Integrand size = 21, antiderivative size = 76 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {1}{2} \left (a^2-3 b^2\right ) x-\frac {2 a b \log (\cos (c+d x))}{d}+\frac {3 b^2 \tan (c+d x)}{2 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^2}{2 d} \] Output:
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 = 1.69 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.13 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {b \left (\frac {\left (-a^2+b^2\right ) \arctan (\tan (c+d x))}{b}+2 a \cos ^2(c+d x)+\left (2 a+\frac {a^2-2 b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\left (2 a+\frac {-a^2+2 b^2}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )+\frac {\left (-a^2+b^2\right ) \sin (2 (c+d x))}{2 b}+2 b \tan (c+d x)\right )}{2 d} \] Input:
Integrate[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]
Output:
(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 |
\(\displaystyle \frac {b \left (\frac {1}{2} \left (\frac {\left (a^2-3 b^2\right ) \arctan (\tan (c+d x))}{b}+2 a \log \left (b^2 \tan ^2(c+d x)+b^2\right )+3 b \tan (c+d x)\right )-\frac {b \tan (c+d x) (a+b \tan (c+d x))^2}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
Input:
Int[Sin[c + d*x]^2*(a + b*Tan[c + d*x])^2,x]
Output:
(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
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 = 1.79 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.43
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(109\) |
default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a b \left (-\frac {\sin \left (d x +c \right )^{2}}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+b^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{\cos \left (d x +c \right )}+\left (\sin \left (d x +c \right )^{3}+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d}\) | \(109\) |
risch | \(2 i x a b +\frac {a^{2} x}{2}-\frac {3 b^{2} x}{2}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} a b}{4 d}+\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{2}}{8 d}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a b}{4 d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2}}{8 d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{2}}{8 d}+\frac {4 i a b c}{d}+\frac {2 i b^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 a b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(174\) |
Input:
int(sin(d*x+c)^2*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/2*cos(d*x+c)*sin(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 = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.33 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2 \, a b \cos \left (d x + c\right )^{3} - 4 \, a b \cos \left (d x + c\right ) \log \left (-\cos \left (d x + c\right )\right ) + {\left ({\left (a^{2} - 3 \, b^{2}\right )} d x - a b\right )} \cos \left (d x + c\right ) - {\left ({\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, b^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \] Input:
integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
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))
\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{2} \sin ^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(sin(d*x+c)**2*(a+b*tan(d*x+c))**2,x)
Output:
Integral((a + b*tan(c + d*x))**2*sin(c + d*x)**2, x)
Time = 0.12 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.08 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {2 \, a b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, b^{2} \tan \left (d x + c\right ) + {\left (a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )} + \frac {2 \, a b - {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1}}{2 \, d} \] Input:
integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
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
Time = 0.18 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.17 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{d} + \frac {b^{2} \tan \left (d x + c\right )}{d} + \frac {{\left (a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )}}{2 \, d} + \frac {2 \, a b - {\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )}{2 \, {\left (\tan \left (d x + c\right )^{2} + 1\right )} d} \] Input:
integrate(sin(d*x+c)^2*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
a*b*log(tan(d*x + c)^2 + 1)/d + b^2*tan(d*x + c)/d + 1/2*(a^2 - 3*b^2)*(d* x + c)/d + 1/2*(2*a*b - (a^2 - b^2)*tan(d*x + c))/((tan(d*x + c)^2 + 1)*d)
Time = 0.82 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {{\cos \left (c+d\,x\right )}^2\,\left (a\,b-\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )\right )+b^2\,\mathrm {tan}\left (c+d\,x\right )+a\,b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )+d\,x\,\left (\frac {a^2}{2}-\frac {3\,b^2}{2}\right )}{d} \] Input:
int(sin(c + d*x)^2*(a + b*tan(c + d*x))^2,x)
Output:
(cos(c + d*x)^2*(a*b - tan(c + d*x)*(a^2/2 - b^2/2)) + b^2*tan(c + d*x) + a*b*log(tan(c + d*x)^2 + 1) + d*x*(a^2/2 - (3*b^2)/2))/d
Time = 0.21 (sec) , antiderivative size = 195, normalized size of antiderivative = 2.57 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) a b -4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a b -4 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a b -2 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a b +\cos \left (d x +c \right ) a^{2} c +\cos \left (d x +c \right ) a^{2} d x -3 \cos \left (d x +c \right ) b^{2} c -3 \cos \left (d x +c \right ) b^{2} d x +\sin \left (d x +c \right )^{3} a^{2}-\sin \left (d x +c \right )^{3} b^{2}-\sin \left (d x +c \right ) a^{2}+3 \sin \left (d x +c \right ) b^{2}}{2 \cos \left (d x +c \right ) d} \] Input:
int(sin(d*x+c)^2*(a+b*tan(d*x+c))^2,x)
Output:
(4*cos(c + d*x)*log(tan((c + d*x)/2)**2 + 1)*a*b - 4*cos(c + d*x)*log(tan( (c + d*x)/2) - 1)*a*b - 4*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b - 2*c os(c + d*x)*sin(c + d*x)**2*a*b + cos(c + d*x)*a**2*c + cos(c + d*x)*a**2* d*x - 3*cos(c + d*x)*b**2*c - 3*cos(c + d*x)*b**2*d*x + sin(c + d*x)**3*a* *2 - sin(c + d*x)**3*b**2 - sin(c + d*x)*a**2 + 3*sin(c + d*x)*b**2)/(2*co s(c + d*x)*d)