\(\int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx\) [554]

Optimal result

Integrand size = 14, antiderivative size = 194 \[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{7/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{7/2} d}-\frac {2 b}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}}-\frac {4 a b}{3 \left (a^2+b^2\right )^2 d (a+b \tan (c+d x))^{3/2}}-\frac {2 b \left (3 a^2-b^2\right )}{\left (a^2+b^2\right )^3 d \sqrt {a+b \tan (c+d x)}} \] Output:

-I*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(7/2)/d+I*arctanh 
((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(7/2)/d-2/5*b/(a^2+b^2)/d/( 
a+b*tan(d*x+c))^(5/2)-4/3*a*b/(a^2+b^2)^2/d/(a+b*tan(d*x+c))^(3/2)-2*b*(3* 
a^2-b^2)/(a^2+b^2)^3/d/(a+b*tan(d*x+c))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.20 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.56 \[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=\frac {i (a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )+(-i a-b) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{5 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{5/2}} \] Input:

Integrate[(a + b*Tan[c + d*x])^(-7/2),x]
 

Output:

(I*(a + I*b)*Hypergeometric2F1[-5/2, 1, -3/2, (a + b*Tan[c + d*x])/(a - I* 
b)] + ((-I)*a - b)*Hypergeometric2F1[-5/2, 1, -3/2, (a + b*Tan[c + d*x])/( 
a + I*b)])/(5*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(5/2))
 

Rubi [A] (warning: unable to verify)

Time = 1.06 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.15, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.929, Rules used = {3042, 3964, 3042, 4012, 3042, 4012, 3042, 4022, 3042, 4020, 25, 73, 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.

\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+b \tan (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3964

\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}}dx}{a^2+b^2}-\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a-b \tan (c+d x)}{(a+b \tan (c+d x))^{5/2}}dx}{a^2+b^2}-\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {\int \frac {a^2-2 b \tan (c+d x) a-b^2}{(a+b \tan (c+d x))^{3/2}}dx}{a^2+b^2}-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a^2+b^2}-\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2-2 b \tan (c+d x) a-b^2}{(a+b \tan (c+d x))^{3/2}}dx}{a^2+b^2}-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a^2+b^2}-\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4012

\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (a^2-3 b^2\right )-b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{a^2+b^2}-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a^2+b^2}-\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {\int \frac {a \left (a^2-3 b^2\right )-b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}}{a^2+b^2}-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}}{a^2+b^2}-\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4022

\(\displaystyle -\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}+\frac {-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}+\frac {-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {1}{2} (a-i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (a+i b)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a^2+b^2}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 4020

\(\displaystyle -\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}+\frac {-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a+i b)^3 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i (a-i b)^3 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a^2+b^2}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}+\frac {-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {i (a-i b)^3 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (a+i b)^3 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}}{a^2+b^2}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}+\frac {-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b)^3 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(a+i b)^3 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}}{a^2+b^2}}{a^2+b^2}}{a^2+b^2}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 b}{5 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{5/2}}+\frac {-\frac {4 a b}{3 d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 b \left (3 a^2-b^2\right )}{d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {(a-i b)^3 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {(a+i b)^3 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}}{a^2+b^2}}{a^2+b^2}}{a^2+b^2}\)

Input:

Int[(a + b*Tan[c + d*x])^(-7/2),x]
 

Output:

(-2*b)/(5*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(5/2)) + ((-4*a*b)/(3*(a^2 + 
b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ((((a + I*b)^3*ArcTan[Tan[c + d*x]/Sq 
rt[a - I*b]])/(Sqrt[a - I*b]*d) + ((a - I*b)^3*ArcTan[Tan[c + d*x]/Sqrt[a 
+ I*b]])/(Sqrt[a + I*b]*d))/(a^2 + b^2) - (2*b*(3*a^2 - b^2))/((a^2 + b^2) 
*d*Sqrt[a + b*Tan[c + d*x]]))/(a^2 + b^2))/(a^2 + b^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

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]
 

rule 221

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]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3964

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n + 1)/(d*(n + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2) 
 Int[(a - b*Tan[c + d*x])*(a + b*Tan[c + d*x])^(n + 1), x], x] /; FreeQ[{a, 
 b, c, d}, x] && NeQ[a^2 + b^2, 0] && LtQ[n, -1]
 

rule 4012

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*((a + b*Tan[e + f*x])^(m + 1)/ 
(f*(m + 1)*(a^2 + b^2))), x] + Simp[1/(a^2 + b^2)   Int[(a + b*Tan[e + f*x] 
)^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a 
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1 
]
 

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(3081\) vs. \(2(166)=332\).

