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]

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

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Rubi [A]  time = 0.18, antiderivative size = 93, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.21, 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]

-1/2*(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)/d

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fricas [A]  time = 0.60, size = 100, normalized size = 1.08 \[ \frac {2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a b^{2} - b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d} \]

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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giac [A]  time = 0.21, size = 85, normalized size = 0.91 \[ \frac {\frac {2 \, a^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} - \frac {2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \]

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

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maple [A]  time = 0.37, size = 95, normalized size = 1.02 \[ -\frac {\sin \left (d x +c \right )}{b d}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{d \left (2 a +2 b \right )}+\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{d \,b^{2} \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{d \left (2 a -2 b \right )} \]

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

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maxima [A]  time = 0.31, size = 82, normalized size = 0.88 \[ \frac {\frac {2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} - \frac {2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \]

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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mupad [B]  time = 12.27, size = 134, normalized size = 1.44 \[ -\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )}-\frac {\sin \left (c+d\,x\right )}{b\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d}-\frac {a^3\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (b^4-a^2\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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