Integrand size = 21, antiderivative size = 249 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {5 \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{3/2}}-\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}+\frac {5 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{3/2}}-\frac {5}{2 b d \sqrt {d \tan (a+b x)}}+\frac {\cos ^2(a+b x)}{2 b d \sqrt {d \tan (a+b x)}} \]
5/8*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b/d^(3/2)*2^(1/2)-5/8*a rctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b/d^(3/2)*2^(1/2)-5/16*ln(d^ (1/2)-2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b/d^(3/2)*2^(1/2)+5 /16*ln(d^(1/2)+2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b/d^(3/2)* 2^(1/2)-5/2/b/d/(d*tan(b*x+a))^(1/2)+1/2*cos(b*x+a)^2/b/d/(d*tan(b*x+a))^( 1/2)
Time = 0.49 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.46 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\frac {\csc (a+b x) \left (-17 \cos (a+b x)+\cos (3 (a+b x))+5 \arcsin (\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}+5 \log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}\right ) \sqrt {d \tan (a+b x)}}{8 b d^2} \]
(Csc[a + b*x]*(-17*Cos[a + b*x] + Cos[3*(a + b*x)] + 5*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Sqrt[Sin[2*(a + b*x)]] + 5*Log[Cos[a + b*x] + Sin[a + b*x ] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]])*Sqrt[d*Tan[a + b*x]])/ (8*b*d^2)
Time = 0.47 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.99, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 3087, 253, 264, 266, 27, 826, 1476, 1082, 217, 1479, 25, 27, 1103}
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.
| \(\displaystyle \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sec (a+b x)^2 (d \tan (a+b x))^{3/2}}dx\) |
| \(\Big \downarrow \) 3087 |
| \(\displaystyle \frac {\int \frac {1}{(d \tan (a+b x))^{3/2} \left (\tan ^2(a+b x)+1\right )^2}d\tan (a+b x)}{b}\) |
| \(\Big \downarrow \) 253 |
| \(\displaystyle \frac {\frac {5}{4} \int \frac {1}{(d \tan (a+b x))^{3/2} \left (\tan ^2(a+b x)+1\right )}d\tan (a+b x)+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {\int \frac {\sqrt {d \tan (a+b x)}}{\tan ^2(a+b x)+1}d\tan (a+b x)}{d^2}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \int \frac {d^3 \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}}{d^3}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \int \frac {d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 826 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \int \frac {\tan (a+b x) d+d}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}+\frac {1}{2} \int \frac {1}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}\right )-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-d \tan (a+b x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (a+b x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\frac {1}{2} \int \frac {d-d \tan (a+b x)}{\tan ^2(a+b x) d^2+d^2}d\sqrt {d \tan (a+b x)}\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d-\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (a+b x)}}{\tan (a+b x) d+d+\sqrt {2} \sqrt {d \tan (a+b x)} \sqrt {d}}d\sqrt {d \tan (a+b x)}}{2 \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {\frac {5}{4} \left (-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )+\frac {1}{2} \left (\frac {\log \left (d \tan (a+b x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (a+b x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (a+b x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (a+b x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{d}-\frac {2}{d \sqrt {d \tan (a+b x)}}\right )+\frac {1}{2 d \left (\tan ^2(a+b x)+1\right ) \sqrt {d \tan (a+b x)}}}{b}\) |
(1/(2*d*Sqrt[d*Tan[a + b*x]]*(1 + Tan[a + b*x]^2)) + (5*((-2*((-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]]/(Sqrt[2]*Sqrt[d]))/2 + (Log[d + d* Tan[a + b*x] - Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[a + b*x]]]/(2*Sqrt[2]*Sqrt[d]) - Log[d + d*Tan[a + b*x] + Sqrt[2]*Sqrt[d]*Sqrt[d*Tan[a + b*x]]]/(2*Sqrt[2] *Sqrt[d]))/2))/d - 2/(d*Sqrt[d*Tan[a + b*x]])))/4)/b
3.3.62.3.1 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_)^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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-(c*x )^(m + 1))*((a + b*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Simp[(m + 2*p + 3)/( 2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, m }, x] && LtQ[p, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s) Int[(r + s*x^2)/(a + b*x^ 4), x], x] - Simp[1/(2*s) Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, 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[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Leaf count of result is larger than twice the leaf count of optimal. \(936\) vs. \(2(189)=378\).