Time = 0.26 (sec) , antiderivative size = 3082, normalized size of antiderivative = 15.89

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3082\)
default	\(\text {Expression too large to display}\)	\(3082\)

Input:

int(1/(a+b*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

-2/5*b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(5/2)-4/3*a*b/(a^2+b^2)^2/d/(a+b*tan(d 
*x+c))^(3/2)+2/d*b^3/(a^2+b^2)^4/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2* 
(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2* 
a)^(1/2))*a^2-3/4/d*b^3/(a^2+b^2)^4*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/ 
2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^ 
(1/2)*a-1/4/d/b/(a^2+b^2)^4*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2 
*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^ 
5+3/4/d*b^3/(a^2+b^2)^4*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^ 
(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-2/d* 
b^3/(a^2+b^2)^4/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^( 
1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2+1/4 
/d/b/(a^2+b^2)^4*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/ 
2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^5-1/d/b/(a^ 
2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2 
)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^5+1/4/d/ 
b/(a^2+b^2)^(9/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)- 
b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^6-5/4/d*b/ 
(a^2+b^2)^(9/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b* 
tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^4-3/d*b/(a^2 
+b^2)^(9/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4720 vs. \(2 (160) = 320\).

Time = 0.21 (sec) , antiderivative size = 4720, normalized size of antiderivative = 24.33 \[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

\[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=\int \frac {1}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(1/(a+b*tan(d*x+c))**(7/2),x)
 

Output:

Integral((a + b*tan(c + d*x))**(-7/2), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*tan(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more 
details)Is
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+b*tan(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[9,19,7]%%%}+%%%{8,[9,17,7]%%%}+%%%{28,[9,15,7]%%%}+ 
%%%{56,[9
 

Mupad [B] (verification not implemented)

Time = 19.46 (sec) , antiderivative size = 5307, normalized size of antiderivative = 27.36 \[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=\text {Too large to display} \] Input:

int(1/(a + b*tan(c + d*x))^(7/2),x)
 

Output:

(log((((a + b*tan(c + d*x))^(1/2)*(16*b^26*d^3 - 96*a^2*b^24*d^3 - 1344*a^ 
4*b^22*d^3 - 5152*a^6*b^20*d^3 - 9648*a^8*b^18*d^3 - 8640*a^10*b^16*d^3 + 
8640*a^14*b^12*d^3 + 9648*a^16*b^10*d^3 + 5152*a^18*b^8*d^3 + 1344*a^20*b^ 
6*d^3 + 96*a^22*b^4*d^3 - 16*a^24*b^2*d^3) - ((-1/(a^7*d^2 + b^7*d^2*1i - 
7*a*b^6*d^2 - a^6*b*d^2*7i - a^2*b^5*d^2*21i + 35*a^3*b^4*d^2 + a^4*b^3*d^ 
2*35i - 21*a^5*b^2*d^2))^(1/2)*(((-1/(a^7*d^2 + b^7*d^2*1i - 7*a*b^6*d^2 - 
 a^6*b*d^2*7i - a^2*b^5*d^2*21i + 35*a^3*b^4*d^2 + a^4*b^3*d^2*35i - 21*a^ 
5*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^32*d^5 + 960*a^3*b^30 
*d^5 + 6720*a^5*b^28*d^5 + 29120*a^7*b^26*d^5 + 87360*a^9*b^24*d^5 + 19219 
2*a^11*b^22*d^5 + 320320*a^13*b^20*d^5 + 411840*a^15*b^18*d^5 + 411840*a^1 
7*b^16*d^5 + 320320*a^19*b^14*d^5 + 192192*a^21*b^12*d^5 + 87360*a^23*b^10 
*d^5 + 29120*a^25*b^8*d^5 + 6720*a^27*b^6*d^5 + 960*a^29*b^4*d^5 + 64*a^31 
*b^2*d^5))/2 - 128*a*b^29*d^4 - 1408*a^3*b^27*d^4 - 6912*a^5*b^25*d^4 - 19 
712*a^7*b^23*d^4 - 35200*a^9*b^21*d^4 - 38016*a^11*b^19*d^4 - 16896*a^13*b 
^17*d^4 + 16896*a^15*b^15*d^4 + 38016*a^17*b^13*d^4 + 35200*a^19*b^11*d^4 
+ 19712*a^21*b^9*d^4 + 6912*a^23*b^7*d^4 + 1408*a^25*b^5*d^4 + 128*a^27*b^ 
3*d^4))/2)*(-1/(a^7*d^2 + b^7*d^2*1i - 7*a*b^6*d^2 - a^6*b*d^2*7i - a^2*b^ 
5*d^2*21i + 35*a^3*b^4*d^2 + a^4*b^3*d^2*35i - 21*a^5*b^2*d^2))^(1/2))/2 - 
 8*b^23*d^2 - 48*a^2*b^21*d^2 - 72*a^4*b^19*d^2 + 192*a^6*b^17*d^2 + 1008* 
a^8*b^15*d^2 + 2016*a^10*b^13*d^2 + 2352*a^12*b^11*d^2 + 1728*a^14*b^9*...
 

