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

Optimal. Leaf size=94 \[ -\frac {2 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{4-2 m}}{d e \left (m^2-5 m+6\right )}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)} \]

[Out]

-2*a^2*(e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c))^(-2+m)/d/e/(2-m)/(3-m)-a*(e*cos(d*x+c))^(4-2*m)*(a+a*sin(d*x+c)
)^(-1+m)/d/e/(3-m)

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Rubi [A]  time = 0.14, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2674, 2673} \[ -\frac {2 a^2 (a \sin (c+d x)+a)^{m-2} (e \cos (c+d x))^{4-2 m}}{d e \left (m^2-5 m+6\right )}-\frac {a (a \sin (c+d x)+a)^{m-1} (e \cos (c+d x))^{4-2 m}}{d e (3-m)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(e*Cos[c + d*x])^(4 - 2*m)*(a + a*Sin[c + d*x])^(-2 + m))/(d*e*(6 - 5*m + m^2)) - (a*(e*Cos[c + d*x])^
(4 - 2*m)*(a + a*Sin[c + d*x])^(-1 + m))/(d*e*(3 - m))

Rule 2673

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rule 2674

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]

Rubi steps

\begin {align*} \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^m \, dx &=-\frac {a (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (3-m)}+\frac {(2 a) \int (e \cos (c+d x))^{3-2 m} (a+a \sin (c+d x))^{-1+m} \, dx}{3-m}\\ &=-\frac {2 a^2 (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-2+m}}{d e \left (6-5 m+m^2\right )}-\frac {a (e \cos (c+d x))^{4-2 m} (a+a \sin (c+d x))^{-1+m}}{d e (3-m)}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 72, normalized size = 0.77 \[ \frac {e^3 \cos ^4(c+d x) ((m-2) \sin (c+d x)+m-4) (a (\sin (c+d x)+1))^m (e \cos (c+d x))^{-2 m}}{d (m-3) (m-2) (\sin (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*Cos[c + d*x]^4*(a*(1 + Sin[c + d*x]))^m*(-4 + m + (-2 + m)*Sin[c + d*x]))/(d*(-3 + m)*(-2 + m)*(e*Cos[c +
d*x])^(2*m)*(1 + Sin[c + d*x])^2)

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fricas [A]  time = 0.46, size = 170, normalized size = 1.81 \[ \frac {{\left ({\left (m - 2\right )} \cos \left (d x + c\right )^{2} + {\left (m - 4\right )} \cos \left (d x + c\right ) + {\left ({\left (m - 2\right )} \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - 2\right )} \left (e \cos \left (d x + c\right )\right )^{-2 \, m + 3} {\left (a \sin \left (d x + c\right ) + a\right )}^{m}}{2 \, d m^{2} - {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right )^{2} - 10 \, d m + {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) + {\left (2 \, d m^{2} - 10 \, d m + {\left (d m^{2} - 5 \, d m + 6 \, d\right )} \cos \left (d x + c\right ) + 12 \, d\right )} \sin \left (d x + c\right ) + 12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="fricas")

[Out]

((m - 2)*cos(d*x + c)^2 + (m - 4)*cos(d*x + c) + ((m - 2)*cos(d*x + c) + 2)*sin(d*x + c) - 2)*(e*cos(d*x + c))
^(-2*m + 3)*(a*sin(d*x + c) + a)^m/(2*d*m^2 - (d*m^2 - 5*d*m + 6*d)*cos(d*x + c)^2 - 10*d*m + (d*m^2 - 5*d*m +
6*d)*cos(d*x + c) + (2*d*m^2 - 10*d*m + (d*m^2 - 5*d*m + 6*d)*cos(d*x + c) + 12*d)*sin(d*x + c) + 12*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.57, size = 0, normalized size = 0.00 \[ \int \left (e \cos \left (d x +c \right )\right )^{3-2 m} \left (a +a \sin \left (d x +c \right )\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.38, size = 351, normalized size = 3.73 \[ \frac {{\left (a^{m} e^{3} {\left (m - 4\right )} - \frac {2 \, a^{m} e^{3} {\left (m - 6\right )} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {a^{m} e^{3} {\left (m + 12\right )} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a^{m} e^{3} {\left (m + 2\right )} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {a^{m} e^{3} {\left (m + 12\right )} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {2 \, a^{m} e^{3} {\left (m - 6\right )} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {a^{m} e^{3} {\left (m - 4\right )} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} e^{\left (-2 \, m \log \left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + m \log \left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )\right )}}{{\left ({\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} + \frac {3 \, {\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, {\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {{\left (m^{2} - 5 \, m + 6\right )} e^{2 \, m} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*cos(d*x+c))^(3-2*m)*(a+a*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

