Integrand size = 21, antiderivative size = 76 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {1}{2} \left (a^2-3 b^2\right ) x-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {3 b^2 \tan (c+d x)}{2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{2 d} \]
1/2*(a^2-3*b^2)*x-2*a*b*ln(cos(d*x+c))/d+3/2*b^2*tan(d*x+c)/d-1/2*cos(d*x+ c)*sin(d*x+c)*(a+b*tan(d*x+c))^2/d
Leaf count is larger than twice the leaf count of optimal. \(162\) vs. \(2(76)=152\).
Time = 2.70 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.13 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {b \left (\frac {\left (-a^2+b^2\right ) \arctan (\tan (c+d x))}{b}+2 a \cos ^2(c+d x)+\left (2 a+\frac {a^2-2 b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\left (2 a+\frac {-a^2+2 b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {\left (-a^2+b^2\right ) \sin (2 (c+d x))}{2 b}+2 b \tan (c+d x)\right )}{2 d} \]
(b*(((-a^2 + b^2)*ArcTan[Tan[c + d*x]])/b + 2*a*Cos[c + d*x]^2 + (2*a + (a ^2 - 2*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]] + (2*a + (-a^2 + 2*b^2)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]] + ((-a^2 + b^2)*Sin[2* (c + d*x)])/(2*b) + 2*b*Tan[c + d*x]))/(2*d)
Time = 0.30 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3999, 531, 25, 27, 657, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (c+d x)^2 (a+b \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 3999 |
\(\displaystyle \frac {b \int \frac {b^2 \tan ^2(c+d x) (a+b \tan (c+d x))^2}{\left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 531 |
\(\displaystyle \frac {b \left (-\frac {\int -\frac {b^2 (a+b \tan (c+d x)) (a+3 b \tan (c+d x))}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 b^2}-\frac {b \tan (c+d x) (a+b \tan (c+d x))^2}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {b \left (\frac {\int \frac {b^2 (a+b \tan (c+d x)) (a+3 b \tan (c+d x))}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 b^2}-\frac {b \tan (c+d x) (a+b \tan (c+d x))^2}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b \left (\frac {1}{2} \int \frac {(a+b \tan (c+d x)) (a+3 b \tan (c+d x))}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))-\frac {b \tan (c+d x) (a+b \tan (c+d x))^2}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
\(\Big \downarrow \) 657 |
\(\displaystyle \frac {b \left (\frac {1}{2} \int \left (\frac {a^2+4 b \tan (c+d x) a-3 b^2}{\tan ^2(c+d x) b^2+b^2}+3\right )d(b \tan (c+d x))-\frac {b \tan (c+d x) (a+b \tan (c+d x))^2}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b \left (\frac {1}{2} \left (\frac {\left (a^2-3 b^2\right ) \arctan (\tan (c+d x))}{b}+2 a \log \left (b^2 \tan ^2(c+d x)+b^2\right )+3 b \tan (c+d x)\right )-\frac {b \tan (c+d x) (a+b \tan (c+d x))^2}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
(b*((((a^2 - 3*b^2)*ArcTan[Tan[c + d*x]])/b + 2*a*Log[b^2 + b^2*Tan[c + d* x]^2] + 3*b*Tan[c + d*x])/2 - (b*Tan[c + d*x]*(a + b*Tan[c + d*x])^2)/(2*( b^2 + b^2*Tan[c + d*x]^2))))/d
3.1.24.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb ol] :> With[{Qx = PolynomialQuotient[x^m, a + b*x^2, x], e = Coeff[Polynomi alRemainder[x^m, a + b*x^2, x], x, 0], f = Coeff[PolynomialRemainder[x^m, a + b*x^2, x], x, 1]}, Simp[(c + d*x)^n*(a*f - b*e*x)*((a + b*x^2)^(p + 1)/( 2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1)) Int[(c + d*x)^(n - 1)*(a + b *x^2)^(p + 1)*ExpandToSum[2*a*b*(p + 1)*(c + d*x)*Qx - a*d*f*n + b*c*e*(2*p + 3) + b*d*e*(n + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGt Q[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && GtQ[n, 1] && IntegerQ[2*p]
