3.69 \(\int \frac{\sin ^2(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=206 \[ -\frac{a^2 b}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac{2 a b \left (a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (a \left (a^2-3 b^2\right ) \tan (c+d x)+b \left (3 a^2-b^2\right )\right )}{2 d \left (a^2+b^2\right )^3}+\frac{b \left (-8 a^2 b^2+3 a^4+b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac{a x \left (-14 a^2 b^2+a^4+9 b^4\right )}{2 \left (a^2+b^2\right )^4} \]

[Out]

(a*(a^4 - 14*a^2*b^2 + 9*b^4)*x)/(2*(a^2 + b^2)^4) + (b*(3*a^4 - 8*a^2*b^2 + b^4)*Log[a*Cos[c + d*x] + b*Sin[c
+ d*x]])/((a^2 + b^2)^4*d) - (a^2*b)/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])^2) - (2*a*b*(a^2 - b^2))/((a^2 +
b^2)^3*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]^2*(b*(3*a^2 - b^2) + a*(a^2 - 3*b^2)*Tan[c + d*x]))/(2*(a^2 +
b^2)^3*d)

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Rubi [A]  time = 0.396246, antiderivative size = 206, 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} \[ -\frac{a^2 b}{2 d \left (a^2+b^2\right )^2 (a+b \tan (c+d x))^2}-\frac{2 a b \left (a^2-b^2\right )}{d \left (a^2+b^2\right )^3 (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (a \left (a^2-3 b^2\right ) \tan (c+d x)+b \left (3 a^2-b^2\right )\right )}{2 d \left (a^2+b^2\right )^3}+\frac{b \left (-8 a^2 b^2+3 a^4+b^4\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )^4}+\frac{a x \left (-14 a^2 b^2+a^4+9 b^4\right )}{2 \left (a^2+b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

(a*(a^4 - 14*a^2*b^2 + 9*b^4)*x)/(2*(a^2 + b^2)^4) + (b*(3*a^4 - 8*a^2*b^2 + b^4)*Log[a*Cos[c + d*x] + b*Sin[c
+ d*x]])/((a^2 + b^2)^4*d) - (a^2*b)/(2*(a^2 + b^2)^2*d*(a + b*Tan[c + d*x])^2) - (2*a*b*(a^2 - b^2))/((a^2 +
b^2)^3*d*(a + b*Tan[c + d*x])) - (Cos[c + d*x]^2*(b*(3*a^2 - b^2) + a*(a^2 - 3*b^2)*Tan[c + d*x]))/(2*(a^2 +
b^2)^3*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))^3} \, dx &=\frac{b \operatorname{Subst}\left (\int \frac{x^2}{(a+x)^3 \left (b^2+x^2\right )^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=-\frac{\cos ^2(c+d x) \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\operatorname{Subst}\left (\int \frac{-\frac{a^4 b^2 \left (a^2-3 b^2\right )}{\left (a^2+b^2\right )^3}+\frac{a^3 b^2 \left (3 a^2+7 b^2\right ) x}{\left (a^2+b^2\right )^3}+\frac{b^2 \left (3 a^4-3 a^2 b^2-2 b^4\right ) x^2}{\left (a^2+b^2\right )^3}+\frac{a b^2 \left (a^2-3 b^2\right ) x^3}{\left (a^2+b^2\right )^3}}{(a+x)^3 \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=-\frac{\cos ^2(c+d x) \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{2 a^2 b^2}{\left (a^2+b^2\right )^2 (a+x)^3}+\frac{4 a b^2 \left (-a^2+b^2\right )}{\left (a^2+b^2\right )^3 (a+x)^2}-\frac{2 \left (3 a^4 b^2-8 a^2 b^4+b^6\right )}{\left (a^2+b^2\right )^4 (a+x)}+\frac{b^2 \left (-a \left (a^4-14 a^2 b^2+9 b^4\right )+2 \left (3 a^4-8 a^2 b^2+b^4\right ) x\right )}{\left (a^2+b^2\right )^4 \left (b^2+x^2\right )}\right ) \, dx,x,b \tan (c+d x)\right )}{2 b d}\\ &=\frac{b \left (3 a^4-8 a^2 b^2+b^4\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac{a^2 b}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}-\frac{2 a b \left (a^2-b^2\right )}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{b \operatorname{Subst}\left (\int \frac{-a \left (a^4-14 a^2 b^2+9 b^4\right )+2 \left (3 a^4-8 a^2 b^2+b^4\right ) x}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^4 d}\\ &=\frac{b \left (3 a^4-8 a^2 b^2+b^4\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac{a^2 b}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}-\frac{2 a b \left (a^2-b^2\right )}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}-\frac{\left (b \left (3 a^4-8 a^2 b^2+b^4\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 )^4 d}+\frac{\left (a b \left (a^4-14 a^2 b^2+9 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 )^4 d}\\ &=\frac{a \left (a^4-14 a^2 b^2+9 b^4\right ) x}{2 \left (a^2+b^2\right )^4}+\frac{b \left (3 a^4-8 a^2 b^2+b^4\right ) \log (\cos (c+d x))}{\left (a^2+b^2\right )^4 d}+\frac{b \left (3 a^4-8 a^2 b^2+b^4\right ) \log (a+b \tan (c+d x))}{\left (a^2+b^2\right )^4 d}-\frac{a^2 b}{2 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^2}-\frac{2 a b \left (a^2-b^2\right )}{\left (a^2+b^2\right )^3 d (a+b \tan (c+d x))}-\frac{\cos ^2(c+d x) \left (b \left (3 a^2-b^2\right )+a \left (a^2-3 b^2\right ) \tan (c+d x)\right )}{2 \left (a^2+b^2\right )^3 d}\\ \end{align*}

