Integrand size = 25, antiderivative size = 221 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {\sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {\sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^2 d \sqrt {\tan (c+d x)}} \]
arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a-b)^(1/2 )/d-arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a+b) ^(1/2)/d+2/15*(15*a^2+2*b^2)*(a+b*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(1/2) -2/5*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-2/15*b*(a+b*tan(d*x+c))^(1/ 2)/a/d/tan(d*x+c)^(3/2)
Time = 1.53 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\frac {15 \sqrt [4]{-1} \sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+15 \sqrt [4]{-1} \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2-a b \tan (c+d x)+\left (15 a^2+2 b^2\right ) \tan ^2(c+d x)\right )}{a^2 \tan ^{\frac {5}{2}}(c+d x)}}{15 d} \]
(15*(-1)^(1/4)*Sqrt[-a + I*b]*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 15*(-1)^(1/4)*Sqrt[a + I*b]*ArcTan[( (-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + ( 2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2 - a*b*Tan[c + d*x] + (15*a^2 + 2*b^2)*T an[c + d*x]^2))/(a^2*Tan[c + d*x]^(5/2)))/(15*d)
Time = 1.41 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.10, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3042, 4051, 27, 3042, 4132, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}
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.
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan (c+d x)^{7/2}}dx\) |
| \(\Big \downarrow \) 4051 |
| \(\displaystyle -\frac {2}{5} \int -\frac {-4 b \tan ^2(c+d x)-5 a \tan (c+d x)+b}{2 \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \int \frac {-4 b \tan ^2(c+d x)-5 a \tan (c+d x)+b}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \int \frac {-4 b \tan (c+d x)^2-5 a \tan (c+d x)+b}{\tan (c+d x)^{5/2} \sqrt {a+b \tan (c+d x)}}dx-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{5} \left (-\frac {2 \int \frac {15 a^2+15 b \tan (c+d x) a+2 b^2+2 b^2 \tan ^2(c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {\int \frac {15 a^2+15 b \tan (c+d x) a+2 b^2+2 b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-\frac {\int \frac {15 a^2+15 b \tan (c+d x) a+2 b^2+2 b^2 \tan (c+d x)^2}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4132 |
| \(\displaystyle \frac {1}{5} \left (-\frac {-\frac {2 \int -\frac {15 \left (a^2 b-a^3 \tan (c+d x)\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{5} \left (-\frac {\frac {15 \int \frac {a^2 b-a^3 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{5} \left (-\frac {\frac {15 \int \frac {a^2 b-a^3 \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\right )-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}\) |
| \(\Big \downarrow \) 4099 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^2 (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^2 (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {1}{2} a^2 (b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a^2 (-b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx\right )}{a}}{3 a}\right )\) |
| \(\Big \downarrow \) 4098 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 (b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}-\frac {a^2 (-b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{2 d}\right )}{a}}{3 a}\right )\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 (b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^2 (-b+i a) \int \frac {1}{\frac {(i a-b) \tan (c+d x)}{a+b \tan (c+d x)}+1}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}\right )}{a}}{3 a}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 (b+i a) \int \frac {1}{1-\frac {(i a+b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}}{d}-\frac {a^2 \sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )}{a}}{3 a}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {-\frac {2 \left (15 a^2+2 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}+\frac {15 \left (\frac {a^2 \sqrt {b+i a} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {a^2 \sqrt {-b+i a} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )}{a}}{3 a}\right )\) |
(-2*Sqrt[a + b*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(5/2)) + ((-2*b*Sqrt[a + b *Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) - ((15*(-((a^2*Sqrt[I*a - b]*Ar cTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d) + (a ^2*Sqrt[I*a + b]*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan [c + d*x]]])/d))/a - (2*(15*a^2 + 2*b^2)*Sqrt[a + b*Tan[c + d*x]])/(a*d*Sq rt[Tan[c + d*x]]))/(3*a))/5
3.7.14.3.1 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[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^n/(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(a^2 + b^2 )) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[a*c *(m + 1) - b*d*n - (b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(m + n + 1)*Tan[e + f*x]^2, 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] && LtQ[m, -1] && GtQ[n, 0] && Int egerQ[2*m]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[A^2/f Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si mp[(A + I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T an[e + f*x]), x], x] + Simp[(A - I*B)/2 Int[(a + b*Tan[e + f*x])^m*(c + d *Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A , B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^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])))
result has leaf size over 500,000. Avoiding possible recursion issues.
Time = 0.52 (sec) , antiderivative size = 1092009, normalized size of antiderivative = 4941.22
\[\text {output too large to display}\]
Leaf count of result is larger than twice the leaf count of optimal. 2779 vs. \(2 (179) = 358\).
Time = 0.50 (sec) , antiderivative size = 2779, normalized size of antiderivative = 12.57 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/120*(15*a^2*d*sqrt((d^2*sqrt(-a^2/d^4) + b)/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^3*b + 2*a*b^3)*d^2 *tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan( d*x + c) + a)*sqrt(tan(d*x + c)) + (a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d* tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^3*b + 4*a*b^3) *d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d*x + c) - (3*a^ 3*b + 4*a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d^2*sqrt(-a^2/d^4) + b)/d^2))/(t an(d*x + c)^2 + 1))*tan(d*x + c)^3 + 15*a^2*d*sqrt((d^2*sqrt(-a^2/d^4) + b )/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^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d ^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + (a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^2 - 4*(a^4*b + 2*a^2*b^3)*d*ta n(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan(d*x + c)^2 - 2*(a^4 + 3*a^2*b^2 + 4*b^4)*d^3*tan(d*x + c) - (3*a^3*b + 4*a*b^3)*d^3)*sqrt(-a^2/d^4))*sqrt((d ^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^3 - 15*a^2 *d*sqrt((d^2*sqrt(-a^2/d^4) + b)/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^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a) *sqrt(tan(d*x + c)) - (a^5*d - (a^5 + 6*a^3*b^2 + 8*a*b^4)*d*tan(d*x + c)^ 2 - 4*(a^4*b + 2*a^2*b^3)*d*tan(d*x + c) - ((a^3*b + 4*a*b^3)*d^3*tan(d...
\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a + b \tan {\left (c + d x \right )}}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {\sqrt {b \tan \left (d x + c\right ) + a}}{\tan \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \]