3.1.65 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx\) [65]
Optimal. Leaf size=114 \[ \frac {2^{\frac {3}{2}+n} c^2 \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2} (3+2 m),\frac {1}{2} (-1-2 n);\frac {1}{2} (5+2 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2+n}}{f (3+2 m)} \]
[Out]
2^(3/2+n)*c^2*cos(f*x+e)^3*hypergeom([3/2+m, -1/2-n],[5/2+m],1/2+1/2*sin(f*x+e))*(1-sin(f*x+e))^(1/2-n)*(a+a*s
in(f*x+e))^m*(c-c*sin(f*x+e))^(-2+n)/f/(3+2*m)
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Rubi [A]
time = 0.22, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {2920, 2824,
2768, 72, 71} \begin {gather*} \frac {c^2 2^{n+\frac {3}{2}} \cos ^3(e+f x) (1-\sin (e+f x))^{\frac {1}{2}-n} (a \sin (e+f x)+a)^m (c-c \sin (e+f x))^{n-2} \, _2F_1\left (\frac {1}{2} (2 m+3),\frac {1}{2} (-2 n-1);\frac {1}{2} (2 m+5);\frac {1}{2} (\sin (e+f x)+1)\right )}{f (2 m+3)} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n,x]
[Out]
(2^(3/2 + n)*c^2*Cos[e + f*x]^3*Hypergeometric2F1[(3 + 2*m)/2, (-1 - 2*n)/2, (5 + 2*m)/2, (1 + Sin[e + f*x])/2
]*(1 - Sin[e + f*x])^(1/2 - n)*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(-2 + n))/(f*(3 + 2*m))
Rule 71
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c
- a*d))^n))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Rule 72
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*(b*((c + d*x)/(b*c - a*d)))^FracPart[n]), Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c -
a*d)), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && !IntegerQ[m] && !IntegerQ[n] &&
(RationalQ[m] || !SimplerQ[n + 1, m + 1])
Rule 2768
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[a^2*(
(g*Cos[e + f*x])^(p + 1)/(f*g*(a + b*Sin[e + f*x])^((p + 1)/2)*(a - b*Sin[e + f*x])^((p + 1)/2))), Subst[Int[(
a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, g, m, p}, x] &&
EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Rule 2824
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[a^IntPart[m]*c^IntPart[m]*(a + b*Sin[e + f*x])^FracPart[m]*((c + d*Sin[e + f*x])^FracPart[m]/Cos[e + f*x]^(2*
FracPart[m])), Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, m, n},
x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && (FractionQ[m] || !FractionQ[n])
Rule 2920
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]
Rubi steps
\begin {align*} \int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^n \, dx &=\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{1+n} \, dx}{a c}\\ &=\left (\cos ^{-2 m}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^m\right ) \int \cos ^{2 (1+m)}(e+f x) (c-c \sin (e+f x))^{-m+n} \, dx\\ &=\frac {\left (c^2 \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{m+\frac {1}{2} (-1-2 (1+m))} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \text {Subst}\left (\int (c-c x)^{-m+\frac {1}{2} (-1+2 (1+m))+n} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (2^{\frac {1}{2}+n} c^3 \cos ^{1-2 m+2 (1+m)}(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-\frac {1}{2}+m+\frac {1}{2} (-1-2 (1+m))+n} \left (\frac {c-c \sin (e+f x)}{c}\right )^{\frac {1}{2}-n} (c+c \sin (e+f x))^{\frac {1}{2} (-1-2 (1+m))}\right ) \text {Subst}\left (\int \left (\frac {1}{2}-\frac {x}{2}\right )^{-m+\frac {1}{2} (-1+2 (1+m))+n} (c+c x)^{\frac {1}{2} (-1+2 (1+m))} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {2^{\frac {3}{2}+n} c^2 \cos ^3(e+f x) \, _2F_1\left (\frac {1}{2} (3+2 m),\frac {1}{2} (-1-2 n);\frac {1}{2} (5+2 m);\frac {1}{2} (1+\sin (e+f x))\right ) (1-\sin (e+f x))^{\frac {1}{2}-n} (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{-2+n}}{f (3+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 = 10.34, size = 2543, normalized size = 22.31 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n,x]
[Out]
(-16*(AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]
+ 8*AppellF1[1/2 + n, -2*m, 2*(2 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]
- 5*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] -
4*AppellF1[1/2 + n, -2*m, 5 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*C
os[(-e + Pi/2 - f*x)/2]*Sin[(-e + Pi/2 - f*x)/2]*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^n*Tan[(-e + Pi/2
- f*x)/4])/(f*(n*(AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2
- f*x)/4]^2] + 8*AppellF1[1/2 + n, -2*m, 2*(2 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2] - 5*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2] - 4*AppellF1[1/2 + n, -2*m, 5 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f
*x)/4]^2])*Csc[(-e + Pi/2 - f*x)/4]*Sec[(-e + Pi/2 - f*x)/4]^3 + (AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 +
