∫ cos5 (c+dx)(A+B sin(c+dx))___________________

3.1012 \(\int \frac{\cos ^5(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=51 \[ \frac{B (a-a \sin (c+d x))^4}{4 a^6 d}-\frac{(A+B) (a-a \sin (c+d x))^3}{3 a^5 d} \]

[Out]

-((A + B)*(a - a*Sin[c + d*x])^3)/(3*a^5*d) + (B*(a - a*Sin[c + d*x])^4)/(4*a^6*d)

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Rubi [A] time = 0.102889, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2836, 43} \[ \frac{B (a-a \sin (c+d x))^4}{4 a^6 d}-\frac{(A+B) (a-a \sin (c+d x))^3}{3 a^5 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

-((A + B)*(a - a*Sin[c + d*x])^3)/(3*a^5*d) + (B*(a - a*Sin[c + d*x])^4)/(4*a^6*d)

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)

*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b

)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d

*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]

&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A] time = 0.0631584, size = 34, normalized size = 0.67 \[ \frac{(\sin (c+d x)-1)^3 (4 A+3 B \sin (c+d x)+B)}{12 a^2 d} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Cos[c + d*x]^5*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

((-1 + Sin[c + d*x])^3*(4*A + B + 3*B*Sin[c + d*x]))/(12*a^2*d)

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Maple [A] time = 0.109, size = 58, normalized size = 1.1 \begin{align*}{\frac{1}{d{a}^{2}} \left ({\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4}}+{\frac{ \left ( A-2\,B \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3}}+{\frac{ \left ( -2\,A+B \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2}}+A\sin \left ( dx+c \right ) \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)

[Out]

1/d/a^2*(1/4*B*sin(d*x+c)^4+1/3*(A-2*B)*sin(d*x+c)^3+1/2*(-2*A+B)*sin(d*x+c)^2+A*sin(d*x+c))

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Maxima [A] time = 1.03872, size = 82, normalized size = 1.61 \begin{align*} \frac{3 \, B \sin \left (d x + c\right )^{4} + 4 \,{\left (A - 2 \, B\right )} \sin \left (d x + c\right )^{3} - 6 \,{\left (2 \, A - B\right )} \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/12*(3*B*sin(d*x + c)^4 + 4*(A - 2*B)*sin(d*x + c)^3 - 6*(2*A - B)*sin(d*x + c)^2 + 12*A*sin(d*x + c))/(a^2*d

)

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Fricas [A] time = 1.45174, size = 161, normalized size = 3.16 \begin{align*} \frac{3 \, B \cos \left (d x + c\right )^{4} + 12 \,{\left (A - B\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left ({\left (A - 2 \, B\right )} \cos \left (d x + c\right )^{2} - 4 \, A + 2 \, B\right )} \sin \left (d x + c\right )}{12 \, a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/12*(3*B*cos(d*x + c)^4 + 12*(A - B)*cos(d*x + c)^2 - 4*((A - 2*B)*cos(d*x + c)^2 - 4*A + 2*B)*sin(d*x + c))/

(a^2*d)

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Sympy [A] time = 85.0489, size = 1545, normalized size = 30.29 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((-2*A*tan(c/2 + d*x/2)**8/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*

tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) + 18*A*tan(c/2 + d*x/2)**7/(9*a**2*d*tan(c/2 +

d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9

*a**2*d) - 44*A*tan(c/2 + d*x/2)**6/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*

tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) + 78*A*tan(c/2 + d*x/2)**5/(9*a**2*d*tan(c/2 +

d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9

*a**2*d) - 84*A*tan(c/2 + d*x/2)**4/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*

tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) + 78*A*tan(c/2 + d*x/2)**3/(9*a**2*d*tan(c/2 +

d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9

*a**2*d) - 44*A*tan(c/2 + d*x/2)**2/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*

tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) + 18*A*tan(c/2 + d*x/2)/(9*a**2*d*tan(c/2 + d*

x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a*

*2*d) - 2*A/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36

*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) - 2*B*tan(c/2 + d*x/2)**8/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*ta

n(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) + 10*B*tan(c/2 +

d*x/2)**6/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*

a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) - 48*B*tan(c/2 + d*x/2)**5/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*ta

n(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) + 60*B*tan(c/2 +

d*x/2)**4/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*

a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) - 48*B*tan(c/2 + d*x/2)**3/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*ta

n(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) + 10*B*tan(c/2 +

d*x/2)**2/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 54*a**2*d*tan(c/2 + d*x/2)**4 + 36*

a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d) - 2*B/(9*a**2*d*tan(c/2 + d*x/2)**8 + 36*a**2*d*tan(c/2 + d*x/2)**6 + 5

4*a**2*d*tan(c/2 + d*x/2)**4 + 36*a**2*d*tan(c/2 + d*x/2)**2 + 9*a**2*d), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)*

*5/(a*sin(c) + a)**2, True))

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Giac [A] time = 1.33772, size = 99, normalized size = 1.94 \begin{align*} \frac{3 \, B \sin \left (d x + c\right )^{4} + 4 \, A \sin \left (d x + c\right )^{3} - 8 \, B \sin \left (d x + c\right )^{3} - 12 \, A \sin \left (d x + c\right )^{2} + 6 \, B \sin \left (d x + c\right )^{2} + 12 \, A \sin \left (d x + c\right )}{12 \, a^{2} d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/12*(3*B*sin(d*x + c)^4 + 4*A*sin(d*x + c)^3 - 8*B*sin(d*x + c)^3 - 12*A*sin(d*x + c)^2 + 6*B*sin(d*x + c)^2

+ 12*A*sin(d*x + c))/(a^2*d)

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