3.4.60 \(\int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) [360]

3.4.60.1 Optimal result

Integrand size = 31, antiderivative size = 188 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}-\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}+\frac {2 a (A b-a B)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (a^2 A-A b^2+2 a b B\right )}{\left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

output

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

3.4.60.2 Mathematica [A] (verified)

Time = 3.87 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.73 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {\frac {3 b \left (-a^2 \left (A \sqrt {-b^2}+b B\right )+b^2 \left (A \sqrt {-b^2}+b B\right )+2 a b \left (A b-\sqrt {-b^2} B\right )\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \sqrt {a-\sqrt {-b^2}}}-\frac {3 b \left (2 a A b^2+a^2 A \sqrt {-b^2}+A \left (-b^2\right )^{3/2}-a^2 b B+b^3 B+2 a b \sqrt {-b^2} B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {-b^2} \sqrt {a+\sqrt {-b^2}}}+\frac {2 a \left (a^2+b^2\right ) (A b-a B)}{(a+b \tan (c+d x))^{3/2}}+\frac {6 b \left (a^2 A-A b^2+2 a b B\right )}{\sqrt {a+b \tan (c+d x)}}}{3 b \left (a^2+b^2\right )^2 d} \]

input

Integrate[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x 
]
 

output

((3*b*(-(a^2*(A*Sqrt[-b^2] + b*B)) + b^2*(A*Sqrt[-b^2] + b*B) + 2*a*b*(A*b 
 - Sqrt[-b^2]*B))*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/ 
(Sqrt[-b^2]*Sqrt[a - Sqrt[-b^2]]) - (3*b*(2*a*A*b^2 + a^2*A*Sqrt[-b^2] + A 
*(-b^2)^(3/2) - a^2*b*B + b^3*B + 2*a*b*Sqrt[-b^2]*B)*ArcTanh[Sqrt[a + b*T 
an[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/(Sqrt[-b^2]*Sqrt[a + Sqrt[-b^2]]) + (2 
*a*(a^2 + b^2)*(A*b - a*B))/(a + b*Tan[c + d*x])^(3/2) + (6*b*(a^2*A - A*b 
^2 + 2*a*b*B))/Sqrt[a + b*Tan[c + d*x]])/(3*b*(a^2 + b^2)^2*d)
 

3.4.60.3 Rubi [A] (warning: unable to verify)

Time = 0.93 (sec) , antiderivative size = 213, normalized size of antiderivative = 1.13, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.355, Rules used = {3042, 4074, 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.

\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4074

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

Int[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(5/2),x]
 

output

(2*a*(A*b - a*B))/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ((-(((a 
 + I*b)^2*(I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d)) 
 + ((a - I*b)^2*(I*A - B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I* 
b]*d))/(a^2 + b^2) + (2*(a^2*A - A*b^2 + 2*a*b*B))/((a^2 + b^2)*d*Sqrt[a + 
 b*Tan[c + d*x]]))/(a^2 + b^2)
 

3.4.60.3.1 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 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]
 

rule 4074

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

3.4.60.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4483\) vs. \(2(164)=328\).

Time = 0.13 (sec) , antiderivative size = 4484, normalized size of antiderivative = 23.85

method	result	size
parts	\(\text {Expression too large to display}\)	\(4484\)
derivativedivides	\(\text {Expression too large to display}\)	\(12841\)
default	\(\text {Expression too large to display}\)	\(12841\)

input

int(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x,method=_RETURNVER 
BOSE)
 

output

A*(2/d*a^2/(a^2+b^2)^2/(a+b*tan(d*x+c))^(1/2)-1/d/(a^2+b^2)^(7/2)/(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^5-3/4/d/(a^2+b^2)^(7/2)*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^4+1/2/d/(a^2+b^2)^3*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^3+1/d/(a^2+b^2)^3/(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^4-1/2/d/(a^2+b^2)^3*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^3-1/d/(a^2+b^2)^3/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arct 
an(((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^4-2/d/(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arc 
tan((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^3+1/d/(a^2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*ar 
ctan((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+3/4/d/(a^2+b^2)^(7/2)*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+1/4/d*b^4/(a^2+b^2)^(7/2)*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)...
 

