3.2.53 \(\int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx\) [153]
Optimal. Leaf size=95 \[ -\frac {F_1\left (1-n;-\frac {1}{2},-\frac {1}{2}-n;2-n;\cos (c+d x),-\cos (c+d x)\right ) \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{\frac {1}{2}-n} \cot (c+d x) (a+a \sec (c+d x))^n}{d (1-n)} \]
[Out]
-AppellF1(1-n,-1/2-n,-1/2,2-n,-cos(d*x+c),cos(d*x+c))*(1+cos(d*x+c))^(1/2-n)*cot(d*x+c)*(a+a*sec(d*x+c))^n*(1-
cos(d*x+c))^(1/2)/d/(1-n)
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Rubi [A]
time = 0.24, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3961, 2953,
3087, 140, 138} \begin {gather*} -\frac {\sqrt {1-\cos (c+d x)} \cot (c+d x) (\cos (c+d x)+1)^{\frac {1}{2}-n} (a \sec (c+d x)+a)^n F_1\left (1-n;-\frac {1}{2},-n-\frac {1}{2};2-n;\cos (c+d x),-\cos (c+d x)\right )}{d (1-n)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^2,x]
[Out]
-((AppellF1[1 - n, -1/2, -1/2 - n, 2 - n, Cos[c + d*x], -Cos[c + d*x]]*Sqrt[1 - Cos[c + d*x]]*(1 + Cos[c + d*x
])^(1/2 - n)*Cot[c + d*x]*(a + a*Sec[c + d*x])^n)/(d*(1 - n)))
Rule 138
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rule 140
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[c^IntPart[n]*((c +
d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 0]
Rule 2953
Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Dist[1/b^2, Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 1)*(a - b*Sin[e + f*x]), x], x] /;
FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && (ILtQ[m, 0] || !IGtQ[n, 0])
Rule 3087
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((c_) + (d_.)*si
n[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[Sqrt[a + b*Sin[e + f*x]]*(Sqrt[c + d*Sin[e + f*x]]/(f*Cos[e +
f*x])), Subst[Int[(a + b*x)^(m - 1/2)*(c + d*x)^(n - 1/2)*(A + B*x)^p, x], x, Sin[e + f*x]], x] /; FreeQ[{a, b
, c, d, e, f, A, B, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]
Rule 3961
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[Sin[e
+ f*x]^FracPart[m]*((a + b*Csc[e + f*x])^FracPart[m]/(b + a*Sin[e + f*x])^FracPart[m]), Int[(g*Cos[e + f*x])^
p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])
Rubi steps
\begin {align*} \int (a+a \sec (c+d x))^n \sin ^2(c+d x) \, dx &=\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^n \sin ^2(c+d x) \, dx\\ &=\frac {\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{-n} (a+a \sec (c+d x))^n\right ) \int (-\cos (c+d x))^{-n} (-a-a \cos (c+d x))^{1+n} (-a+a \cos (c+d x)) \, dx}{a^2}\\ &=-\frac {\left ((-\cos (c+d x))^n (-a-a \cos (c+d x))^{\frac {1}{2}-n} \sqrt {-a+a \cos (c+d x)} \csc (c+d x) (a+a \sec (c+d x))^n\right ) \text {Subst}\left (\int (-x)^{-n} (-a-a x)^{\frac {1}{2}+n} \sqrt {-a+a x} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {\left ((-\cos (c+d x))^n (1+\cos (c+d x))^{-\frac {1}{2}-n} (-a-a \cos (c+d x)) \sqrt {-a+a \cos (c+d x)} \csc (c+d x) (a+a \sec (c+d x))^n\right ) \text {Subst}\left (\int (-x)^{-n} (1+x)^{\frac {1}{2}+n} \sqrt {-a+a x} \, dx,x,\cos (c+d x)\right )}{a^2 d}\\ &=-\frac {\left ((-\cos (c+d x))^n (1+\cos (c+d x))^{-\frac {1}{2}-n} (-a-a \cos (c+d x)) (-a+a \cos (c+d x)) \csc (c+d x) (a+a \sec (c+d x))^n\right ) \text {Subst}\left (\int \sqrt {1-x} (-x)^{-n} (1+x)^{\frac {1}{2}+n} \, dx,x,\cos (c+d x)\right )}{a^2 d \sqrt {1-\cos (c+d x)}}\\ &=-\frac {F_1\left (1-n;-\frac {1}{2},-\frac {1}{2}-n;2-n;\cos (c+d x),-\cos (c+d x)\right ) \sqrt {1-\cos (c+d x)} (1+\cos (c+d x))^{\frac {1}{2}-n} \cot (c+d x) (a+a \sec (c+d x))^n}{d (1-n)}\\ \end {align*}
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Mathematica [C] Result contains complex when optimal does not.
