3.74 \(\int \cos ^2(e+f x) (a+a \sin (e+f x))^m (c-c \sin (e+f x))^{3/2} \, dx\)

Optimal. Leaf size=172 \[ \frac {64 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7) \left (4 m^2+16 m+15\right ) \sqrt {c-c \sin (e+f x)}}+\frac {16 c \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{a f \left (4 m^2+24 m+35\right )}+\frac {2 \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7)} \]

[Out]

2*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^(3/2)/a/f/(7+2*m)+64*c^2*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m
)/a/f/(8*m^3+60*m^2+142*m+105)/(c-c*sin(f*x+e))^(1/2)+16*c*cos(f*x+e)*(a+a*sin(f*x+e))^(1+m)*(c-c*sin(f*x+e))^
(1/2)/a/f/(4*m^2+24*m+35)

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Rubi [A]  time = 0.47, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {2841, 2740, 2738} \[ \frac {64 c^2 \cos (e+f x) (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7) \left (4 m^2+16 m+15\right ) \sqrt {c-c \sin (e+f x)}}+\frac {16 c \cos (e+f x) \sqrt {c-c \sin (e+f x)} (a \sin (e+f x)+a)^{m+1}}{a f \left (4 m^2+24 m+35\right )}+\frac {2 \cos (e+f x) (c-c \sin (e+f x))^{3/2} (a \sin (e+f x)+a)^{m+1}}{a f (2 m+7)} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

(64*c^2*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m))/(a*f*(7 + 2*m)*(15 + 16*m + 4*m^2)*Sqrt[c - c*Sin[e + f*x]]
) + (16*c*Cos[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*Sqrt[c - c*Sin[e + f*x]])/(a*f*(35 + 24*m + 4*m^2)) + (2*C
os[e + f*x]*(a + a*Sin[e + f*x])^(1 + m)*(c - c*Sin[e + f*x])^(3/2))/(a*f*(7 + 2*m))

Rule 2738

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[
(-2*b*Cos[e + f*x]*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]]), x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[n, -2^(-1)]

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2841

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))^{3/2} \, dx &=\frac {\int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{5/2} \, dx}{a c}\\ &=\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2}}{a f (7+2 m)}+\frac {8 \int (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2} \, dx}{a (7+2 m)}\\ &=\frac {16 c \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)}}{a f \left (35+24 m+4 m^2\right )}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2}}{a f (7+2 m)}+\frac {(32 c) \int (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)} \, dx}{a \left (35+24 m+4 m^2\right )}\\ &=\frac {64 c^2 \cos (e+f x) (a+a \sin (e+f x))^{1+m}}{a f (3+2 m) \left (35+24 m+4 m^2\right ) \sqrt {c-c \sin (e+f x)}}+\frac {16 c \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {c-c \sin (e+f x)}}{a f \left (35+24 m+4 m^2\right )}+\frac {2 \cos (e+f x) (a+a \sin (e+f x))^{1+m} (c-c \sin (e+f x))^{3/2}}{a f (7+2 m)}\\ \end {align*}

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Mathematica [A]  time = 3.10, size = 149, normalized size = 0.87 \[ -\frac {c \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3 (a (\sin (e+f x)+1))^m \left (4 \left (4 m^2+24 m+27\right ) \sin (e+f x)+\left (4 m^2+16 m+15\right ) \cos (2 (e+f x))-12 m^2-80 m-157\right )}{f (2 m+3) (2 m+5) (2 m+7) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]^2*(a + a*Sin[e + f*x])^m*(c - c*Sin[e + f*x])^(3/2),x]

[Out]

-((c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(a*(1 + Sin[e + f*x]))^m*Sqrt[c - c*Sin[e + f*x]]*(-157 - 80*m -
12*m^2 + (15 + 16*m + 4*m^2)*Cos[2*(e + f*x)] + 4*(27 + 24*m + 4*m^2)*Sin[e + f*x]))/(f*(3 + 2*m)*(5 + 2*m)*(7
+ 2*m)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])))

