3.499 \(\int (b \sec (e+f x))^n \sin ^2(e+f x) \, dx\)
Optimal. Leaf size=73 \[ -\frac {b \sin (e+f x) (b \sec (e+f x))^{n-1} \, _2F_1\left (-\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]
[Out]
-b*hypergeom([-1/2, 1/2-1/2*n],[3/2-1/2*n],cos(f*x+e)^2)*(b*sec(f*x+e))^(-1+n)*sin(f*x+e)/f/(1-n)/(sin(f*x+e)^
2)^(1/2)
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Rubi [A] time = 0.08, antiderivative size = 73, normalized size of antiderivative = 1.00,
number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used =
{2632, 2576} \[ -\frac {b \sin (e+f x) (b \sec (e+f x))^{n-1} \, _2F_1\left (-\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right )}{f (1-n) \sqrt {\sin ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[(b*Sec[e + f*x])^n*Sin[e + f*x]^2,x]
[Out]
-((b*Hypergeometric2F1[-1/2, (1 - n)/2, (3 - n)/2, Cos[e + f*x]^2]*(b*Sec[e + f*x])^(-1 + n)*Sin[e + f*x])/(f*
(1 - n)*Sqrt[Sin[e + f*x]^2]))
Rule 2576
Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b^(2*IntPar
t[(n - 1)/2] + 1)*(b*Sin[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Cos[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/
2, (1 - n)/2, (3 + m)/2, Cos[e + f*x]^2])/(a*f*(m + 1)*(Sin[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a,
b, e, f, m, n}, x] && SimplerQ[n, m]
Rule 2632
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a^2*(a*Sec[e
+ f*x])^(m - 1)*(b*Csc[e + f*x])^(n + 1)*(a*Cos[e + f*x])^(m - 1)*(b*Sin[e + f*x])^(n + 1))/b^2, Int[1/((a*Co
s[e + f*x])^m*(b*Sin[e + f*x])^n), x], x] /; FreeQ[{a, b, e, f, m, n}, x]
Rubi steps
\begin {align*} \int (b \sec (e+f x))^n \sin ^2(e+f x) \, dx &=\left (b^2 (b \cos (e+f x))^{-1+n} (b \sec (e+f x))^{-1+n}\right ) \int (b \cos (e+f x))^{-n} \sin ^2(e+f x) \, dx\\ &=-\frac {b \, _2F_1\left (-\frac {1}{2},\frac {1-n}{2};\frac {3-n}{2};\cos ^2(e+f x)\right ) (b \sec (e+f x))^{-1+n} \sin (e+f x)}{f (1-n) \sqrt {\sin ^2(e+f x)}}\\ \end {align*}
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Mathematica [C] time = 18.63, size = 4143, normalized size = 56.75 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(b*Sec[e + f*x])^n*Sin[e + f*x]^2,x]
[Out]
(24*(Sec[(e + f*x)/2]^2)^(-3 + n)*(b*Sec[e + f*x])^n*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Sin[e + f*x]^2*Tan[(e
+ f*x)/2]*((AppellF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2)/(3*Appe
llF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-2 + n)*AppellF1[3/2, n, 3 - n, 5/2, T
an[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x
)/2]^2])*Tan[(e + f*x)/2]^2) - AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]/(3*Appell
F1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-3 + n)*AppellF1[3/2, n, 4 - n, 5/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/
2]^2])*Tan[(e + f*x)/2]^2)))/(f*(12*(Sec[(e + f*x)/2]^2)^(-2 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*((Appell
F1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2)/(3*AppellF1[1/2, n, 2 - n,
3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-2 + n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/2]^2,
-Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f
*x)/2]^2) - AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]/(3*AppellF1[1/2, n, 3 - n, 3
/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-3 + n)*AppellF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]^2, -T
an[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x
)/2]^2)) + 24*(-3 + n)*(Sec[(e + f*x)/2]^2)^(-3 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Tan[(e + f*x)/2]^2*((
AppellF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2)/(3*AppellF1[1/2, n,
2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-2 + n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/
2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[
(e + f*x)/2]^2) - AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]/(3*AppellF1[1/2, n, 3
- n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-3 + n)*AppellF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]
^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e
+ f*x)/2]^2)) + 24*(Sec[(e + f*x)/2]^2)^(-3 + n)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^n*Tan[(e + f*x)/2]*((Appel
lF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/(3*Appel
lF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-2 + n)*AppellF1[3/2, n, 3 - n, 5/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)
/2]^2])*Tan[(e + f*x)/2]^2) + (Sec[(e + f*x)/2]^2*(-1/3*((2 - n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/2]
^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2]) + (n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f
*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3))/(3*AppellF1[1/2, n, 2 - n, 3/2, Tan[(e
+ f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-2 + n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*
x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) -
