Integrand size = 21, antiderivative size = 139 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {1}{2} \left (a^4-18 a^2 b^2+5 b^4\right ) x-\frac {4 a b \left (a^2-2 b^2\right ) \log (\cos (c+d x))}{d}+\frac {b^2 \left (18 a^2-5 b^2\right ) \tan (c+d x)}{2 d}+\frac {4 a b^3 \tan ^2(c+d x)}{d}+\frac {5 b^4 \tan ^3(c+d x)}{6 d}-\frac {\cos (c+d x) \sin (c+d x) (a+b \tan (c+d x))^4}{2 d} \]
1/2*(a^4-18*a^2*b^2+5*b^4)*x-4*a*b*(a^2-2*b^2)*ln(cos(d*x+c))/d+1/2*b^2*(1 8*a^2-5*b^2)*tan(d*x+c)/d+4*a*b^3*tan(d*x+c)^2/d+5/6*b^4*tan(d*x+c)^3/d-1/ 2*cos(d*x+c)*sin(d*x+c)*(a+b*tan(d*x+c))^4/d
Time = 6.31 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.89 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {b \left (-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \arctan (\tan (c+d x))}{2 b}+2 a (a-b) (a+b) \cos ^2(c+d x)+\frac {1}{2} \left (4 a^3-8 a b^2+\frac {a^4-12 a^2 b^2+3 b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}-b \tan (c+d x)\right )+\frac {1}{2} \left (4 a^3-8 a b^2-\frac {a^4-12 a^2 b^2+3 b^4}{\sqrt {-b^2}}\right ) \log \left (\sqrt {-b^2}+b \tan (c+d x)\right )-\frac {\left (a^4-6 a^2 b^2+b^4\right ) \cos (c+d x) \sin (c+d x)}{2 b}+2 b \left (3 a^2-b^2\right ) \tan (c+d x)+2 a b^2 \tan ^2(c+d x)+\frac {1}{3} b^3 \tan ^3(c+d x)\right )}{d} \]
(b*(-1/2*((a^4 - 6*a^2*b^2 + b^4)*ArcTan[Tan[c + d*x]])/b + 2*a*(a - b)*(a + b)*Cos[c + d*x]^2 + ((4*a^3 - 8*a*b^2 + (a^4 - 12*a^2*b^2 + 3*b^4)/Sqrt [-b^2])*Log[Sqrt[-b^2] - b*Tan[c + d*x]])/2 + ((4*a^3 - 8*a*b^2 - (a^4 - 1 2*a^2*b^2 + 3*b^4)/Sqrt[-b^2])*Log[Sqrt[-b^2] + b*Tan[c + d*x]])/2 - ((a^4 - 6*a^2*b^2 + b^4)*Cos[c + d*x]*Sin[c + d*x])/(2*b) + 2*b*(3*a^2 - b^2)*T an[c + d*x] + 2*a*b^2*Tan[c + d*x]^2 + (b^3*Tan[c + d*x]^3)/3))/d
Time = 0.35 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.13, 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.
| \(\displaystyle \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^2 (a+b \tan (c+d x))^4dx\) |
| \(\Big \downarrow \) 3999 |
| \(\displaystyle \frac {b \int \frac {b^2 \tan ^2(c+d x) (a+b \tan (c+d x))^4}{\left (\tan ^2(c+d x) b^2+b^2\right )^2}d(b \tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 531 |
| \(\displaystyle \frac {b \left (-\frac {\int -\frac {b^2 (a+b \tan (c+d x))^3 (a+5 b \tan (c+d x))}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 b^2}-\frac {b \tan (c+d x) (a+b \tan (c+d x))^4}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b \left (\frac {\int \frac {b^2 (a+b \tan (c+d x))^3 (a+5 b \tan (c+d x))}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{2 b^2}-\frac {b \tan (c+d x) (a+b \tan (c+d x))^4}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \int \frac {(a+b \tan (c+d x))^3 (a+5 b \tan (c+d x))}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))-\frac {b \tan (c+d x) (a+b \tan (c+d x))^4}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 657 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \int \left (18 a^2+16 b \tan (c+d x) a-5 b^2+5 b^2 \tan ^2(c+d x)+\frac {a^4-18 b^2 a^2+8 b \left (a^2-2 b^2\right ) \tan (c+d x) a+5 b^4}{\tan ^2(c+d x) b^2+b^2}\right )d(b \tan (c+d x))-\frac {b \tan (c+d x) (a+b \tan (c+d x))^4}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {b \left (\frac {1}{2} \left (b \left (18 a^2-5 b^2\right ) \tan (c+d x)+4 a \left (a^2-2 b^2\right ) \log \left (b^2 \tan ^2(c+d x)+b^2\right )+\frac {\left (a^4-18 a^2 b^2+5 b^4\right ) \arctan (\tan (c+d x))}{b}+8 a b^2 \tan ^2(c+d x)+\frac {5}{3} b^3 \tan ^3(c+d x)\right )-\frac {b \tan (c+d x) (a+b \tan (c+d x))^4}{2 \left (b^2 \tan ^2(c+d x)+b^2\right )}\right )}{d}\) |
