\(\int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx\) [1332]

Optimal result

Integrand size = 27, antiderivative size = 93 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\log (1-\sin (c+d x))}{2 (a+b) d}-\frac {\log (1+\sin (c+d x))}{2 (a-b) d}+\frac {a^3 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right ) d}-\frac {\sin (c+d x)}{b d} \]

[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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {2916, 12, 1643} \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a^3 \log (a+b \sin (c+d x))}{b^2 d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}-\frac {\log (\sin (c+d x)+1)}{2 d (a-b)}-\frac {\sin (c+d x)}{b d} \]

[In]

Int[(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)*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 1643

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 2916

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/b)*x)^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*} \text {integral}& = \frac {b \text {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 {\text {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 {\text {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] (verified)

Time = 0.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {\log (1-\sin (c+d x))}{a+b}+\frac {\log (1+\sin (c+d x))}{a-b}-\frac {2 a^3 \log (a+b \sin (c+d x))}{b^2 \left (a^2-b^2\right )}+\frac {2 \sin (c+d x)}{b}}{2 d} \]

[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

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {-\frac {\sin \left (d x +c \right )}{b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\)	\(87\)
default	\(\frac {-\frac {\sin \left (d x +c \right )}{b}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}+\frac {a^{3} \ln \left (a +b \sin \left (d x +c \right )\right )}{b^{2} \left (a +b \right ) \left (a -b \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}}{d}\)	\(87\)
norman	\(\frac {-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b d}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {a^{3} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{b^{2} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{\left (a +b \right ) d}-\frac {a \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{2} d}\)	\(173\)
parallelrisch	\(\frac {-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a \,b^{2}+\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b^{3}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a \,b^{2}-\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b^{3}-a^{3} \ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\ln \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a \,b^{2}+a^{3} \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\sin \left (d x +c \right ) a^{2} b +b^{3} \sin \left (d x +c \right )}{\left (a^{2}-b^{2}\right ) b^{2} d}\)	\(176\)
risch	\(\frac {i a x}{b^{2}}+\frac {i {\mathrm e}^{i \left (d x +c \right )}}{2 b d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 b d}+\frac {i x}{a -b}+\frac {i c}{d \left (a -b \right )}+\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}-\frac {2 i a^{3} x}{b^{2} \left (a^{2}-b^{2}\right )}-\frac {2 i a^{3} c}{b^{2} d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}+\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{b^{2} d \left (a^{2}-b^{2}\right )}\)	\(234\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a b^{2} + b^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a b^{2} - b^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{2} b^{2} - b^{4}\right )} d} \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a^{3} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b} - \frac {2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.91 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, a^{3} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b^{2} - b^{4}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b} - \frac {2 \, \sin \left (d x + c\right )}{b}}{2 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 12.15 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.44 \[ \int \frac {\sin ^2(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{d\,\left (a+b\right )}-\frac {\sin \left (c+d\,x\right )}{b\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{d\,\left (a-b\right )}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{b^2\,d}-\frac {a^3\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (b^4-a^2\,b^2\right )} \]

[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))