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3.1013 \(\int \frac{\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=66 \[ -\frac{(A-B) \sin (c+d x)}{a^2 d}+\frac{2 (A-B) \log (\sin (c+d x)+1)}{a^2 d}-\frac{B (a-a \sin (c+d x))^2}{2 a^4 d} \]

[Out]

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

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Rubi [A]  time = 0.107344, antiderivative size = 66, 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, 77} \[ -\frac{(A-B) \sin (c+d x)}{a^2 d}+\frac{2 (A-B) \log (\sin (c+d x)+1)}{a^2 d}-\frac{B (a-a \sin (c+d x))^2}{2 a^4 d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

(2*(A - B)*Log[1 + Sin[c + d*x]])/(a^2*d) - ((A - B)*Sin[c + d*x])/(a^2*d) - (B*(a - a*Sin[c + d*x])^2)/(2*a^4
*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,
 0]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) (A+B \sin (c+d x))}{(a+a \sin (c+d x))^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a-x) \left (A+\frac{B x}{a}\right )}{a+x} \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-A+B+\frac{B (a-x)}{a}+\frac{2 a (A-B)}{a+x}\right ) \, dx,x,a \sin (c+d x)\right )}{a^3 d}\\ &=\frac{2 (A-B) \log (1+\sin (c+d x))}{a^2 d}-\frac{(A-B) \sin (c+d x)}{a^2 d}-\frac{B (a-a \sin (c+d x))^2}{2 a^4 d}\\ \end{align*}

Mathematica [A]  time = 0.0931061, size = 51, normalized size = 0.77 \[ -\frac{2 (A-2 B) \sin (c+d x)-4 (A-B) \log (\sin (c+d x)+1)+B \sin ^2(c+d x)+B}{2 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Sin[c + d*x]))/(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.109, size = 85, normalized size = 1.3 \begin{align*} -{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d{a}^{2}}}-{\frac{A\sin \left ( dx+c \right ) }{d{a}^{2}}}+2\,{\frac{B\sin \left ( dx+c \right ) }{d{a}^{2}}}+2\,{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{d{a}^{2}}}-2\,{\frac{B\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.02946, size = 73, normalized size = 1.11 \begin{align*} \frac{\frac{4 \,{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2}} - \frac{B \sin \left (d x + c\right )^{2} + 2 \,{\left (A - 2 \, B\right )} \sin \left (d x + c\right )}{a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.47416, size = 126, normalized size = 1.91 \begin{align*} \frac{B \cos \left (d x + c\right )^{2} + 4 \,{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (A - 2 \, B\right )} \sin \left (d x + c\right )}{2 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 27.5817, size = 1428, normalized size = 21.64 \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)**3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))**2,x)

[Out]

Piecewise((28*A*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/
2 + d*x/2)**2 + 7*a**2*d) + 56*A*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(7*a**2*d*tan(c/2 + d*x/2)**4 +
 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 28*A*log(tan(c/2 + d*x/2) + 1)/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14
*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 14*A*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(7*a**2*d*tan(
c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 28*A*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/
2)**2/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 14*A*log(tan(c/2 + d*x/2)**2
 + 1)/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 6*A*tan(c/2 + d*x/2)**4/(7*a
**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 14*A*tan(c/2 + d*x/2)**3/(7*a**2*d*tan
(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 12*A*tan(c/2 + d*x/2)**2/(7*a**2*d*tan(c/2 + d*
x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 14*A*tan(c/2 + d*x/2)/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14
*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 6*A/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 +
7*a**2*d) - 28*B*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**4/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c
/2 + d*x/2)**2 + 7*a**2*d) - 56*B*log(tan(c/2 + d*x/2) + 1)*tan(c/2 + d*x/2)**2/(7*a**2*d*tan(c/2 + d*x/2)**4
+ 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 28*B*log(tan(c/2 + d*x/2) + 1)/(7*a**2*d*tan(c/2 + d*x/2)**4 + 1
4*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 14*B*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x/2)**4/(7*a**2*d*tan
(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 28*B*log(tan(c/2 + d*x/2)**2 + 1)*tan(c/2 + d*x
/2)**2/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 14*B*log(tan(c/2 + d*x/2)**
2 + 1)/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 10*B*tan(c/2 + d*x/2)**4/(7
*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 28*B*tan(c/2 + d*x/2)**3/(7*a**2*d*t
an(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 34*B*tan(c/2 + d*x/2)**2/(7*a**2*d*tan(c/2 +
d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) + 28*B*tan(c/2 + d*x/2)/(7*a**2*d*tan(c/2 + d*x/2)**4 +
14*a**2*d*tan(c/2 + d*x/2)**2 + 7*a**2*d) - 10*B/(7*a**2*d*tan(c/2 + d*x/2)**4 + 14*a**2*d*tan(c/2 + d*x/2)**2
 + 7*a**2*d), Ne(d, 0)), (x*(A + B*sin(c))*cos(c)**3/(a*sin(c) + a)**2, True))

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Giac [A]  time = 1.42122, size = 124, normalized size = 1.88 \begin{align*} -\frac{\frac{4 \,{\left (A - B\right )} \log \left (\frac{{\left | a \sin \left (d x + c\right ) + a \right |}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left | a \right |}}\right )}{a^{2}} + \frac{{\left (a \sin \left (d x + c\right ) + a\right )}^{2}{\left (B + \frac{2 \,{\left (A a^{2} - 3 \, B a^{2}\right )}}{{\left (a \sin \left (d x + c\right ) + a\right )} a}\right )}}{a^{4}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sin(d*x+c))/(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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

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