\(\int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx\) [14]

Optimal result

Integrand size = 19, antiderivative size = 49 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a x}{2}-\frac {b \log (\cos (c+d x))}{d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))}{2 d} \]

[Out]

1/2*a*x-b*ln(cos(d*x+c))/d-1/2*cos(d*x+c)*sin(d*x+c)*(a+b*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {833, 649, 209, 266} \[ \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {\sin (c+d x) \cos (c+d x) (a+b \tan (c+d x))}{2 d}+\frac {a x}{2}-\frac {b \log (\cos (c+d x))}{d} \]

[In]

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

[Out]

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

Rule 209

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 649

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[(-a)*c]

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(d + e*x)^(m
- 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Dist[1/(2*a*c*(p + 1)),
Int[(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^2*f*(2*p + 3) + e*(a*e*g*m - c*
d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ
[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.14 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a (c+d x)}{2 d}-\frac {b \left (-\frac {1}{2} \cos ^2(c+d x)+\log (\cos (c+d x))\right )}{d}-\frac {a \sin (2 (c+d x))}{4 d} \]

[In]

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

[Out]

(a*(c + d*x))/(2*d) - (b*(-1/2*Cos[c + d*x]^2 + Log[Cos[c + d*x]]))/d - (a*Sin[2*(c + d*x)])/(4*d)

Maple [A] (verified)

Time = 0.59 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(52\)
default	\(\frac {a \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )}{d}\)	\(52\)
risch	\(i b x +\frac {a x}{2}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} b}{8 d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a}{8 d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} b}{8 d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a}{8 d}+\frac {2 i b c}{d}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(99\)

[In]

int(sin(d*x+c)^2*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.96 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a d x + b \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, b \log \left (-\cos \left (d x + c\right )\right )}{2 \, d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

integrate(sin(d*x+c)**2*(a+b*tan(d*x+c)),x)

[Out]

Integral((a + b*tan(c + d*x))*sin(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.06 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {{\left (d x + c\right )} a + b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - \frac {a \tan \left (d x + c\right ) - b}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 373 vs. \(2 (45) = 90\).

Time = 0.38 (sec) , antiderivative size = 373, normalized size of antiderivative = 7.61 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2 \, a d x \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + 2 \, a d x \tan \left (d x\right )^{2} + 2 \, a d x \tan \left (c\right )^{2} + b \tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) \tan \left (d x\right )^{2} + 2 \, a \tan \left (d x\right )^{2} \tan \left (c\right ) - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) \tan \left (c\right )^{2} + 2 \, a \tan \left (d x\right ) \tan \left (c\right )^{2} + 2 \, a d x - b \tan \left (d x\right )^{2} - 4 \, b \tan \left (d x\right ) \tan \left (c\right ) - b \tan \left (c\right )^{2} - 2 \, b \log \left (\frac {4 \, {\left (\tan \left (d x\right )^{2} \tan \left (c\right )^{2} - 2 \, \tan \left (d x\right ) \tan \left (c\right ) + 1\right )}}{\tan \left (d x\right )^{2} \tan \left (c\right )^{2} + \tan \left (d x\right )^{2} + \tan \left (c\right )^{2} + 1}\right ) - 2 \, a \tan \left (d x\right ) - 2 \, a \tan \left (c\right ) + b}{4 \, {\left (d \tan \left (d x\right )^{2} \tan \left (c\right )^{2} + d \tan \left (d x\right )^{2} + d \tan \left (c\right )^{2} + d\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.02 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x)) \, dx=\frac {\frac {b\,{\cos \left (c+d\,x\right )}^2}{2}-\frac {a\,\sin \left (c+d\,x\right )\,\cos \left (c+d\,x\right )}{2}+\frac {b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}+\frac {a\,d\,x}{2}}{d} \]

[In]

int(sin(c + d*x)^2*(a + b*tan(c + d*x)),x)

[Out]

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