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fricas [A]  time = 0.48, size = 243, normalized size = 1.41 \[ \frac {2 \, {\left ({\left (4 \, c m^{2} + 16 \, c m + 15 \, c\right )} \cos \left (f x + e\right )^{4} + {\left (4 \, c m^{2} + 32 \, c m + 39 \, c\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (2 \, c m - c\right )} \cos \left (f x + e\right )^{2} + 32 \, c \cos \left (f x + e\right ) - {\left ({\left (4 \, c m^{2} + 16 \, c m + 15 \, c\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left (2 \, c m + 3 \, c\right )} \cos \left (f x + e\right )^{2} - 32 \, c \cos \left (f x + e\right ) - 64 \, c\right )} \sin \left (f x + e\right ) + 64 \, c\right )} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (a \sin \left (f x + e\right ) + a\right )}^{m}}{8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + {\left (8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + 105 \, f\right )} \cos \left (f x + e\right ) - {\left (8 \, f m^{3} + 60 \, f m^{2} + 142 \, f m + 105 \, f\right )} \sin \left (f x + e\right ) + 105 \, f} \]

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))^(3/2),x, algorithm="fricas")

[Out]

2*((4*c*m^2 + 16*c*m + 15*c)*cos(f*x + e)^4 + (4*c*m^2 + 32*c*m + 39*c)*cos(f*x + e)^3 + 8*(2*c*m - c)*cos(f*x
+ e)^2 + 32*c*cos(f*x + e) - ((4*c*m^2 + 16*c*m + 15*c)*cos(f*x + e)^3 - 8*(2*c*m + 3*c)*cos(f*x + e)^2 - 32*
c*cos(f*x + e) - 64*c)*sin(f*x + e) + 64*c)*sqrt(-c*sin(f*x + e) + c)*(a*sin(f*x + e) + a)^m/(8*f*m^3 + 60*f*m
^2 + 142*f*m + (8*f*m^3 + 60*f*m^2 + 142*f*m + 105*f)*cos(f*x + e) - (8*f*m^3 + 60*f*m^2 + 142*f*m + 105*f)*si
n(f*x + e) + 105*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} \cos \left (f x + e\right )^{2}\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

integrate((-c*sin(f*x + e) + c)^(3/2)*(a*sin(f*x + e) + a)^m*cos(f*x + e)^2, x)

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maple [F]  time = 0.74, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{2}\left (f x +e \right )\right ) \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c -c \sin \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

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))^(3/2),x)

[Out]

int(cos(f*x+e)^2*(a+a*sin(f*x+e))^m*(c-c*sin(f*x+e))^(3/2),x)

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maxima [B]  time = 0.52, size = 423, normalized size = 2.46 \[ -\frac {2 \, {\left ({\left (4 \, m^{2} + 32 \, m + 71\right )} a^{m} c^{\frac {3}{2}} - \frac {{\left (4 \, m^{2} - 105\right )} a^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {{\left (12 \, m^{2} + 64 \, m - 91\right )} a^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (12 \, m^{2} + 32 \, m + 245\right )} a^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {{\left (12 \, m^{2} + 32 \, m + 245\right )} a^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {{\left (12 \, m^{2} + 64 \, m - 91\right )} a^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac {{\left (4 \, m^{2} - 105\right )} a^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {{\left (4 \, m^{2} + 32 \, m + 71\right )} a^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}}\right )} e^{\left (2 \, m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left (8 \, m^{3} + 60 \, m^{2} + 142 \, m + \frac {2 \, {\left (8 \, m^{3} + 60 \, m^{2} + 142 \, m + 105\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {{\left (8 \, m^{3} + 60 \, m^{2} + 142 \, m + 105\right )} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 105\right )} f {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac {3}{2}}} \]

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))^(3/2),x, algorithm="maxima")

[Out]

