3.1332 \(\int \frac{\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=93 \[ \frac{a^3 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac{\sin (c+d x)}{b d} \]

[Out]

-Log[1 - Sin[c + d*x]]/(2*(a + b)*d) - Log[1 + Sin[c + d*x]]/(2*(a - b)*d) + (a^3*Log[a + b*Sin[c + d*x]])/(b^
2*(a^2 - b^2)*d) - Sin[c + d*x]/(b*d)

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Rubi [A]  time = 0.180854, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2837, 12, 1629} \[ \frac{a^3 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)}-\frac{\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac{\sin (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Log[1 - Sin[c + d*x]]/(2*(a + b)*d) - Log[1 + Sin[c + d*x]]/(2*(a - b)*d) + (a^3*Log[a + b*Sin[c + d*x]])/(b^
2*(a^2 - b^2)*d) - Sin[c + d*x]/(b*d)

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

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

Rule 1629

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*
Pq*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

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

Mathematica [A]  time = 0.210526, size = 83, normalized size = 0.89 \[ -\frac{-\frac{2 a^3 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )}+\frac{\log (1-\sin (c+d x))}{a+b}+\frac{\log (\sin (c+d x)+1)}{a-b}+\frac{2 \sin (c+d x)}{b}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[1 - Sin[c + d*x]]/(a + b) + Log[1 + Sin[c + d*x]]/(a - b) - (2*a^3*Log[a + b*Sin[c + d*x]])/(b^2*(a^2 -
b^2)) + (2*Sin[c + d*x])/b)/(2*d)

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Maple [A]  time = 0.064, size = 95, normalized size = 1. \begin{align*} -{\frac{\sin \left ( dx+c \right ) }{bd}}+{\frac{{a}^{3}\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d{b}^{2} \left ( a+b \right ) \left ( a-b \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{d \left ( 2\,a+2\,b \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{d \left ( 2\,a-2\,b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sin(d*x+c)/b/d+1/d/b^2*a^3/(a+b)/(a-b)*ln(a+b*sin(d*x+c))-1/d/(2*a+2*b)*ln(sin(d*x+c)-1)-1/d/(2*a-2*b)*ln(1+s
in(d*x+c))

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Maxima [A]  time = 0.987271, size = 111, normalized size = 1.19 \begin{align*} \frac{\frac{2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b^{2} - b^{4}} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} - \frac{2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.20683, size = 223, normalized size = 2.4 \begin{align*} \frac{2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left (a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a b^{2} - b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (a^{2} b^{2} - b^{4}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [A]  time = 1.25744, size = 115, normalized size = 1.24 \begin{align*} \frac{\frac{2 \, a^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac{\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} - \frac{2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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