Mathematica [A]  time = 3.95863, size = 316, normalized size = 1.53 \[ -\frac{b \left (\frac{a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \sin (2 (c+d x))}{2 b}+\left (3 a^2-b^2\right ) \left (a^2+b^2\right ) \cos ^2(c+d x)+\frac{a \left (a^2-3 b^2\right ) \left (a^2+b^2\right ) \tan ^{-1}(\tan (c+d x))}{b}+\frac{a^2 \left (a^2+b^2\right )^2}{(a+b \tan (c+d x))^2}+\frac{4 \left (a^5-a b^4\right )}{a+b \tan (c+d x)}+\left (-8 a^2 b^2-\frac{-8 a^3 b^2+a^5+3 a b^4}{\sqrt{-b^2}}+3 a^4+b^4\right ) \log \left (\sqrt{-b^2}-b \tan (c+d x)\right )-2 \left (-8 a^2 b^2+3 a^4+b^4\right ) \log (a+b \tan (c+d x))+\left (-8 a^2 b^2+\frac{-8 a^3 b^2+a^5+3 a b^4}{\sqrt{-b^2}}+3 a^4+b^4\right ) \log \left (\sqrt{-b^2}+b \tan (c+d x)\right )\right )}{2 d \left (a^2+b^2\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^2/(a + b*Tan[c + d*x])^3,x]

[Out]

-(b*((a*(a^2 - 3*b^2)*(a^2 + b^2)*ArcTan[Tan[c + d*x]])/b + (3*a^2 - b^2)*(a^2 + b^2)*Cos[c + d*x]^2 + (3*a^4
- 8*a^2*b^2 + b^4 - (a^5 - 8*a^3*b^2 + 3*a*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] - 2*(3*a^4 - 8*a^
2*b^2 + b^4)*Log[a + b*Tan[c + d*x]] + (3*a^4 - 8*a^2*b^2 + b^4 + (a^5 - 8*a^3*b^2 + 3*a*b^4)/Sqrt[-b^2])*Log[
Sqrt[-b^2] + b*Tan[c + d*x]] + (a*(a^2 - 3*b^2)*(a^2 + b^2)*Sin[2*(c + d*x)])/(2*b) + (a^2*(a^2 + b^2)^2)/(a +
b*Tan[c + d*x])^2 + (4*(a^5 - a*b^4))/(a + b*Tan[c + d*x])))/(2*(a^2 + b^2)^4*d)

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Maple [B]  time = 0.115, size = 542, normalized size = 2.6 \begin{align*} \text{result too large to display} \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))^3,x)

[Out]

-1/2/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*tan(d*x+c)*a^5+1/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*tan(d*x+c)*a^3*b^2+3/2/d/(
a^2+b^2)^4/(1+tan(d*x+c)^2)*tan(d*x+c)*a*b^4-3/2/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*a^4*b-1/d/(a^2+b^2)^4/(1+tan(d
*x+c)^2)*a^2*b^3+1/2/d/(a^2+b^2)^4/(1+tan(d*x+c)^2)*b^5-3/2/d/(a^2+b^2)^4*ln(1+tan(d*x+c)^2)*a^4*b+4/d/(a^2+b^
2)^4*ln(1+tan(d*x+c)^2)*a^2*b^3-1/2/d/(a^2+b^2)^4*ln(1+tan(d*x+c)^2)*b^5-7/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a^
3*b^2+9/2/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a*b^4+1/2/d/(a^2+b^2)^4*arctan(tan(d*x+c))*a^5-1/2*a^2*b/(a^2+b^2)^
2/d/(a+b*tan(d*x+c))^2+3/d*b/(a^2+b^2)^4*ln(a+b*tan(d*x+c))*a^4-8/d*b^3/(a^2+b^2)^4*ln(a+b*tan(d*x+c))*a^2+1/d
*b^5/(a^2+b^2)^4*ln(a+b*tan(d*x+c))-2/d*b*a^3/(a^2+b^2)^3/(a+b*tan(d*x+c))+2/d*b^3/(a^2+b^2)^3*a/(a+b*tan(d*x+
c))

