Integrand size = 25, antiderivative size = 229 \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}} \]
I*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/(I*a-b)^ (1/2)-I*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d/( I*a+b)^(1/2)+2/15*(15*a^2-8*b^2)*(a+b*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^( 1/2)-2/5*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2)+8/15*b*(a+b*tan(d*x+c ))^(1/2)/a^2/d/tan(d*x+c)^(3/2)
Time = 2.92 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {-\frac {15 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {15 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2+4 a b \tan (c+d x)+\left (15 a^2-8 b^2\right ) \tan ^2(c+d x)\right )}{a^3 \tan ^{\frac {5}{2}}(c+d x)}}{15 d} \]
((-15*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqr t[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + (15*(-1)^(1/4)*ArcTan[((-1)^(1/4) *Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b ] + (2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2 + 4*a*b*Tan[c + d*x] + (15*a^2 - 8 *b^2)*Tan[c + d*x]^2))/(a^3*Tan[c + d*x]^(5/2)))/(15*d)
Time = 0.97 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4052, 27, 3042, 4132, 27, 3042, 4133, 27, 3042, 4058, 613, 104, 218, 221}
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.
\(\displaystyle \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^{7/2} \sqrt {a+b \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4052 |
\(\displaystyle -\frac {2 \int \frac {4 b \tan ^2(c+d x)+5 a \tan (c+d x)+4 b}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\int \frac {4 b \tan ^2(c+d x)+5 a \tan (c+d x)+4 b}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \frac {4 b \tan (c+d x)^2+5 a \tan (c+d x)+4 b}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4132 |
\(\displaystyle -\frac {-\frac {2 \int -\frac {15 a^2-8 b^2-8 b^2 \tan ^2(c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {\int \frac {15 a^2-8 b^2-8 b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {\int \frac {15 a^2-8 b^2-8 b^2 \tan (c+d x)^2}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4133 |
\(\displaystyle -\frac {\frac {-\frac {2 \int \frac {15 a^3 \sqrt {\tan (c+d x)}}{2 \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {\frac {-15 a^2 \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\frac {-15 a^2 \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4058 |
\(\displaystyle -\frac {\frac {-\frac {15 a^2 \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\) |
\(\Big \downarrow \) 613 |
\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\frac {1}{2} \int \frac {1}{\sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)-\frac {1}{2} \int \frac {1}{(i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)\right )}{d}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}-\int \frac {1}{i-\frac {(a+i b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}+\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}\right )}{d}}{3 a}}{5 a}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}-\frac {i \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}\right )}{d}}{3 a}}{5 a}\) |
(-2*Sqrt[a + b*Tan[c + d*x]])/(5*a*d*Tan[c + d*x]^(5/2)) - ((-8*b*Sqrt[a + b*Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) + ((-15*a^2*((I*ArcTan[(Sqrt[ I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a - b] - (I *ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqr t[I*a + b]))/d - (2*(15*a^2 - 8*b^2)*Sqrt[a + b*Tan[c + d*x]])/(a*d*Sqrt[T an[c + d*x]]))/(3*a))/(5*a)
3.7.39.3.1 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_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
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_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(e_.)*(x_)]/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sym bol] :> Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] + x)), x ], x] - Simp[e/(2*b) Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] - x)), x ], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 /((m + 1)*(a^2 + b^2)*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / ; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ erQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S imp[ff/f Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, 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_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* (m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d )*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ [b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 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*b^2 + a^2*C)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d) *(a^2 + b^2)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Sim p[A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2)) - a*C*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - b*C)*Tan[e + f*x] - d*(A*b^2 + a^2*C)*( m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m , -1] && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.65 (sec) , antiderivative size = 947049, normalized size of antiderivative = 4135.59
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 4103 vs. \(2 (183) = 366\).
Time = 0.68 (sec) , antiderivative size = 4103, normalized size of antiderivative = 17.92 \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \]
-1/120*(15*a^3*d*sqrt(((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)* d^4)) - b)/((a^2 + b^2)*d^2))*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b ^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*d^2*tan(d*x + c) - (a^6 + 4*a^4*b^2 + 7*a^2*b^4 + 4*b^6)*d^2)*sqrt(-a^2/((a^4 + 2*a^2* b^2 + b^4)*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((a^6 + 7* a^4*b^2 + 12*a^2*b^4)*d*tan(d*x + c)^2 + 2*(a^5*b + a^3*b^3 - 4*a*b^5)*d*t an(d*x + c) - (a^6 + 3*a^4*b^2 + 4*a^2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*t an(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-a^2/((a^4 + 2*a^2 *b^2 + b^4)*d^4)))*sqrt(((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4 )*d^4)) - b)/((a^2 + b^2)*d^2)))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 + 15 *a^3*d*sqrt(((a^2 + b^2)*d^2*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - b) /((a^2 + b^2)*d^2))*log(-(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a* b^4)*tan(d*x + c) + (2*(a^5*b + 3*a^3*b^3 + 2*a*b^5)*d^2*tan(d*x + c) - (a ^6 + 4*a^4*b^2 + 7*a^2*b^4 + 4*b^6)*d^2)*sqrt(-a^2/((a^4 + 2*a^2*b^2 + b^4 )*d^4)))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((a^6 + 7*a^4*b^2 + 12*a^2*b^4)*d*tan(d*x + c)^2 + 2*(a^5*b + a^3*b^3 - 4*a*b^5)*d*tan(d*x + c) - (a^6 + 3*a^4*b^2 + 4*a^2*b^4)*d - 2*((a^4*b^3 + 5*a^2*b^5 + 4*b^7)*d^ 3*tan(d*x + c)^2 + (a^7 + 6*a^5*b^2 + 13*a^3*b^4 + 8*a*b^6)*d^3*tan(d*x + c) + (a^6*b + 3*a^4*b^3 + 2*a^2*b^5)*d^3)*sqrt(-a^2/((a^4 + 2*a^2*b^2 +...
\[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}} \,d x \]