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

Optimal. Leaf size=148 \[ -\frac{a^2 b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (\left (a^2-b^2\right ) \tan (c+d x)+2 a b\right )}{2 d \left (a^2+b^2\right )^2}+\frac{2 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (-6 a^2 b^2+a^4+b^4\right )}{2 \left (a^2+b^2\right )^3} \]

[Out]

((a^4 - 6*a^2*b^2 + b^4)*x)/(2*(a^2 + b^2)^3) + (2*a*b*(a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2
+ b^2)^3*d) - (a^2*b)/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]^2*(2*a*b + (a^2 - b^2)*Tan[c + d
*x]))/(2*(a^2 + b^2)^2*d)

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Rubi [A]  time = 0.294904, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {3516, 1647, 1629, 635, 203, 260} \[ -\frac{a^2 b}{d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (\left (a^2-b^2\right ) \tan (c+d x)+2 a b\right )}{2 d \left (a^2+b^2\right )^2}+\frac{2 a b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^3}+\frac{x \left (-6 a^2 b^2+a^4+b^4\right )}{2 \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^4 - 6*a^2*b^2 + b^4)*x)/(2*(a^2 + b^2)^3) + (2*a*b*(a^2 - b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2
+ b^2)^3*d) - (a^2*b)/((a^2 + b^2)^2*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]^2*(2*a*b + (a^2 - b^2)*Tan[c + d
*x]))/(2*(a^2 + b^2)^2*d)

Rule 3516

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

Rule 1647

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(d +
e*x)^m*Pq, a + c*x^2, x], f = Coeff[PolynomialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 0], g = Coeff[Polyn
omialRemainder[(d + e*x)^m*Pq, a + c*x^2, x], x, 1]}, Simp[((a*g - c*f*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1))
, x] + Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*ExpandToSum[(2*a*c*(p + 1)*Q)/(d + e*x)^m +
(c*f*(2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[Pq, x] && NeQ[c*d^2 + a*e^2, 0] &
& LtQ[p, -1] && ILtQ[m, 0]

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 635

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 203

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

Rule 260

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]

Rubi steps

\begin{align*} \int \frac{\sin ^2(c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{x^2}{(a+x)^2 \left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\cos ^2(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{a^2 b^2 \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2}+\frac{2 a b^2 x}{a^2+b^2}+\frac{b^2 \left (a^2-b^2\right ) x^2}{\left (a^2+b^2\right )^2}}{(a+x)^2 \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=-\frac{\cos ^2(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{2 a^2 b^2}{\left (a^2+b^2\right )^2 (a+x)^2}+\frac{4 a b^2 \left (-a^2+b^2\right )}{\left (a^2+b^2\right )^3 (a+x)}+\frac{b^2 \left (-a^4+6 a^2 b^2-b^4+4 a \left (a^2-b^2\right ) x\right )}{\left (a^2+b^2\right )^3 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=\frac{2 a b \left (a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{a^2 b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}-\frac{b \operatorname{Subst}\left (\int \frac{-a^4+6 a^2 b^2-b^4+4 a \left (a^2-b^2\right ) x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{2 a b \left (a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{a^2 b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}-\frac{\left (2 a b \left (a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{\left (a^2+b^2\right )^3 d}+\frac{\left (b \left (a^4-6 a^2 b^2+b^4\right )\right ) \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ &=\frac{\left (a^4-6 a^2 b^2+b^4\right ) x}{2 \left (a^2+b^2\right )^3}+\frac{2 a b \left (a^2-b^2\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^3 d}+\frac{2 a b \left (a^2-b^2\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^3 d}-\frac{a^2 b}{\left (a^2+b^2\right )^2 d (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 3.25967, size = 246, normalized size = 1.66 \[ -\frac{b \left (\frac{(a-b) (a+b) \left (a^2+b^2\right ) \sin (2 (c+d x))}{2 b}+2 a \left (a^2+b^2\right ) \cos ^2(c+d x)+\frac{\left (a^2-b^2\right ) \left (a^2+b^2\right ) \tan ^{-1}(\tan (c+d x))}{b}+\frac{2 a^2 \left (a^2+b^2\right )}{a+b \tan (c+d x)}+a \left (\frac{3 a b^2-a^3}{\sqrt{-b^2}}+2 a^2-2 b^2\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )+a \left (\frac{a^3-3 a b^2}{\sqrt{-b^2}}+2 a^2-2 b^2\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )-4 a (a-b) (a+b) \log (a+b \tan (c+d x))\right )}{2 d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(((a^2 - b^2)*(a^2 + b^2)*ArcTan[Tan[c + d*x]])/b + 2*a*(a^2 + b^2)*Cos[c + d*x]^2 + a*(2*a^2 - 2*b^2 + (-
a^3 + 3*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 4*a*(a - b)*(a + b)*Log[a + b*Tan[c + d*x]] + a*
(2*a^2 - 2*b^2 + (a^3 - 3*a*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + ((a - b)*(a + b)*(a^2 + b^2)*S
in[2*(c + d*x)])/(2*b) + (2*a^2*(a^2 + b^2))/(a + b*Tan[c + d*x])))/(2*(a^2 + b^2)^3*d)