time = 15.30, size = 4297, normalized size = 45.23 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^2,x]
[Out]
(2^(3 + n)*Cos[(c + d*x)/2]^5*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*Sin[(c + d*x)/2]*(C
os[2*(c + d*x)]*(-1/4*(1 + Sec[c + d*x])^n - ((1 + Sec[c + d*x])^n*Sin[c + d*x]^2)/2 - ((1 + Sec[c + d*x])^n*S
in[c + d*x]^4)/4) + (I/4)*(1 + Sec[c + d*x])^n*Sin[2*(c + d*x)] + (I/2)*(1 + Sec[c + d*x])^n*Sin[c + d*x]^2*Si
n[2*(c + d*x)] + (I/4)*(1 + Sec[c + d*x])^n*Sin[c + d*x]^4*Sin[2*(c + d*x)] + Cos[c + d*x]^4*(-1/4*(Cos[2*(c +
d*x)]*(1 + Sec[c + d*x])^n) + (I/4)*(1 + Sec[c + d*x])^n*Sin[2*(c + d*x)]) + Cos[c + d*x]^3*((-I)*Cos[2*(c +
d*x)]*(1 + Sec[c + d*x])^n*Sin[c + d*x] - (1 + Sec[c + d*x])^n*Sin[c + d*x]*Sin[2*(c + d*x)]) + Cos[c + d*x]^2
*(Cos[2*(c + d*x)]*((1 + Sec[c + d*x])^n/2 + (3*(1 + Sec[c + d*x])^n*Sin[c + d*x]^2)/2) - (I/2)*(1 + Sec[c + d
*x])^n*Sin[2*(c + d*x)] - ((3*I)/2)*(1 + Sec[c + d*x])^n*Sin[c + d*x]^2*Sin[2*(c + d*x)]) + Cos[c + d*x]*(Cos[
2*(c + d*x)]*(I*(1 + Sec[c + d*x])^n*Sin[c + d*x] + I*(1 + Sec[c + d*x])^n*Sin[c + d*x]^3) + (1 + Sec[c + d*x]
)^n*Sin[c + d*x]*Sin[2*(c + d*x)] + (1 + Sec[c + d*x])^n*Sin[c + d*x]^3*Sin[2*(c + d*x)]))*((3*AppellF1[1/2, n
, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2)/(3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*
x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-2*AppellF1[3/2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*Ap
pellF1[3/2, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2) - AppellF1[1/2, n, 3,
3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]/(AppellF1[1/2, n, 3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2] + (2*(-3*AppellF1[3/2, n, 4, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3, 5/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/3)))/(d*(1 + Sec[c + d*x])^n*(2^(2 + n)*Cos[(c
+ d*x)/2]^6*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*((3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*
x)/2]^2]*Sec[(c + d*x)/2]^2)/(3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-2*Appe
llF1[3/2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2, 5/2, Tan[(c + d*x)/2
]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2) - AppellF1[1/2, n, 3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2
]^2]/(AppellF1[1/2, n, 3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2, n, 4, 5/2, Tan[
(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
)*Tan[(c + d*x)/2]^2)/3)) - 5*2^(2 + n)*Cos[(c + d*x)/2]^4*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*Sin[(c + d*x)/2
]^2*((3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2)/(3*AppellF1[1/2,
n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-2*AppellF1[3/2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan
[(c + d*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2
) - AppellF1[1/2, n, 3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]/(AppellF1[1/2, n, 3, 3/2, Tan[(c + d*x)/
2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2, n, 4, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*Appe
llF1[3/2, 1 + n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/3)) + 2^(3 + n)*Cos[(c
+ d*x)/2]^5*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*Sin[(c + d*x)/2]*((3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]
^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]^2,
-Tan[(c + d*x)/2]^2] + 2*(-2*AppellF1[3/2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/
2, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2) + (3*Sec[(c + d*x)/2]^2*((-2*A
ppellF1[3/2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3 + (n*A
ppellF1[3/2, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3))/
(3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(-2*AppellF1[3/2, n, 3, 5/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/2, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Ta
n[(c + d*x)/2]^2) - (-(AppellF1[3/2, n, 4, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Ta
n[(c + d*x)/2]) + (n*AppellF1[3/2, 1 + n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*
Tan[(c + d*x)/2])/3)/(AppellF1[1/2, n, 3, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + (2*(-3*AppellF1[3/2,
n, 4, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/2, 1 + n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan
[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)/3) - (3*AppellF1[1/2, n, 2, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2
]*Sec[(c + d*x)/2]^2*(2*(-2*AppellF1[3/2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + n*AppellF1[3/2
, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] + 3*((-2*Appell
F1[3/2, n, 3, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3 + (n*Appell
F1[3/2, 1 + n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan...
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Maple [F]
time = 0.25, size = 0, normalized size = 0.00 \[\int \left (a +a \sec \left (d x +c \right )\right )^{n} \left (\sin ^{2}\left (d x +c \right )\right )\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^n*sin(d*x+c)^2,x)
[Out]
int((a+a*sec(d*x+c))^n*sin(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((a+a*sec(d*x+c))^n*sin(d*x+c)^2,x, algorithm="maxima")
[Out]
integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^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((a+a*sec(d*x+c))^n*sin(d*x+c)^2,x, algorithm="fricas")
[Out]
integral(-(cos(d*x + c)^2 - 1)*(a*sec(d*x + c) + a)^n, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \sin ^{2}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))**n*sin(d*x+c)**2,x)
[Out]
Integral((a*(sec(c + d*x) + 1))**n*sin(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((a+a*sec(d*x+c))^n*sin(d*x+c)^2,x, algorithm="giac")
[Out]
integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^2, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\sin \left (c+d\,x\right )}^2\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(c + d*x)^2*(a + a/cos(c + d*x))^n,x)
[Out]
int(sin(c + d*x)^2*(a + a/cos(c + d*x))^n, x)
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