Integrand size = 36, antiderivative size = 237 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {(a-b) B \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a-b) B \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 a^{3/2} B \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} \left (a^2+b^2\right ) d}+\frac {(a+b) B \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a+b) B \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]
-1/2*(a-b)*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)-1/2*( a-b)*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1/2)+1/4*(a+b)*B* ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)/d*2^(1/2)-1/4*(a+b)*B* ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(a^2+b^2)/d*2^(1/2)+2*a^(3/2)*B* arctan(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/(a^2+b^2)/d/b^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.18 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {B \left (3 a \left (2 \sqrt {2} \sqrt {b} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \sqrt {2} \sqrt {b} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+8 \sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )+\sqrt {2} \sqrt {b} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\sqrt {2} \sqrt {b} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )+8 b^{3/2} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},1,\frac {7}{4},-\tan ^2(c+d x)\right ) \tan ^{\frac {3}{2}}(c+d x)\right )}{12 \sqrt {b} \left (a^2+b^2\right ) d} \]
(B*(3*a*(2*Sqrt[2]*Sqrt[b]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*Sqrt [2]*Sqrt[b]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + 8*Sqrt[a]*ArcTan[(Sqr t[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]] + Sqrt[2]*Sqrt[b]*Log[1 - Sqrt[2]*Sqrt[T an[c + d*x]] + Tan[c + d*x]] - Sqrt[2]*Sqrt[b]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]) + 8*b^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Tan[c + d*x]^2]*Tan[c + d*x]^(3/2)))/(12*Sqrt[b]*(a^2 + b^2)*d)
Time = 0.76 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.86, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.472, Rules used = {2011, 3042, 4056, 25, 3042, 4017, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}
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.
\(\displaystyle \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle B \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \frac {\tan (c+d x)^{3/2}}{a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4056 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}+\frac {\int -\frac {a-b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {\int \frac {a-b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {\int \frac {a-b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 4017 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \int \frac {a-b \tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 1482 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \int \frac {\tan (c+d x)+1}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 1476 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {1}{2} \int \frac {1}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} \int \frac {1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 1082 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\tan (c+d x)-1}d\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 217 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a+b) \int \frac {1-\tan (c+d x)}{\tan ^2(c+d x)+1}d\sqrt {\tan (c+d x)}+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 1479 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a+b) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a+b) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a+b) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\tan (c+d x)}}{\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\tan (c+d x)}+1}{\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1}d\sqrt {\tan (c+d x)}\right )+\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 1103 |
\(\displaystyle B \left (\frac {a^2 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}-\frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 4117 |
\(\displaystyle B \left (\frac {a^2 \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}-\frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 73 |
\(\displaystyle B \left (\frac {2 a^2 \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}-\frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
\(\Big \downarrow \) 218 |
\(\displaystyle B \left (\frac {2 a^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}-\frac {2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\right )\) |
B*((2*a^(3/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[b]*(a^2 + b^2)*d) - (2*(((a - b)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2] ) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((a + b)*(-1/2*Lo g[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2] *Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d))
3.5.23.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 2*(d/e), 2]}, Simp[e/(2*c) Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ e/(2*c) Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ -2*(d/e), 2]}, Simp[e/(2*c*q) Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Simp[e/(2*c*q) Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ a*c, 2]}, Simp[(d*q + a*e)/(2*a*c) Int[(q + c*x^2)/(a + c*x^4), x], x] + Simp[(d*q - a*e)/(2*a*c) Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a , c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- a)*c]
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ )]], x_Symbol] :> Simp[2/f Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & & NeQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2) Int[Simp[a^2*c - b^2*c + 2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e + f*x ]], x], x] + Simp[(b*c - a*d)^2/(c^2 + d^2) Int[(1 + Tan[e + f*x]^2)/(Sqr t[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e , f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[A/f Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Time = 0.05 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.95
method | result | size |
derivativedivides | \(\frac {B \left (\frac {2 a^{2} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}\right )}{d}\) | \(226\) |
default | \(\frac {B \left (\frac {2 a^{2} \arctan \left (\frac {b \left (\sqrt {\tan }\left (d x +c \right )\right )}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}+\frac {-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {b \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{a^{2}+b^{2}}\right )}{d}\) | \(226\) |
1/d*B*(2*a^2/(a^2+b^2)/(a*b)^(1/2)*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))+ 2/(a^2+b^2)*(-1/8*a*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1 -2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2) )+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/8*b*2^(1/2)*(ln((1-2^(1/2)*tan( d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan (1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))))
Leaf count of result is larger than twice the leaf count of optimal. 1565 vs. \(2 (199) = 398\).