(b*(-1/2*(b*Tan[c + d*x]*(a + b*Tan[c + d*x])^4)/(b^2 + b^2*Tan[c + d*x]^2 ) + (((a^4 - 18*a^2*b^2 + 5*b^4)*ArcTan[Tan[c + d*x]])/b + 4*a*(a^2 - 2*b^ 2)*Log[b^2 + b^2*Tan[c + d*x]^2] + b*(18*a^2 - 5*b^2)*Tan[c + d*x] + 8*a*b ^2*Tan[c + d*x]^2 + (5*b^3*Tan[c + d*x]^3)/3)/2))/d
3.1.42.3.1 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 = 3.53 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.80
| method | result | size |
| derivativedivides | \(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+4 a \,b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(250\) |
| default | \(\frac {a^{4} \left (-\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{2}-\ln \left (\cos \left (d x +c \right )\right )\right )+6 a^{2} b^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+4 a \,b^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{2 \cos \left (d x +c \right )^{2}}+\frac {\left (\sin ^{4}\left (d x +c \right )\right )}{2}+\sin ^{2}\left (d x +c \right )+2 \ln \left (\cos \left (d x +c \right )\right )\right )+b^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )}{d}\) | \(250\) |
| risch | \(\frac {i {\mathrm e}^{2 i \left (d x +c \right )} b^{4}}{8 d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} b^{4}}{8 d}+\frac {x \,a^{4}}{2}-9 x \,a^{2} b^{2}+\frac {5 x \,b^{4}}{2}+\frac {{\mathrm e}^{2 i \left (d x +c \right )} a^{3} b}{2 d}-\frac {{\mathrm e}^{2 i \left (d x +c \right )} a \,b^{3}}{2 d}-8 i x a \,b^{3}+4 i x \,a^{3} b +\frac {i {\mathrm e}^{2 i \left (d x +c \right )} a^{4}}{8 d}+\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a^{3} b}{2 d}-\frac {{\mathrm e}^{-2 i \left (d x +c \right )} a \,b^{3}}{2 d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )} a^{2} b^{2}}{4 d}-\frac {16 i b^{3} a c}{d}+\frac {8 i b \,a^{3} c}{d}-\frac {i {\mathrm e}^{-2 i \left (d x +c \right )} a^{4}}{8 d}+\frac {3 i {\mathrm e}^{-2 i \left (d x +c \right )} a^{2} b^{2}}{4 d}-\frac {2 i b^{2} \left (-18 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+12 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-36 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}+12 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-18 a^{2}+7 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {4 b \,a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {8 b^{3} a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(406\) |
1/d*(a^4*(-1/2*sin(d*x+c)*cos(d*x+c)+1/2*d*x+1/2*c)+4*a^3*b*(-1/2*sin(d*x+ c)^2-ln(cos(d*x+c)))+6*a^2*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)+4*a*b^3*(1/2*sin(d*x+c)^6/cos(d*x+c) ^2+1/2*sin(d*x+c)^4+sin(d*x+c)^2+2*ln(cos(d*x+c)))+b^4*(1/3*sin(d*x+c)^7/c os(d*x+c)^3-4/3*sin(d*x+c)^7/cos(d*x+c)-4/3*(sin(d*x+c)^5+5/4*sin(d*x+c)^3 +15/8*sin(d*x+c))*cos(d*x+c)+5/2*d*x+5/2*c))
Time = 0.27 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.34 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {12 \, {\left (a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{5} + 12 \, a b^{3} \cos \left (d x + c\right ) - 24 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\cos \left (d x + c\right )\right ) - 3 \, {\left (2 \, a^{3} b - 2 \, a b^{3} - {\left (a^{4} - 18 \, a^{2} b^{2} + 5 \, b^{4}\right )} d x\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} - 2 \, b^{4} - 2 \, {\left (18 \, a^{2} b^{2} - 7 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]
