 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2
]^2]*Tan[(a + b*x)/2]^2))/(b*m*(2 + m)*(Cos[a + b*x]*Sec[(a + b*x)/2]^2)^m*((2^(-1 + m)*(Sec[(a + b*x)/2]^2)^(
2*m)*(Cos[(a + b*x)/2]*(-Sin[(a + b*x)/2] + Sin[(3*(a + b*x))/2]))^(-1 + m)*(Cos[(a + b*x)/2]*(-1/2*Cos[(a + b
*x)/2] + (3*Cos[(3*(a + b*x))/2])/2) - (Sin[(a + b*x)/2]*(-Sin[(a + b*x)/2] + Sin[(3*(a + b*x))/2]))/2)*((2 +
m)*AppellF1[m/2, -m, 2*m, 1 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - m*AppellF1[1 + m/2, -m, 1 + 2*m,
2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Tan[(a + b*x)/2]^2))/((2 + m)*(Cos[a + b*x]*Sec[(a + b*x)/2
]^2)^m) + (2^m*(Sec[(a + b*x)/2]^2)^(2*m)*(Cos[(a + b*x)/2]*(-Sin[(a + b*x)/2] + Sin[(3*(a + b*x))/2]))^m*Tan[
(a + b*x)/2]*((2 + m)*AppellF1[m/2, -m, 2*m, 1 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2] - m*AppellF1[1
+ m/2, -m, 1 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Tan[(a + b*x)/2]^2))/((2 + m)*(Cos[a + b
*x]*Sec[(a + b*x)/2]^2)^m) - (2^(-1 + m)*(Sec[(a + b*x)/2]^2)^(2*m)*(Cos[a + b*x]*Sec[(a + b*x)/2]^2)^(-1 - m)
*(Cos[(a + b*x)/2]*(-Sin[(a + b*x)/2] + Sin[(3*(a + b*x))/2]))^m*(-(Sec[(a + b*x)/2]^2*Sin[a + b*x]) + Cos[a +
b*x]*Sec[(a + b*x)/2]^2*Tan[(a + b*x)/2])*((2 + m)*AppellF1[m/2, -m, 2*m, 1 + m/2, Tan[(a + b*x)/2]^2, -Tan[(
a + b*x)/2]^2] - m*AppellF1[1 + m/2, -m, 1 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Tan[(a + b
*x)/2]^2))/(2 + m) + (2^(-1 + m)*(Sec[(a + b*x)/2]^2)^(2*m)*(Cos[(a + b*x)/2]*(-Sin[(a + b*x)/2] + Sin[(3*(a +
b*x))/2]))^m*(-(m*AppellF1[1 + m/2, -m, 1 + 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Sec[(a + b
*x)/2]^2*Tan[(a + b*x)/2]) + (2 + m)*(-1/2*(m^2*AppellF1[1 + m/2, 1 - m, 2*m, 2 + m/2, Tan[(a + b*x)/2]^2, -Ta
n[(a + b*x)/2]^2]*Sec[(a + b*x)/2]^2*Tan[(a + b*x)/2])/(1 + m/2) - (m^2*AppellF1[1 + m/2, -m, 1 + 2*m, 2 + m/2
, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Sec[(a + b*x)/2]^2*Tan[(a + b*x)/2])/(1 + m/2)) - m*Tan[(a + b*x)/2
]^2*(-(((1 + m/2)*m*AppellF1[2 + m/2, 1 - m, 1 + 2*m, 3 + m/2, Tan[(a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Sec[(a
+ b*x)/2]^2*Tan[(a + b*x)/2])/(2 + m/2)) - ((1 + m/2)*(1 + 2*m)*AppellF1[2 + m/2, -m, 2 + 2*m, 3 + m/2, Tan[(
a + b*x)/2]^2, -Tan[(a + b*x)/2]^2]*Sec[(a + b*x)/2]^2*Tan[(a + b*x)/2])/(2 + m/2))))/(m*(2 + m)*(Cos[a + b*x]
*Sec[(a + b*x)/2]^2)^m)))
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Maple [F]
time = 0.24, size = 0, normalized size = 0.00 \[\int \left (\csc ^{3}\left (x b +a \right )\right ) \left (\sin ^{m}\left (2 x b +2 a \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(b*x+a)^3*sin(2*b*x+2*a)^m,x)
[Out]
int(csc(b*x+a)^3*sin(2*b*x+2*a)^m,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(csc(b*x+a)^3*sin(2*b*x+2*a)^m,x, algorithm="maxima")
[Out]
integrate(sin(2*b*x + 2*a)^m*csc(b*x + 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(csc(b*x+a)^3*sin(2*b*x+2*a)^m,x, algorithm="fricas")
[Out]
integral(sin(2*b*x + 2*a)^m*csc(b*x + a)^3, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sin ^{m}{\left (2 a + 2 b x \right )} \csc ^{3}{\left (a + b x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(b*x+a)**3*sin(2*b*x+2*a)**m,x)
[Out]
Integral(sin(2*a + 2*b*x)**m*csc(a + b*x)**3, 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(csc(b*x+a)^3*sin(2*b*x+2*a)^m,x, algorithm="giac")
[Out]
integrate(sin(2*b*x + 2*a)^m*csc(b*x + a)^3, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (2\,a+2\,b\,x\right )}^m}{{\sin \left (a+b\,x\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(2*a + 2*b*x)^m/sin(a + b*x)^3,x)
[Out]
int(sin(2*a + 2*b*x)^m/sin(a + b*x)^3, x)
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