n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 8*AppellF1[1/2 + n, -2*m, 2*(2 + m + n), 3/2 +
n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 5*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n
, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*AppellF1[1/2 + n, -2*m, 5 + 2*(m + n), 3/2 + n,
Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Csc[(-e + Pi/2 - f*x)/2]*Sec[(-e + Pi/2 - f*x)/4]^2
*Sec[(-e + Pi/2 - f*x)/2] + 2*(m + n)*(AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4
]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 8*AppellF1[1/2 + n, -2*m, 2*(2 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]
^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 5*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^
2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*AppellF1[1/2 + n, -2*m, 5 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2
, -Tan[(-e + Pi/2 - f*x)/4]^2])*Sec[(-e + Pi/2 - f*x)/4]^2*Sec[(-e + Pi/2 - f*x)/2]*Tan[(-e + Pi/2 - f*x)/4] -
4*m*(AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]
+ 8*AppellF1[1/2 + n, -2*m, 2*(2 + m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]
- 5*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] -
4*AppellF1[1/2 + n, -2*m, 5 + 2*(m + n), 3/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*S
ec[(-e + Pi/2 - f*x)/2]^2*Tan[(-e + Pi/2 - f*x)/4] + 32*m*(AppellF1[1/2 + n, -2*m, 2*(1 + m + n), 3/2 + n, Tan
[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 8*AppellF1[1/2 + n, -2*m, 2*(2 + m + n), 3/2 + n, Tan[
(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 5*AppellF1[1/2 + n, -2*m, 3 + 2*(m + n), 3/2 + n, Tan[(
-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] - 4*AppellF1[1/2 + n, -2*m, 5 + 2*(m + n), 3/2 + n, Tan[(-
e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2])*Sec[e + f*x]^2*Sin[(-e + Pi/2 - f*x)/4]^2*Tan[(-e + Pi/2 -
f*x)/4] - (2*(m*(1 + 2*n)*AppellF1[3/2 + n, 1 - 2*m, 2*(1 + m + n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan
[(-e + Pi/2 - f*x)/4]^2] - 5*(1/2 + n)*(2*m*AppellF1[3/2 + n, 1 - 2*m, 3 + 2*(m + n), 5/2 + n, Tan[(-e + Pi/2
- f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (3 + 2*(m + n))*AppellF1[3/2 + n, -2*m, 2*(2 + m + n), 5/2 + n, Ta
n[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]) - 4*(1/2 + n)*(2*m*AppellF1[3/2 + n, 1 - 2*m, 5 + 2*(m
+ n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + (5 + 2*(m + n))*AppellF1[3/2 + n, -
2*m, 2*(3 + m + n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2]) + (1 + m + n)*(1 + 2*n)
*AppellF1[3/2 + n, -2*m, 3 + 2*m + 2*n, 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 - f*x)/4]^2] + 16
*(1/2 + n)*(m*AppellF1[3/2 + n, 1 - 2*m, 2*(2 + m + n), 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e + Pi/2 -
f*x)/4]^2] + (2 + m + n)*AppellF1[3/2 + n, -2*m, 5 + 2*m + 2*n, 5/2 + n, Tan[(-e + Pi/2 - f*x)/4]^2, -Tan[(-e
+ Pi/2 - f*x)/4]^2]))*Csc[(-e + Pi/2 - f*x)/2]*Sec[(-e + Pi/2 - f*x)/4]^2*Sec[(-e + Pi/2 - f*x)/2]*Tan[(-e +
Pi/2 - f*x)/4]^2)/(3/2 + n)))
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Maple [F]
time = 0.12, size = 0, normalized size = 0.00 \[\int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{n}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x)
[Out]
int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,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(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="maxima")
[Out]
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n*cos(f*x + e)^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(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="fricas")
[Out]
integral((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (- c \left (\sin {\left (e + f x \right )} - 1\right )\right )^{n} \cos ^{2}{\left (e + f x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(f*x+e)**2*(a+a*sin(f*x+e))**m*(c-c*sin(f*x+e))**n,x)
[Out]
Integral((a*(sin(e + f*x) + 1))**m*(-c*(sin(e + f*x) - 1))**n*cos(e + f*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(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^n,x, algorithm="giac")
[Out]
integrate((a*sin(f*x + e) + a)^m*(-c*sin(f*x + e) + c)^n*cos(f*x + e)^2, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (e+f\,x\right )}^2\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^n \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n,x)
[Out]
int(cos(e + f*x)^2*(a + a*sin(e + f*x))^m*(c - c*sin(e + f*x))^n, x)
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