1/6*(12*(a^3*b - a*b^3)*cos(d*x + c)^5 + 12*a*b^3*cos(d*x + c) - 24*(a^3*b - 2*a*b^3)*cos(d*x + c)^3*log(-cos(d*x + c)) - 3*(2*a^3*b - 2*a*b^3 - (a^ 4 - 18*a^2*b^2 + 5*b^4)*d*x)*cos(d*x + c)^3 - (3*(a^4 - 6*a^2*b^2 + b^4)*c os(d*x + c)^4 - 2*b^4 - 2*(18*a^2*b^2 - 7*b^4)*cos(d*x + c)^2)*sin(d*x + c ))/(d*cos(d*x + c)^3)
\[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{4} \sin ^{2}{\left (c + d x \right )}\, dx \]
Time = 0.70 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.11 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\frac {2 \, b^{4} \tan \left (d x + c\right )^{3} + 12 \, a b^{3} \tan \left (d x + c\right )^{2} + 3 \, {\left (a^{4} - 18 \, a^{2} b^{2} + 5 \, b^{4}\right )} {\left (d x + c\right )} + 12 \, {\left (a^{3} b - 2 \, a b^{3}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 12 \, {\left (3 \, a^{2} b^{2} - b^{4}\right )} \tan \left (d x + c\right ) + \frac {3 \, {\left (4 \, a^{3} b - 4 \, a b^{3} - {\left (a^{4} - 6 \, a^{2} b^{2} + b^{4}\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{2} + 1}}{6 \, d} \]
1/6*(2*b^4*tan(d*x + c)^3 + 12*a*b^3*tan(d*x + c)^2 + 3*(a^4 - 18*a^2*b^2 + 5*b^4)*(d*x + c) + 12*(a^3*b - 2*a*b^3)*log(tan(d*x + c)^2 + 1) + 12*(3* a^2*b^2 - b^4)*tan(d*x + c) + 3*(4*a^3*b - 4*a*b^3 - (a^4 - 6*a^2*b^2 + b^ 4)*tan(d*x + c))/(tan(d*x + c)^2 + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 3651 vs. \(2 (131) = 262\).
Time = 2.57 (sec) , antiderivative size = 3651, normalized size of antiderivative = 26.27 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=\text {Too large to display} \]
1/6*(3*a^4*d*x*tan(d*x)^5*tan(c)^5 - 54*a^2*b^2*d*x*tan(d*x)^5*tan(c)^5 + 15*b^4*d*x*tan(d*x)^5*tan(c)^5 - 12*a^3*b*log(4*(tan(d*x)^2*tan(c)^2 - 2*t an(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan (d*x)^5*tan(c)^5 + 24*a*b^3*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)^5*tan(c) ^5 + 3*a^4*d*x*tan(d*x)^5*tan(c)^3 - 54*a^2*b^2*d*x*tan(d*x)^5*tan(c)^3 + 15*b^4*d*x*tan(d*x)^5*tan(c)^3 - 9*a^4*d*x*tan(d*x)^4*tan(c)^4 + 162*a^2*b ^2*d*x*tan(d*x)^4*tan(c)^4 - 45*b^4*d*x*tan(d*x)^4*tan(c)^4 + 3*a^4*d*x*ta n(d*x)^3*tan(c)^5 - 54*a^2*b^2*d*x*tan(d*x)^3*tan(c)^5 + 15*b^4*d*x*tan(d* x)^3*tan(c)^5 + 6*a^3*b*tan(d*x)^5*tan(c)^5 + 6*a*b^3*tan(d*x)^5*tan(c)^5 - 12*a^3*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)^5*tan(c)^3 + 24*a*b^3*log (4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + ta n(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^5*tan(c)^3 + 36*a^3*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)^4*tan(c)^4 - 72*a*b^3*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))*ta n(d*x)^4*tan(c)^4 + 3*a^4*tan(d*x)^5*tan(c)^4 - 54*a^2*b^2*tan(d*x)^5*tan( c)^4 + 15*b^4*tan(d*x)^5*tan(c)^4 - 12*a^3*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...
Time = 4.98 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.16 \[ \int \sin ^2(c+d x) (a+b \tan (c+d x))^4 \, dx=x\,\left (\frac {a^4}{2}-9\,a^2\,b^2+\frac {5\,b^4}{2}\right )-\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (4\,a\,b^3-2\,a^3\,b\right )}{d}-\frac {{\cos \left (c+d\,x\right )}^2\,\left (\mathrm {tan}\left (c+d\,x\right )\,\left (\frac {a^4}{2}-3\,a^2\,b^2+\frac {b^4}{2}\right )+2\,a\,b^3-2\,a^3\,b\right )}{d}+\frac {b^4\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}-\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,b^4-6\,a^2\,b^2\right )}{d}+\frac {2\,a\,b^3\,{\mathrm {tan}\left (c+d\,x\right )}^2}{d} \]