(-1/3*((3 - n)*AppellF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e
+ f*x)/2]) + (n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*
Tan[(e + f*x)/2])/3)/(3*AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-3 + n)*Ap
pellF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(
e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) - (AppellF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -
Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*(2*((-2 + n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e +
f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2*
Tan[(e + f*x)/2] + 3*(-1/3*((2 - n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[
(e + f*x)/2]^2*Tan[(e + f*x)/2]) + (n*AppellF1[3/2, 1 + n, 2 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3) + 2*Tan[(e + f*x)/2]^2*((-2 + n)*((-3*(3 - n)*AppellF1[5/2, n, 4 - n
, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 + (3*n*AppellF1[5/2, 1
+ n, 3 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5) + n*((-3*(2
- n)*AppellF1[5/2, 1 + n, 3 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x
)/2])/5 + (3*(1 + n)*AppellF1[5/2, 2 + n, 2 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2
]^2*Tan[(e + f*x)/2])/5))))/(3*AppellF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-2
+ n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2
, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)^2 + (AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x
)/2]^2, -Tan[(e + f*x)/2]^2]*(2*((-3 + n)*AppellF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
] + n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2*Tan[(e + f
*x)/2] + 3*(-1/3*((3 - n)*AppellF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/
2]^2*Tan[(e + f*x)/2]) + (n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e +
f*x)/2]^2*Tan[(e + f*x)/2])/3) + 2*Tan[(e + f*x)/2]^2*((-3 + n)*((-3*(4 - n)*AppellF1[5/2, n, 5 - n, 7/2, Tan
[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 + (3*n*AppellF1[5/2, 1 + n, 4 - n
, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5) + n*((-3*(3 - n)*Appel
lF1[5/2, 1 + n, 4 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 +
(3*(1 + n)*AppellF1[5/2, 2 + n, 3 - n, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e
+ f*x)/2])/5))))/(3*AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-3 + n)*Appel
lF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)^2) + 24*n*(Sec[(e + f*x)/2]^2)^(-3 + n)*(Cos[(e + f*x)/2
]^2*Sec[e + f*x])^(-1 + n)*Tan[(e + f*x)/2]*((AppellF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/
2]^2]*Sec[(e + f*x)/2]^2)/(3*AppellF1[1/2, n, 2 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-2 +
n)*AppellF1[3/2, n, 3 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2 - n, 5/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2) - AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^
2, -Tan[(e + f*x)/2]^2]/(3*AppellF1[1/2, n, 3 - n, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*((-3 + n)
*AppellF1[3/2, n, 4 - n, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3 - n, 5/2, Ta
n[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))*(-(Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]
) + Cos[(e + f*x)/2]^2*Sec[e + f*x]*Tan[e + f*x])))
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fricas [F] time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (\cos \left (f x + e\right )^{2} - 1\right )} \left (b \sec \left (f x + e\right )\right )^{n}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*sec(f*x+e))^n*sin(f*x+e)^2,x, algorithm="fricas")
[Out]
integral(-(cos(f*x + e)^2 - 1)*(b*sec(f*x + e))^n, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*sec(f*x+e))^n*sin(f*x+e)^2,x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e))^n*sin(f*x + e)^2, x)
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maple [F] time = 1.34, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x +e \right )\right )^{n} \left (\sin ^{2}\left (f x +e \right )\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*sec(f*x+e))^n*sin(f*x+e)^2,x)
[Out]
int((b*sec(f*x+e))^n*sin(f*x+e)^2,x)
________________________________________________________________________________________
maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (f x + e\right )\right )^{n} \sin \left (f x + e\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*sec(f*x+e))^n*sin(f*x+e)^2,x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e))^n*sin(f*x + e)^2, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^2\,{\left (\frac {b}{\cos \left (e+f\,x\right )}\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(e + f*x)^2*(b/cos(e + f*x))^n,x)
[Out]
int(sin(e + f*x)^2*(b/cos(e + f*x))^n, x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (e + f x \right )}\right )^{n} \sin ^{2}{\left (e + f x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*sec(f*x+e))**n*sin(f*x+e)**2,x)
[Out]
Integral((b*sec(e + f*x))**n*sin(e + f*x)**2, x)
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