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Maxima [B]  time = 1.84901, size = 625, normalized size = 3.03 \begin{align*} \frac{\frac{{\left (a^{5} - 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )}{\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac{2 \,{\left (3 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{{\left (3 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{8 \, a^{4} b - 4 \, a^{2} b^{3} +{\left (5 \, a^{3} b^{2} - 7 \, a b^{4}\right )} \tan \left (d x + c\right )^{3} +{\left (7 \, a^{4} b - 6 \, a^{2} b^{3} - b^{5}\right )} \tan \left (d x + c\right )^{2} +{\left (a^{5} + 7 \, a^{3} b^{2} - 6 \, a b^{4}\right )} \tan \left (d x + c\right )}{a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} b^{6} +{\left (a^{6} b^{2} + 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{4} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\right )} \tan \left (d x + c\right )^{3} +{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )} \tan \left (d x + c\right )^{2} + 2 \,{\left (a^{7} b + 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} + a b^{7}\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))^3,x, algorithm="maxima")

[Out]

1/2*((a^5 - 14*a^3*b^2 + 9*a*b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 2*(3*a^4*b - 8*a
^2*b^3 + b^5)*log(b*tan(d*x + c) + a)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (3*a^4*b - 8*a^2*b^3 +
b^5)*log(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (8*a^4*b - 4*a^2*b^3 + (5*a^3*
b^2 - 7*a*b^4)*tan(d*x + c)^3 + (7*a^4*b - 6*a^2*b^3 - b^5)*tan(d*x + c)^2 + (a^5 + 7*a^3*b^2 - 6*a*b^4)*tan(d
*x + c))/(a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6 + (a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*tan(d*x + c)^4 + 2*(
a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c)^3 + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*tan(d*
x + c)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^7)*tan(d*x + c)))/d

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Fricas [B]  time = 2.48319, size = 1152, normalized size = 5.59 \begin{align*} \frac{13 \, a^{4} b^{3} - 8 \, a^{2} b^{5} - b^{7} - 2 \,{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )} \cos \left (d x + c\right )^{4} + 2 \,{\left (a^{5} b^{2} - 14 \, a^{3} b^{4} + 9 \, a b^{6}\right )} d x -{\left (a^{6} b + 23 \, a^{4} b^{3} - 21 \, a^{2} b^{5} - 3 \, b^{7} - 2 \,{\left (a^{7} - 15 \, a^{5} b^{2} + 23 \, a^{3} b^{4} - 9 \, a b^{6}\right )} d x\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{4} b^{3} - 8 \, a^{2} b^{5} + b^{7} +{\left (3 \, a^{6} b - 11 \, a^{4} b^{3} + 9 \, a^{2} b^{5} - b^{7}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, a^{5} b^{2} - 8 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\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 ) - 2 \,{\left ({\left (a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (4 \, a^{5} b^{2} - 3 \, a^{3} b^{4} + 3 \, a b^{6} +{\left (a^{6} b - 14 \, a^{4} b^{3} + 9 \, a^{2} b^{5}\right )} d x\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \,{\left ({\left (a^{10} + 3 \, a^{8} b^{2} + 2 \, a^{6} b^{4} - 2 \, a^{4} b^{6} - 3 \, a^{2} b^{8} - b^{10}\right )} d \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{9} b + 4 \, a^{7} b^{3} + 6 \, a^{5} b^{5} + 4 \, a^{3} b^{7} + a b^{9}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right ) +{\left (a^{8} b^{2} + 4 \, a^{6} b^{4} + 6 \, a^{4} b^{6} + 4 \, a^{2} b^{8} + b^{10}\right )} d\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))^3,x, algorithm="fricas")

[Out]

