\(\int \frac {\sin ^5(c+d x)}{(a+b \sin ^2(c+d x))^2} \, dx\) [54]

Optimal result

Integrand size = 23, antiderivative size = 102 \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {a (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 b^{5/2} (a+b)^{3/2} d}-\frac {\cos (c+d x)}{b^2 d}-\frac {a^2 \cos (c+d x)}{2 b^2 (a+b) d \left (a+b-b \cos ^2(c+d x)\right )} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 1.69 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.69 \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {a (3 a+4 b) \arctan \left (\frac {\sqrt {b}-i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+\frac {a (3 a+4 b) \arctan \left (\frac {\sqrt {b}+i \sqrt {a} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a-b}}\right )}{(-a-b)^{3/2}}+2 \sqrt {b} \cos (c+d x) \left (-1-\frac {a^2}{(a+b) (2 a+b-b \cos (2 (c+d x)))}\right )}{2 b^{5/2} d} \] Input:

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

Output:

((a*(3*a + 4*b)*ArcTan[(Sqrt[b] - I*Sqrt[a]*Tan[(c + d*x)/2])/Sqrt[-a - b] 
])/(-a - b)^(3/2) + (a*(3*a + 4*b)*ArcTan[(Sqrt[b] + I*Sqrt[a]*Tan[(c + d* 
x)/2])/Sqrt[-a - b]])/(-a - b)^(3/2) + 2*Sqrt[b]*Cos[c + d*x]*(-1 - a^2/(( 
a + b)*(2*a + b - b*Cos[2*(c + d*x)]))))/(2*b^(5/2)*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3665, 300, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3665

\(\displaystyle -\frac {\int \frac {\left (1-\cos ^2(c+d x)\right )^2}{\left (-b \cos ^2(c+d x)+a+b\right )^2}d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 300

\(\displaystyle -\frac {\int \left (\frac {1}{b^2}-\frac {a (a+2 b)-2 a b \cos ^2(c+d x)}{b^2 \left (-b \cos ^2(c+d x)+a+b\right )^2}\right )d\cos (c+d x)}{d}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\frac {a^2 \cos (c+d x)}{2 b^2 (a+b) \left (a-b \cos ^2(c+d x)+b\right )}-\frac {a (3 a+4 b) \text {arctanh}\left (\frac {\sqrt {b} \cos (c+d x)}{\sqrt {a+b}}\right )}{2 b^{5/2} (a+b)^{3/2}}+\frac {\cos (c+d x)}{b^2}}{d}\)

Input:

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

Output:

-((-1/2*(a*(3*a + 4*b)*ArcTanh[(Sqrt[b]*Cos[c + d*x])/Sqrt[a + b]])/(b^(5/ 
2)*(a + b)^(3/2)) + Cos[c + d*x]/b^2 + (a^2*Cos[c + d*x])/(2*b^2*(a + b)*( 
a + b - b*Cos[c + d*x]^2)))/d)
 

Defintions of rubi rules used

rule 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3665

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f 
Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + 
 f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
 

Maple [A] (verified)

Time = 1.28 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(\frac {-\frac {\cos \left (d x +c \right )}{b^{2}}+\frac {a \left (-\frac {a \cos \left (d x +c \right )}{2 \left (a +b \right ) \left (a +b -b \cos \left (d x +c \right )^{2}\right )}+\frac {\left (3 a +4 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{b^{2}}}{d}\)	\(90\)
default	\(\frac {-\frac {\cos \left (d x +c \right )}{b^{2}}+\frac {a \left (-\frac {a \cos \left (d x +c \right )}{2 \left (a +b \right ) \left (a +b -b \cos \left (d x +c \right )^{2}\right )}+\frac {\left (3 a +4 b \right ) \operatorname {arctanh}\left (\frac {b \cos \left (d x +c \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{b^{2}}}{d}\)	\(90\)
risch	\(-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 d \,b^{2}}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 d \,b^{2}}+\frac {a^{2} \left ({\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )}\right )}{b^{2} \left (a +b \right ) d \left (b \,{\mathrm e}^{4 i \left (d x +c \right )}-4 a \,{\mathrm e}^{2 i \left (d x +c \right )}-2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+b \right )}+\frac {3 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d \,b^{2}}+\frac {i a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{\sqrt {-a b -b^{2}}\, \left (a +b \right ) d b}-\frac {3 i a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) d \,b^{2}}-\frac {i a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (d x +c \right )}}{\sqrt {-a b -b^{2}}}+1\right )}{\sqrt {-a b -b^{2}}\, \left (a +b \right ) d b}\)	\(377\)

Input:

int(sin(d*x+c)^5/(a+b*sin(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(-1/b^2*cos(d*x+c)+1/b^2*a*(-1/2*a/(a+b)*cos(d*x+c)/(a+b-b*cos(d*x+c)^ 
2)+1/2*(3*a+4*b)/(a+b)/((a+b)*b)^(1/2)*arctanh(b*cos(d*x+c)/((a+b)*b)^(1/2 
))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (93) = 186\).