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g*x)^n/(a + c*x^ 2)), x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IntegersQ[n]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[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]
Time = 1.17 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.43
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(109\) |
default | \(\frac {a^{2} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(109\) |
risch | \(2 i a b x +\frac {a^{2} x}{2}-\frac {3 b^{2} x}{2}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} a b}{4 d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a b}{4 d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {4 i a b c}{d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(174\) |
1/d*(a^2*(-1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+2*a*b*(-1/2*sin(d*x+c) ^2-ln(cos(d*x+c)))+b^2*(sin(d*x+c)^5/cos(d*x+c)+(sin(d*x+c)^3+3/2*sin(d*x+ c))*cos(d*x+c)-3/2*d*x-3/2*c))
Time = 0.26 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.33 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2 \, a b \cos \left (d x + c\right )^{3} - 4 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + {\left ({\left (a^{2} - 3 \, b^{2}\right )} d x - a b\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]
1/2*(2*a*b*cos(d*x + c)^3 - 4*a*b*cos(d*x + c)*log(-cos(d*x + c)) + ((a^2 - 3*b^2)*d*x - a*b)*cos(d*x + c) - ((a^2 - b^2)*cos(d*x + c)^2 - 2*b^2)*si n(d*x + c))/(d*cos(d*x + c))
\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sin ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, b^{2} \tan \left (d x + c\right ) + {\left (a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, a b - {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \]
1/2*(2*a*b*log(tan(d*x + c)^2 + 1) + 2*b^2*tan(d*x + c) + (a^2 - 3*b^2)*(d *x + c) + (2*a*b - (a^2 - b^2)*tan(d*x + c))/(tan(d*x + c)^2 + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 981 vs. \(2 (70) = 140\).
Time = 0.62 (sec) , antiderivative size = 981, normalized size of antiderivative = 12.91 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\text {Too large to display} \]
1/2*(a^2*d*x*tan(d*x)^3*tan(c)^3 - 3*b^2*d*x*tan(d*x)^3*tan(c)^3 - 2*a*b*l og(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)^3*tan(c)^3 + a^2*d*x*tan(d*x)^3*tan(c ) - 3*b^2*d*x*tan(d*x)^3*tan(c) - a^2*d*x*tan(d*x)^2*tan(c)^2 + 3*b^2*d*x* tan(d*x)^2*tan(c)^2 + a^2*d*x*tan(d*x)*tan(c)^3 - 3*b^2*d*x*tan(d*x)*tan(c )^3 + a*b*tan(d*x)^3*tan(c)^3 - 2*a*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 )^3*tan(c) + 2*a*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(ta n(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + a^2* tan(d*x)^3*tan(c)^2 - 3*b^2*tan(d*x)^3*tan(c)^2 - 2*a*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)*tan(c)^3 + a^2*tan(d*x)^2*tan(c)^3 - 3*b^2*tan(d*x)^2* tan(c)^3 - a^2*d*x*tan(d*x)^2 + 3*b^2*d*x*tan(d*x)^2 + a^2*d*x*tan(d*x)*ta n(c) - 3*b^2*d*x*tan(d*x)*tan(c) - a*b*tan(d*x)^3*tan(c) - a^2*d*x*tan(c)^ 2 + 3*b^2*d*x*tan(c)^2 - 5*a*b*tan(d*x)^2*tan(c)^2 - a*b*tan(d*x)*tan(c)^3 + 2*a*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*t an(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2 - 2*b^2*tan(d*x)^3 - 2*a* 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)*tan(c) - 2*a^2*tan(d*x)^2*tan(c) + 2*a*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*...
Time = 4.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\cos \left (c+d\,x\right )}^2\,\left (a\,b-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )\right )+b^2\,\mathrm {tan}\left (c+d\,x\right )+a\,b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )+d\,x\,\left (\frac {a^2}{2}-\frac {3\,b^2}{2}\right )}{d} \]