(a^m*e^3*(m - 4) - 2*a^m*e^3*(m - 6)*sin(d*x + c)/(cos(d*x + c) + 1) - a^m*e^3*(m + 12)*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2 + 4*a^m*e^3*(m + 2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - a^m*e^3*(m + 12)*sin(d*x + c)^4/(cos(d
*x + c) + 1)^4 - 2*a^m*e^3*(m - 6)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + a^m*e^3*(m - 4)*sin(d*x + c)^6/(cos(d
*x + c) + 1)^6)*e^(-2*m*log(-sin(d*x + c)/(cos(d*x + c) + 1) + 1) + m*log(sin(d*x + c)^2/(cos(d*x + c) + 1)^2
+ 1))/(((m^2 - 5*m + 6)*e^(2*m) + 3*(m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*(m^2 - 5*m
+ 6)*e^(2*m)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + (m^2 - 5*m + 6)*e^(2*m)*sin(d*x + c)^6/(cos(d*x + c) + 1)^
6)*d)

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mupad [B]  time = 8.77, size = 241, normalized size = 2.56 \[ \frac {e^3\,{\left (a\,\left (\sin \left (c+d\,x\right )+1\right )\right )}^m\,\left (14\,m-24\,\sin \left (c+d\,x\right )-36\,\sin \left (3\,c+3\,d\,x\right )-12\,\sin \left (5\,c+5\,d\,x\right )+24\,{\sin \left (2\,c+2\,d\,x\right )}^2-4\,{\sin \left (3\,c+3\,d\,x\right )}^2+8\,m\,\sin \left (c+d\,x\right )-17\,m\,\left (2\,{\sin \left (c+d\,x\right )}^2-1\right )+12\,m\,\sin \left (3\,c+3\,d\,x\right )+4\,m\,\sin \left (5\,c+5\,d\,x\right )-2\,m\,\left (2\,{\sin \left (2\,c+2\,d\,x\right )}^2-1\right )+m\,\left (2\,{\sin \left (3\,c+3\,d\,x\right )}^2-1\right )+132\,{\sin \left (c+d\,x\right )}^2-128\right )}{8\,d\,{\left (-e\,\left (2\,{\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\right )}^{2\,m}\,\left (m^2-5\,m+6\right )\,\left (12\,{\sin \left (c+d\,x\right )}^2+15\,\sin \left (c+d\,x\right )-\sin \left (3\,c+3\,d\,x\right )+4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e^3*(a*(sin(c + d*x) + 1))^m*(14*m - 24*sin(c + d*x) - 36*sin(3*c + 3*d*x) - 12*sin(5*c + 5*d*x) + 24*sin(2*c
+ 2*d*x)^2 - 4*sin(3*c + 3*d*x)^2 + 8*m*sin(c + d*x) - 17*m*(2*sin(c + d*x)^2 - 1) + 12*m*sin(3*c + 3*d*x) +
4*m*sin(5*c + 5*d*x) - 2*m*(2*sin(2*c + 2*d*x)^2 - 1) + m*(2*sin(3*c + 3*d*x)^2 - 1) + 132*sin(c + d*x)^2 - 12
8))/(8*d*(-e*(2*sin(c/2 + (d*x)/2)^2 - 1))^(2*m)*(m^2 - 5*m + 6)*(15*sin(c + d*x) - sin(3*c + 3*d*x) + 12*sin(
c + d*x)^2 + 4))

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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((e*cos(d*x+c))**(3-2*m)*(a+a*sin(d*x+c))**m,x)

[Out]

Timed out

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