3.4.60.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7209 vs. \(2 (158) = 316\).

Time = 2.26 (sec) , antiderivative size = 7209, normalized size of antiderivative = 38.35 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm= 
"fricas")
 

output

Too large to include
 

3.4.60.6 Sympy [F]

\[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan {\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

input

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**(5/2),x)
 

output

Integral((A + B*tan(c + d*x))*tan(c + d*x)/(a + b*tan(c + d*x))**(5/2), x)
 

3.4.60.7 Maxima [F(-1)]

Timed out. \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm= 
"maxima")
 

output

Timed out
 

3.4.60.8 Giac [F(-1)]

Timed out. \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(5/2),x, algorithm= 
"giac")
 

output

Timed out
 

3.4.60.9 Mupad [B] (verification not implemented)

Time = 26.56 (sec) , antiderivative size = 9464, normalized size of antiderivative = 50.34 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

int((tan(c + d*x)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^(5/2),x)
 

output

(log(((((320*B^4*a^2*b^8*d^4 - 16*B^4*b^10*d^4 - 1760*B^4*a^4*b^6*d^4 + 16 
00*B^4*a^6*b^4*d^4 - 400*B^4*a^8*b^2*d^4)^(1/2) - 4*B^2*a^5*d^2 + 40*B^2*a 
^3*b^2*d^2 - 20*B^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a 
^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*((a + b*tan(c + d*x))^ 
(1/2)*(320*B^2*a^4*b^14*d^3 - 16*B^2*b^18*d^3 + 1024*B^2*a^6*b^12*d^3 + 14 
40*B^2*a^8*b^10*d^3 + 1024*B^2*a^10*b^8*d^3 + 320*B^2*a^12*b^6*d^3 - 16*B^ 
2*a^16*b^2*d^3) + ((((320*B^4*a^2*b^8*d^4 - 16*B^4*b^10*d^4 - 1760*B^4*a^4 
*b^6*d^4 + 1600*B^4*a^6*b^4*d^4 - 400*B^4*a^8*b^2*d^4)^(1/2) - 4*B^2*a^5*d 
^2 + 40*B^2*a^3*b^2*d^2 - 20*B^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b 
^8*d^4 + 10*a^4*b^6*d^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(896*B*a^ 
6*b^15*d^4 - 32*B*b^21*d^4 - 160*B*a^2*b^19*d^4 - 128*B*a^4*b^17*d^4 - ((( 
(320*B^4*a^2*b^8*d^4 - 16*B^4*b^10*d^4 - 1760*B^4*a^4*b^6*d^4 + 1600*B^4*a 
^6*b^4*d^4 - 400*B^4*a^8*b^2*d^4)^(1/2) - 4*B^2*a^5*d^2 + 40*B^2*a^3*b^2*d 
^2 - 20*B^2*a*b^4*d^2)/(a^10*d^4 + b^10*d^4 + 5*a^2*b^8*d^4 + 10*a^4*b^6*d 
^4 + 10*a^6*b^4*d^4 + 5*a^8*b^2*d^4))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64 
*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18*d^5 + 7680*a^7*b^16*d^5 + 1 
3440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 13440*a^13*b^10*d^5 + 7680*a^15* 
b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 + 64*a^21*b^2*d^5))/4 + 313 
6*B*a^8*b^13*d^4 + 4928*B*a^10*b^11*d^4 + 4480*B*a^12*b^9*d^4 + 2432*B*a^1 
4*b^7*d^4 + 736*B*a^16*b^5*d^4 + 96*B*a^18*b^3*d^4))/4))/4 - 96*B^3*a^3...