Time = 19.77 (sec) , antiderivative size = 937, normalized size of antiderivative = 3.76
1/16/b*csc(b*x+a)*(4*cos(b*x+a)^2*sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+ a)/(cos(b*x+a)+1)^2)^(1/2)-20*sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/( cos(b*x+a)+1)^2)^(1/2)+10*cos(b*x+a)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+ a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1)/(-1+cos(b*x+a)))+10*co s(b*x+a)*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1) ^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))-5*cos(b*x+a)*ln(-(cot(b*x+a)*cos( b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)- 3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-2*cos( b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(-1+cos(b*x+a)))+5*cos(b*x+a)*ln((2*sin(b* x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+cs c(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-cot(b*x+a)*cos(b*x+a)+2*cot(b*x+a) +2*cos(b*x+a)+sin(b*x+a)-csc(b*x+a)-2)/(-1+cos(b*x+a)))-10*arctan((sin(b*x +a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(cos(b*x+a)+1)^2)^(1/2)-cos(b*x+a)+1)/ (-1+cos(b*x+a)))-10*arctan((sin(b*x+a)*2^(1/2)*(-cos(b*x+a)*sin(b*x+a)/(co s(b*x+a)+1)^2)^(1/2)+cos(b*x+a)-1)/(-1+cos(b*x+a)))+5*ln(-(cot(b*x+a)*cos( b*x+a)-2*cot(b*x+a)+2*sin(b*x+a)*(-cot(b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)- 3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+cot(b*x+a)-csc(b*x+a))^(1/2)-2*cos( b*x+a)-sin(b*x+a)+csc(b*x+a)+2)/(-1+cos(b*x+a)))-5*ln((2*sin(b*x+a)*(-cot( b*x+a)^3+3*cot(b*x+a)^2*csc(b*x+a)-3*cot(b*x+a)*csc(b*x+a)^2+csc(b*x+a)^3+ cot(b*x+a)-csc(b*x+a))^(1/2)-cot(b*x+a)*cos(b*x+a)+2*cot(b*x+a)+2*cos(b...
Result contains complex when optimal does not.
Time = 0.43 (sec) , antiderivative size = 1034, normalized size of antiderivative = 4.15 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\text {Too large to display} \]
1/32*(5*b*d^2*(-1/(b^4*d^6))^(1/4)*log(-1/2*cos(b*x + a)*sin(b*x + a) + 1/ 2*(b^3*d^4*(-1/(b^4*d^6))^(3/4)*cos(b*x + a)^2 - b*d*(-1/(b^4*d^6))^(1/4)* cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1/4*(2*b^2* d^3*cos(b*x + a)^2 - b^2*d^3)*sqrt(-1/(b^4*d^6)))*sin(b*x + a) - 5*b*d^2*( -1/(b^4*d^6))^(1/4)*log(-1/2*cos(b*x + a)*sin(b*x + a) - 1/2*(b^3*d^4*(-1/ (b^4*d^6))^(3/4)*cos(b*x + a)^2 - b*d*(-1/(b^4*d^6))^(1/4)*cos(b*x + a)*si n(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1/4*(2*b^2*d^3*cos(b*x + a )^2 - b^2*d^3)*sqrt(-1/(b^4*d^6)))*sin(b*x + a) - 5*I*b*d^2*(-1/(b^4*d^6)) ^(1/4)*log(-1/2*cos(b*x + a)*sin(b*x + a) + 1/2*(I*b^3*d^4*(-1/(b^4*d^6))^ (3/4)*cos(b*x + a)^2 + I*b*d*(-1/(b^4*d^6))^(1/4)*cos(b*x + a)*sin(b*x + a ))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 1/4*(2*b^2*d^3*cos(b*x + a)^2 - b^2 *d^3)*sqrt(-1/(b^4*d^6)))*sin(b*x + a) + 5*I*b*d^2*(-1/(b^4*d^6))^(1/4)*lo g(-1/2*cos(b*x + a)*sin(b*x + a) + 1/2*(-I*b^3*d^4*(-1/(b^4*d^6))^(3/4)*co s(b*x + a)^2 - I*b*d*(-1/(b^4*d^6))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt( d*sin(b*x + a)/cos(b*x + a)) - 1/4*(2*b^2*d^3*cos(b*x + a)^2 - b^2*d^3)*sq rt(-1/(b^4*d^6)))*sin(b*x + a) - 5*b*d^2*(-1/(b^4*d^6))^(1/4)*log(2*(b^3*d ^4*(-1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) - b*d*(-1/(b^4*d^6))^(1/ 4)*cos(b*x + a)^2)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*sin(b*x + a) + 5 *b*d^2*(-1/(b^4*d^6))^(1/4)*log(-2*(b^3*d^4*(-1/(b^4*d^6))^(3/4)*cos(b*x + a)*sin(b*x + a) - b*d*(-1/(b^4*d^6))^(1/4)*cos(b*x + a)^2)*sqrt(d*sin(...
\[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {\cos ^{2}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.51 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.82 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {\frac {10 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {10 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {5 \, \sqrt {2} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {5 \, \sqrt {2} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right )}{\sqrt {d}} + \frac {8 \, {\left (5 \, d^{2} \tan \left (b x + a\right )^{2} + 4 \, d^{2}\right )}}{\left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} + \sqrt {d \tan \left (b x + a\right )} d^{2}}}{16 \, b d} \]
-1/16*(10*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) + 10*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d))/sqrt(d) - 5*sqrt(2)*log(d*tan(b*x + a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d)/sqrt(d) + 5*sqrt(2)*log(d*tan(b *x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d)/sqrt(d) + 8*(5*d^2*tan (b*x + a)^2 + 4*d^2)/((d*tan(b*x + a))^(5/2) + sqrt(d*tan(b*x + a))*d^2))/ (b*d)
Time = 0.42 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.01 \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=-\frac {\frac {10 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} + \frac {10 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{b d^{2}} - \frac {5 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}} + \frac {5 \, \sqrt {2} {\left | d \right |}^{\frac {3}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{b d^{2}} + \frac {8 \, {\left (5 \, d^{2} \tan \left (b x + a\right )^{2} + 4 \, d^{2}\right )}}{{\left (\sqrt {d \tan \left (b x + a\right )} d^{2} \tan \left (b x + a\right )^{2} + \sqrt {d \tan \left (b x + a\right )} d^{2}\right )} b}}{16 \, d} \]
-1/16*(10*sqrt(2)*abs(d)^(3/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d)))/(b*d^2) + 10*sqrt(2)*abs(d)^(3/2)*ar ctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs (d)))/(b*d^2) - 5*sqrt(2)*abs(d)^(3/2)*log(d*tan(b*x + a) + sqrt(2)*sqrt(d *tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d^2) + 5*sqrt(2)*abs(d)^(3/2)*log (d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d ^2) + 8*(5*d^2*tan(b*x + a)^2 + 4*d^2)/((sqrt(d*tan(b*x + a))*d^2*tan(b*x + a)^2 + sqrt(d*tan(b*x + a))*d^2)*b))/d
Timed out. \[ \int \frac {\cos ^2(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx=\int \frac {{\cos \left (a+b\,x\right )}^2}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}} \,d x \]