3.458 \(\int \frac{\cos ^2(c+d x)}{(a+b \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=115 \[ \frac{\tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{a \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2} \]

[Out]

ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]/((a^2 - b^2)^(3/2)*d) - Cos[c + d*x]/(2*b*d*(a + b*Sin[c + d*
x])^2) + (a*Cos[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.127862, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2693, 2754, 12, 2660, 618, 204} \[ \frac{\tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac{a \cos (c+d x)}{2 b d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\cos (c+d x)}{2 b d (a+b \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]]/((a^2 - b^2)^(3/2)*d) - Cos[c + d*x]/(2*b*d*(a + b*Sin[c + d*
x])^2) + (a*Cos[c + d*x])/(2*b*(a^2 - b^2)*d*(a + b*Sin[c + d*x]))

Rule 2693

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

Rule 2754

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.305711, size = 93, normalized size = 0.81 \[ \frac{\frac{2 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+\frac{\cos (c+d x) (a \sin (c+d x)+b)}{(a+b \sin (c+d x))^2}}{2 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (Cos[c + d*x]*(b + a*Sin[c + d*x]))/(a
+ b*Sin[c + d*x])^2)/(2*(a - b)*(a + b)*d)

________________________________________________________________________________________

Maple [B]  time = 0.095, size = 443, normalized size = 3.9 \begin{align*} -{\frac{a}{d \left ({a}^{2}-{b}^{2} \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{-2}}+2\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{2}}{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2}a \left ({a}^{2}-{b}^{2} \right ) }}+{\frac{b}{d \left ({a}^{2}-{b}^{2} \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{-2}}+2\,{\frac{{b}^{3} \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2} \left ({a}^{2}-{b}^{2} \right ){a}^{2}}}+{\frac{a}{d \left ({a}^{2}-{b}^{2} \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{-2}}+2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ){b}^{2}}{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{2}a \left ({a}^{2}-{b}^{2} \right ) }}+{\frac{b}{d \left ({a}^{2}-{b}^{2} \right ) } \left ( \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) ^{-2}}+{\frac{1}{d}\arctan \left ({\frac{1}{2} \left ( 2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b \right ){\frac{1}{\sqrt{{a}^{2}-{b}^{2}}}}} \right ) \left ({a}^{2}-{b}^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3+2/d/(tan(1/2*d*x+1/2
*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)^3*b^2+1/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2
*d*x+1/2*c)*b+a)^2*b/(a^2-b^2)*tan(1/2*d*x+1/2*c)^2+2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*b^
3/(a^2-b^2)/a^2*tan(1/2*d*x+1/2*c)^2+1/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2*a/(a^2-b^2)*tan(1
/2*d*x+1/2*c)+2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/a/(a^2-b^2)*tan(1/2*d*x+1/2*c)*b^2+1/d/(
tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)^2/(a^2-b^2)*b+1/d/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*d*x
+1/2*c)+2*b)/(a^2-b^2)^(1/2))

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 3.63783, size = 1107, normalized size = 9.63 \begin{align*} \left [-\frac{2 \,{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) -{\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{4 \,{\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (d x + c\right ) -{\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} d\right )}}, -\frac{{\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \,{\left ({\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - 2 \,{\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d \sin \left (d x + c\right ) -{\left (a^{6} - a^{4} b^{2} - a^{2} b^{4} + b^{6}\right )} d\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(2*(a^3 - a*b^2)*cos(d*x + c)*sin(d*x + c) - (b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)*sqrt(
-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x +
c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)) + 2*(a^2*b - b^3
)*cos(d*x + c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c)^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d*sin(d*x + c) -
(a^6 - a^4*b^2 - a^2*b^4 + b^6)*d), -1/2*((a^3 - a*b^2)*cos(d*x + c)*sin(d*x + c) + (b^2*cos(d*x + c)^2 - 2*a
*b*sin(d*x + c) - a^2 - b^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c))) + (a
^2*b - b^3)*cos(d*x + c))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d*cos(d*x + c)^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d*sin(
d*x + c) - (a^6 - a^4*b^2 - a^2*b^4 + b^6)*d)]

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a+b*sin(d*x+c))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.14356, size = 279, normalized size = 2.43 \begin{align*} \frac{\frac{\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} - \frac{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2} b}{{\left (a^{4} - a^{2} b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(a^2 - b^2)^
(3/2) - (a^3*tan(1/2*d*x + 1/2*c)^3 - 2*a*b^2*tan(1/2*d*x + 1/2*c)^3 - a^2*b*tan(1/2*d*x + 1/2*c)^2 - 2*b^3*ta
n(1/2*d*x + 1/2*c)^2 - a^3*tan(1/2*d*x + 1/2*c) - 2*a*b^2*tan(1/2*d*x + 1/2*c) - a^2*b)/((a^4 - a^2*b^2)*(a*ta
n(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x + 1/2*c) + a)^2))/d