3.1340 \(\int \frac{\sin (c+d x) \tan ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=133 \[ \frac{2 a^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{3/2}}-\frac{b \tan (c+d x)}{d \left (a^2-b^2\right )}+\frac{a \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac{a^2 x}{b \left (a^2-b^2\right )}+\frac{b x}{a^2-b^2} \]

[Out]

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

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Rubi [A]  time = 0.177964, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2902, 2606, 8, 3473, 2735, 2660, 618, 204} \[ \frac{2 a^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b d \left (a^2-b^2\right )^{3/2}}-\frac{b \tan (c+d x)}{d \left (a^2-b^2\right )}+\frac{a \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac{a^2 x}{b \left (a^2-b^2\right )}+\frac{b x}{a^2-b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2902

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

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 2735

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

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

Mathematica [A]  time = 0.769586, size = 152, normalized size = 1.14 \[ \frac{\frac{2 a^3 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b (a-b \sin (c+d x))-\left (a^2-b^2\right ) (c+d x) \cos (c+d x)}{(a-b) (a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.093, size = 138, normalized size = 1. \begin{align*} -2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{bd}}+2\,{\frac{{a}^{3}}{d \left ( a-b \right ) \left ( a+b \right ) b\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+16\,{\frac{1}{d \left ( 16\,a-16\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) }}-16\,{\frac{1}{d \left ( 16\,a+16\,b \right ) \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/b*arctan(tan(1/2*d*x+1/2*c))+2/d/(a-b)/(a+b)*a^3/b/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b
)/(a^2-b^2)^(1/2))+16/d/(16*a-16*b)/(tan(1/2*d*x+1/2*c)+1)-16/d/(16*a+16*b)/(tan(1/2*d*x+1/2*c)-1)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62385, size = 822, normalized size = 6.18 \begin{align*} \left [\frac{\sqrt{-a^{2} + b^{2}} a^{3} \cos \left (d x + c\right ) \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 \, a^{3} b - 2 \, a b^{3} - 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) - 2 \,{\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}, -\frac{\sqrt{a^{2} - b^{2}} a^{3} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) - a^{3} b + a b^{3} +{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (a^{2} b^{2} - b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \cos \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.19808, size = 177, normalized size = 1.33 \begin{align*} \frac{\frac{2 \,{\left (\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 )\right )} a^{3}}{{\left (a^{2} b - b^{3}\right )} \sqrt{a^{2} - b^{2}}} - \frac{d x + c}{b} + \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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