Time = 0.32 (sec) , antiderivative size = 3156, normalized size of antiderivative = 13.32 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]
[1/2*((a^2 + b^2)*d*sqrt((2*B^2*a*b + (a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(B ^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^ 6 + b^8)*d^4)))/((a^4 + 2*a^2*b^2 + b^4)*d^2))*log(((a^4*b + 2*a^2*b^3 + b ^5)*d^3*sqrt(-(B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^ 4*b^4 + 4*a^2*b^6 + b^8)*d^4)) + (B^2*a^3 - B^2*a*b^2)*d)*sqrt((2*B^2*a*b + (a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/(( a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (B^3*a^2 - B^3*b^2)*sqrt(tan(d*x + c))) - (a^2 + b^2)*d*sqrt( (2*B^2*a*b + (a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt(-(B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/((a^4 + 2 *a^2*b^2 + b^4)*d^2))*log(-((a^4*b + 2*a^2*b^3 + b^5)*d^3*sqrt(-(B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8) *d^4)) + (B^2*a^3 - B^2*a*b^2)*d)*sqrt((2*B^2*a*b + (a^4 + 2*a^2*b^2 + b^4 )*d^2*sqrt(-(B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b^2 + 6*a^4* b^4 + 4*a^2*b^6 + b^8)*d^4)))/((a^4 + 2*a^2*b^2 + b^4)*d^2)) - (B^3*a^2 - B^3*b^2)*sqrt(tan(d*x + c))) - (a^2 + b^2)*d*sqrt((2*B^2*a*b - (a^4 + 2*a^ 2*b^2 + b^4)*d^2*sqrt(-(B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4)/((a^8 + 4*a^6*b ^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)))/((a^4 + 2*a^2*b^2 + b^4)*d^2))*lo g(((a^4*b + 2*a^2*b^3 + b^5)*d^3*sqrt(-(B^4*a^4 - 2*B^4*a^2*b^2 + B^4*b^4) /((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d^4)) - (B^2*a^3 - B^...
\[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=B \int \frac {\tan ^{\frac {3}{2}}{\left (c + d x \right )}}{a + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {8 \, B a^{2} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} - \frac {{\left (2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a + b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a + b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} B}{a^{2} + b^{2}}}{4 \, d} \]
1/4*(8*B*a^2*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^2 + b^2)*sqrt(a*b) ) - (2*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c))) ) + 2*sqrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c))) ) + sqrt(2)*(a + b)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - s qrt(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))*B/(a^2 + b^2))/d
Timed out. \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Timed out} \]
Time = 39.45 (sec) , antiderivative size = 16878, normalized size of antiderivative = 71.22 \[ \int \frac {\tan ^{\frac {3}{2}}(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\text {Too large to display} \]
(log(((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b^2)^2*((4*(-B^ 4*b^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a *b^5*d^2)/(d^4*(a^2 + b^2)^4))^(1/2) + (768*B*a^2*b^4*(a^2 + b^2))/d)*((4* (-B^4*b^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B ^2*a*b^5*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (64*B^2*a*b^2*tan(c + d*x)^( 1/2)*(2*a^8 + 15*b^8 - 17*a^2*b^6 + 51*a^4*b^4 + 21*a^6*b^2))/(d^2*(a^2 + b^2)^2))*((4*(-B^4*b^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b ^3*d^2 + 16*B^2*a*b^5*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (32*B^3*a*b^3*( 4*a^8 + b^8 - 77*a^2*b^6 + 47*a^4*b^4 + 33*a^6*b^2))/(d^3*(a^2 + b^2)^3))* ((4*(-B^4*b^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a*b^5*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (16*B^4*b^3*tan(c + d*x) ^(1/2)*(a^10 - 2*b^10 - 4*a^2*b^8 - 27*a^4*b^6 + 15*a^6*b^4 + 9*a^8*b^2))/ (d^4*(a^2 + b^2)^4))*((4*(-B^4*b^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a*b^5*d^2)/(d^4*(a^2 + b^2)^4))^(1/2))/4 + (8* B^5*a^2*b^4*(a^6 + 10*b^6 + 27*a^2*b^4 + 10*a^4*b^2))/(d^5*(a^2 + b^2)^4)) *(((192*B^4*a^2*b^10*d^4 - 16*B^4*b^12*d^4 - 608*B^4*a^4*b^8*d^4 + 192*B^4 *a^6*b^6*d^4 - 16*B^4*a^8*b^4*d^4)^(1/2) - 16*B^2*a^3*b^3*d^2 + 16*B^2*a*b ^5*d^2)/(a^8*d^4 + b^8*d^4 + 4*a^2*b^6*d^4 + 6*a^4*b^4*d^4 + 4*a^6*b^2*d^4 ))^(1/2))/4 + (log(((((((((128*b^3*tan(c + d*x)^(1/2)*(a^2 - b^2)*(a^2 + b ^2)^2*(-(4*(-B^4*b^4*d^4*(a^4 + b^4 - 6*a^2*b^2)^2)^(1/2) + 16*B^2*a^3*...