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Maple [B]  time = 0.089, size = 352, normalized size = 2.4 \begin{align*} -{\frac{\tan \left ( dx+c \right ){a}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}+{\frac{\tan \left ( dx+c \right ){b}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{b{a}^{3}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{{b}^{3}a}{d \left ({a}^{2}+{b}^{2} \right ) ^{3} \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{3}b}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{{b}^{3}a\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{4}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-{\frac{b{a}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ^{2}d \left ( a+b\tan \left ( dx+c \right ) \right ) }}+2\,{\frac{b{a}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}}-2\,{\frac{{b}^{3}a\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d/(a^2+b^2)^3/(1+tan(d*x+c)^2)*tan(d*x+c)*a^4+1/2/d/(a^2+b^2)^3/(1+tan(d*x+c)^2)*tan(d*x+c)*b^4-1/d/(a^2+
b^2)^3/(1+tan(d*x+c)^2)*b*a^3-1/d/(a^2+b^2)^3/(1+tan(d*x+c)^2)*b^3*a-1/d/(a^2+b^2)^3*ln(1+tan(d*x+c)^2)*a^3*b+
1/d/(a^2+b^2)^3*b^3*a*ln(1+tan(d*x+c)^2)-3/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^2*b^2+1/2/d/(a^2+b^2)^3*arctan(t
an(d*x+c))*b^4+1/2/d/(a^2+b^2)^3*arctan(tan(d*x+c))*a^4-a^2*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))+2/d*b*a^3/(a^2+b^
2)^3*ln(a+b*tan(d*x+c))-2/d*b^3/(a^2+b^2)^3*a*ln(a+b*tan(d*x+c))

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Maxima [B]  time = 1.55602, size = 396, normalized size = 2.68 \begin{align*} \frac{\frac{{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{4 \,{\left (a^{3} b - a b^{3}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{4 \, a^{2} b +{\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{3} + a b^{2}\right )} \tan \left (d x + c\right )}{a^{5} + 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )^{3} +{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^4 - 6*a^2*b^2 + b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 4*(a^3*b - a*b^3)*log(b*tan(d*x +
c) + a)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(a^3*b - a*b^3)*log(tan(d*x + c)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*
a^2*b^4 + b^6) - (4*a^2*b + (3*a^2*b - b^3)*tan(d*x + c)^2 + (a^3 + a*b^2)*tan(d*x + c))/(a^5 + 2*a^3*b^2 + a*
b^4 + (a^4*b + 2*a^2*b^3 + b^5)*tan(d*x + c)^3 + (a^5 + 2*a^3*b^2 + a*b^4)*tan(d*x + c)^2 + (a^4*b + 2*a^2*b^3
+ b^5)*tan(d*x + c)))/d

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Fricas [B]  time = 2.2491, size = 640, normalized size = 4.32 \begin{align*} -\frac{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right )^{3} +{\left (a^{2} b^{3} - b^{5} -{\left (a^{5} - 6 \, a^{3} b^{2} + a b^{4}\right )} d x\right )} \cos \left (d x + c\right ) - 2 \,{\left ({\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (d x + c\right ) +{\left (a^{3} b^{2} - a b^{4}\right )} \sin \left (d x + c\right )\right )} \log \left (2 \, a b \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + b^{2}\right ) -{\left (3 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b - 6 \, a^{2} b^{3} + b^{5}\right )} d x -{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} d \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((a^4*b + 2*a^2*b^3 + b^5)*cos(d*x + c)^3 + (a^2*b^3 - b^5 - (a^5 - 6*a^3*b^2 + a*b^4)*d*x)*cos(d*x + c)
- 2*((a^4*b - a^2*b^3)*cos(d*x + c) + (a^3*b^2 - a*b^4)*sin(d*x + c))*log(2*a*b*cos(d*x + c)*sin(d*x + c) + (a
^2 - b^2)*cos(d*x + c)^2 + b^2) - (3*a^3*b^2 + a*b^4 + (a^4*b - 6*a^2*b^3 + b^5)*d*x - (a^5 + 2*a^3*b^2 + a*b^
4)*cos(d*x + c)^2)*sin(d*x + c))/((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*d*cos(d*x + c) + (a^6*b + 3*a^4*b^3 +
3*a^2*b^5 + b^7)*d*sin(d*x + c))

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.17682, size = 355, normalized size = 2.4 \begin{align*} \frac{\frac{{\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )}{\left (d x + c\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{2 \,{\left (a^{3} b - a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{4 \,{\left (a^{3} b^{2} - a b^{4}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{3 \, a^{2} b \tan \left (d x + c\right )^{2} - b^{3} \tan \left (d x + c\right )^{2} + a^{3} \tan \left (d x + c\right ) + a b^{2} \tan \left (d x + c\right ) + 4 \, a^{2} b}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \tan \left (d x + c\right )^{3} + a \tan \left (d x + c\right )^{2} + b \tan \left (d x + c\right ) + a\right )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a^4 - 6*a^2*b^2 + b^4)*(d*x + c)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - 2*(a^3*b - a*b^3)*log(tan(d*x + c
)^2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 4*(a^3*b^2 - a*b^4)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 3*a^4
*b^3 + 3*a^2*b^5 + b^7) - (3*a^2*b*tan(d*x + c)^2 - b^3*tan(d*x + c)^2 + a^3*tan(d*x + c) + a*b^2*tan(d*x + c)
+ 4*a^2*b)/((a^4 + 2*a^2*b^2 + b^4)*(b*tan(d*x + c)^3 + a*tan(d*x + c)^2 + b*tan(d*x + c) + a)))/d