Reduce [F]

\[ \int \frac {1}{(a+b \tan (c+d x))^{7/2}} \, dx=\frac {-2 \sqrt {a +\tan \left (d x +c \right ) b}-5 \left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{4} b^{4}+4 \tan \left (d x +c \right )^{3} a \,b^{3}+6 \tan \left (d x +c \right )^{2} a^{2} b^{2}+4 \tan \left (d x +c \right ) a^{3} b +a^{4}}d x \right ) \tan \left (d x +c \right )^{3} b^{4} d -15 \left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{4} b^{4}+4 \tan \left (d x +c \right )^{3} a \,b^{3}+6 \tan \left (d x +c \right )^{2} a^{2} b^{2}+4 \tan \left (d x +c \right ) a^{3} b +a^{4}}d x \right ) \tan \left (d x +c \right )^{2} a \,b^{3} d -15 \left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{4} b^{4}+4 \tan \left (d x +c \right )^{3} a \,b^{3}+6 \tan \left (d x +c \right )^{2} a^{2} b^{2}+4 \tan \left (d x +c \right ) a^{3} b +a^{4}}d x \right ) \tan \left (d x +c \right ) a^{2} b^{2} d -5 \left (\int \frac {\sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2}}{\tan \left (d x +c \right )^{4} b^{4}+4 \tan \left (d x +c \right )^{3} a \,b^{3}+6 \tan \left (d x +c \right )^{2} a^{2} b^{2}+4 \tan \left (d x +c \right ) a^{3} b +a^{4}}d x \right ) a^{3} b d}{5 b d \left (\tan \left (d x +c \right )^{3} b^{3}+3 \tan \left (d x +c \right )^{2} a \,b^{2}+3 \tan \left (d x +c \right ) a^{2} b +a^{3}\right )} \] Input:

int(1/(a+b*tan(d*x+c))^(7/2),x)
 

Output:

( - 2*sqrt(tan(c + d*x)*b + a) - 5*int((sqrt(tan(c + d*x)*b + a)*tan(c + d 
*x)**2)/(tan(c + d*x)**4*b**4 + 4*tan(c + d*x)**3*a*b**3 + 6*tan(c + d*x)* 
*2*a**2*b**2 + 4*tan(c + d*x)*a**3*b + a**4),x)*tan(c + d*x)**3*b**4*d - 1 
5*int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)**4*b**4 + 4 
*tan(c + d*x)**3*a*b**3 + 6*tan(c + d*x)**2*a**2*b**2 + 4*tan(c + d*x)*a** 
3*b + a**4),x)*tan(c + d*x)**2*a*b**3*d - 15*int((sqrt(tan(c + d*x)*b + a) 
*tan(c + d*x)**2)/(tan(c + d*x)**4*b**4 + 4*tan(c + d*x)**3*a*b**3 + 6*tan 
(c + d*x)**2*a**2*b**2 + 4*tan(c + d*x)*a**3*b + a**4),x)*tan(c + d*x)*a** 
2*b**2*d - 5*int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)* 
*4*b**4 + 4*tan(c + d*x)**3*a*b**3 + 6*tan(c + d*x)**2*a**2*b**2 + 4*tan(c 
 + d*x)*a**3*b + a**4),x)*a**3*b*d)/(5*b*d*(tan(c + d*x)**3*b**3 + 3*tan(c 
 + d*x)**2*a*b**2 + 3*tan(c + d*x)*a**2*b + a**3))