1/4*(13*a^4*b^3 - 8*a^2*b^5 - b^7 - 2*(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*cos(d*x + c)^4 + 2*(a^5*b^2 - 14*a
^3*b^4 + 9*a*b^6)*d*x - (a^6*b + 23*a^4*b^3 - 21*a^2*b^5 - 3*b^7 - 2*(a^7 - 15*a^5*b^2 + 23*a^3*b^4 - 9*a*b^6)
*d*x)*cos(d*x + c)^2 + 2*(3*a^4*b^3 - 8*a^2*b^5 + b^7 + (3*a^6*b - 11*a^4*b^3 + 9*a^2*b^5 - b^7)*cos(d*x + c)^
2 + 2*(3*a^5*b^2 - 8*a^3*b^4 + a*b^6)*cos(d*x + c)*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) - 2*((a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*cos(d*x + c)^3 - 2*(4*a^5*b^2 - 3*a^3*b^
4 + 3*a*b^6 + (a^6*b - 14*a^4*b^3 + 9*a^2*b^5)*d*x)*cos(d*x + c))*sin(d*x + c))/((a^10 + 3*a^8*b^2 + 2*a^6*b^4
- 2*a^4*b^6 - 3*a^2*b^8 - b^10)*d*cos(d*x + c)^2 + 2*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*d*co
s(d*x + c)*sin(d*x + c) + (a^8*b^2 + 4*a^6*b^4 + 6*a^4*b^6 + 4*a^2*b^8 + b^10)*d)

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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))**3,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.23699, size = 651, normalized size = 3.16 \begin{align*} \frac{\frac{{\left (a^{5} - 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )}{\left (d x + c\right )}}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} - \frac{{\left (3 \, a^{4} b - 8 \, a^{2} b^{3} + b^{5}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}} + \frac{2 \,{\left (3 \, a^{4} b^{2} - 8 \, a^{2} b^{4} + b^{6}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{8} b + 4 \, a^{6} b^{3} + 6 \, a^{4} b^{5} + 4 \, a^{2} b^{7} + b^{9}} + \frac{3 \, a^{4} b \tan \left (d x + c\right )^{2} - 8 \, a^{2} b^{3} \tan \left (d x + c\right )^{2} + b^{5} \tan \left (d x + c\right )^{2} - a^{5} \tan \left (d x + c\right ) + 2 \, a^{3} b^{2} \tan \left (d x + c\right ) + 3 \, a b^{4} \tan \left (d x + c\right ) - 10 \, a^{2} b^{3} + 2 \, b^{5}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )}{\left (\tan \left (d x + c\right )^{2} + 1\right )}} - \frac{9 \, a^{4} b^{3} \tan \left (d x + c\right )^{2} - 24 \, a^{2} b^{5} \tan \left (d x + c\right )^{2} + 3 \, b^{7} \tan \left (d x + c\right )^{2} + 22 \, a^{5} b^{2} \tan \left (d x + c\right ) - 48 \, a^{3} b^{4} \tan \left (d x + c\right ) + 2 \, a b^{6} \tan \left (d x + c\right ) + 14 \, a^{6} b - 22 \, a^{4} b^{3}}{{\left (a^{8} + 4 \, a^{6} b^{2} + 6 \, a^{4} b^{4} + 4 \, a^{2} b^{6} + b^{8}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{2}}}{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))^3,x, algorithm="giac")

[Out]

1/2*((a^5 - 14*a^3*b^2 + 9*a*b^4)*(d*x + c)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) - (3*a^4*b - 8*a^2
*b^3 + b^5)*log(tan(d*x + c)^2 + 1)/(a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) + 2*(3*a^4*b^2 - 8*a^2*b^4
+ b^6)*log(abs(b*tan(d*x + c) + a))/(a^8*b + 4*a^6*b^3 + 6*a^4*b^5 + 4*a^2*b^7 + b^9) + (3*a^4*b*tan(d*x + c)
^2 - 8*a^2*b^3*tan(d*x + c)^2 + b^5*tan(d*x + c)^2 - a^5*tan(d*x + c) + 2*a^3*b^2*tan(d*x + c) + 3*a*b^4*tan(d
*x + c) - 10*a^2*b^3 + 2*b^5)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(tan(d*x + c)^2 + 1)) - (9*a^4*
b^3*tan(d*x + c)^2 - 24*a^2*b^5*tan(d*x + c)^2 + 3*b^7*tan(d*x + c)^2 + 22*a^5*b^2*tan(d*x + c) - 48*a^3*b^4*t
an(d*x + c) + 2*a*b^6*tan(d*x + c) + 14*a^6*b - 22*a^4*b^3)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*(
b*tan(d*x + c) + a)^2))/d