-2*((4*m^2 + 32*m + 71)*a^m*c^(3/2) - (4*m^2 - 105)*a^m*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) - (12*m^2 + 64
*m - 91)*a^m*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (12*m^2 + 32*m + 245)*a^m*c^(3/2)*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + (12*m^2 + 32*m + 245)*a^m*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (12*m^2 + 64*m -
91)*a^m*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - (4*m^2 - 105)*a^m*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) +
1)^6 + (4*m^2 + 32*m + 71)*a^m*c^(3/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7)*e^(2*m*log(sin(f*x + e)/(cos(f*x
+ e) + 1) + 1) - m*log(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))/((8*m^3 + 60*m^2 + 142*m + 2*(8*m^3 + 60*m^2
+ 142*m + 105)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + (8*m^3 + 60*m^2 + 142*m + 105)*sin(f*x + e)^4/(cos(f*x +
e) + 1)^4 + 105)*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2))

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mupad [B]  time = 13.97, size = 528, normalized size = 3.07 \[ -\frac {\sqrt {c-c\,\sin \left (e+f\,x\right )}\,\left (\frac {c\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (m^2\,4{}\mathrm {i}+m\,16{}\mathrm {i}+15{}\mathrm {i}\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}-\frac {c\,{\mathrm {e}}^{e\,7{}\mathrm {i}+f\,x\,7{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (4\,m^2+16\,m+15\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}-\frac {c\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (4\,m^2+48\,m+63\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}-\frac {c\,{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (12\,m^2+80\,m-35\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {c\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (m^2\,4{}\mathrm {i}+m\,48{}\mathrm {i}+63{}\mathrm {i}\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {c\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (m^2\,12{}\mathrm {i}+m\,80{}\mathrm {i}-35{}\mathrm {i}\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}-\frac {c\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (12\,m^2+112\,m+525\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}+\frac {c\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,\left (m^2\,12{}\mathrm {i}+m\,112{}\mathrm {i}+525{}\mathrm {i}\right )}{4\,f\,\left (8\,m^3+60\,m^2+142\,m+105\right )}\right )}{{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\left (m^3\,8{}\mathrm {i}+m^2\,60{}\mathrm {i}+m\,142{}\mathrm {i}+105{}\mathrm {i}\right )}{8\,m^3+60\,m^2+142\,m+105}} \]

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))^(3/2),x)

[Out]

-((c - c*sin(e + f*x))^(1/2)*((c*(a + a*sin(e + f*x))^m*(m*16i + m^2*4i + 15i))/(4*f*(142*m + 60*m^2 + 8*m^3 +
105)) - (c*exp(e*7i + f*x*7i)*(a + a*sin(e + f*x))^m*(16*m + 4*m^2 + 15))/(4*f*(142*m + 60*m^2 + 8*m^3 + 105)
) - (c*exp(e*1i + f*x*1i)*(a + a*sin(e + f*x))^m*(48*m + 4*m^2 + 63))/(4*f*(142*m + 60*m^2 + 8*m^3 + 105)) - (
c*exp(e*5i + f*x*5i)*(a + a*sin(e + f*x))^m*(80*m + 12*m^2 - 35))/(4*f*(142*m + 60*m^2 + 8*m^3 + 105)) + (c*ex
p(e*6i + f*x*6i)*(a + a*sin(e + f*x))^m*(m*48i + m^2*4i + 63i))/(4*f*(142*m + 60*m^2 + 8*m^3 + 105)) + (c*exp(
e*2i + f*x*2i)*(a + a*sin(e + f*x))^m*(m*80i + m^2*12i - 35i))/(4*f*(142*m + 60*m^2 + 8*m^3 + 105)) - (c*exp(e
*3i + f*x*3i)*(a + a*sin(e + f*x))^m*(112*m + 12*m^2 + 525))/(4*f*(142*m + 60*m^2 + 8*m^3 + 105)) + (c*exp(e*4
i + f*x*4i)*(a + a*sin(e + f*x))^m*(m*112i + m^2*12i + 525i))/(4*f*(142*m + 60*m^2 + 8*m^3 + 105))))/(exp(e*4i
+ f*x*4i) - (exp(e*3i + f*x*3i)*(m*142i + m^2*60i + m^3*8i + 105i))/(142*m + 60*m^2 + 8*m^3 + 105))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(3/2),x)

[Out]

Timed out

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