Time = 0.11 (sec) , antiderivative size = 427, normalized size of antiderivative = 4.19 \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\left [-\frac {4 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, a^{3} + 7 \, a^{2} b + 4 \, a b^{2} - {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a b + b^{2}} \log \left (\frac {b \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a b + b^{2}} \cos \left (d x + c\right ) + a + b}{b \cos \left (d x + c\right )^{2} - a - b}\right ) - 2 \, {\left (3 \, a^{3} b + 7 \, a^{2} b^{2} + 6 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (d x + c\right )}{4 \, {\left ({\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} d\right )}}, -\frac {2 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{3} - {\left (3 \, a^{3} + 7 \, a^{2} b + 4 \, a b^{2} - {\left (3 \, a^{2} b + 4 \, a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a b - b^{2}} \arctan \left (\frac {\sqrt {-a b - b^{2}} \cos \left (d x + c\right )}{a + b}\right ) - {\left (3 \, a^{3} b + 7 \, a^{2} b^{2} + 6 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (d x + c\right )}{2 \, {\left ({\left (a^{2} b^{4} + 2 \, a b^{5} + b^{6}\right )} d \cos \left (d x + c\right )^{2} - {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} d\right )}}\right ] \] Input:

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

Output:

[-1/4*(4*(a^2*b^2 + 2*a*b^3 + b^4)*cos(d*x + c)^3 + (3*a^3 + 7*a^2*b + 4*a 
*b^2 - (3*a^2*b + 4*a*b^2)*cos(d*x + c)^2)*sqrt(a*b + b^2)*log((b*cos(d*x 
+ c)^2 + 2*sqrt(a*b + b^2)*cos(d*x + c) + a + b)/(b*cos(d*x + c)^2 - a - b 
)) - 2*(3*a^3*b + 7*a^2*b^2 + 6*a*b^3 + 2*b^4)*cos(d*x + c))/((a^2*b^4 + 2 
*a*b^5 + b^6)*d*cos(d*x + c)^2 - (a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*d), 
 -1/2*(2*(a^2*b^2 + 2*a*b^3 + b^4)*cos(d*x + c)^3 - (3*a^3 + 7*a^2*b + 4*a 
*b^2 - (3*a^2*b + 4*a*b^2)*cos(d*x + c)^2)*sqrt(-a*b - b^2)*arctan(sqrt(-a 
*b - b^2)*cos(d*x + c)/(a + b)) - (3*a^3*b + 7*a^2*b^2 + 6*a*b^3 + 2*b^4)* 
cos(d*x + c))/((a^2*b^4 + 2*a*b^5 + b^6)*d*cos(d*x + c)^2 - (a^3*b^3 + 3*a 
^2*b^4 + 3*a*b^5 + b^6)*d)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\text {Timed out} \] Input:

integrate(sin(d*x+c)**5/(a+b*sin(d*x+c)**2)**2,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.28 \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {2 \, a^{2} \cos \left (d x + c\right )}{a^{2} b^{2} + 2 \, a b^{3} + b^{4} - {\left (a b^{3} + b^{4}\right )} \cos \left (d x + c\right )^{2}} + \frac {{\left (3 \, a + 4 \, b\right )} a \log \left (\frac {b \cos \left (d x + c\right ) - \sqrt {{\left (a + b\right )} b}}{b \cos \left (d x + c\right ) + \sqrt {{\left (a + b\right )} b}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {4 \, \cos \left (d x + c\right )}{b^{2}}}{4 \, d} \] Input:

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

Output:

-1/4*(2*a^2*cos(d*x + c)/(a^2*b^2 + 2*a*b^3 + b^4 - (a*b^3 + b^4)*cos(d*x 
+ c)^2) + (3*a + 4*b)*a*log((b*cos(d*x + c) - sqrt((a + b)*b))/(b*cos(d*x 
+ c) + sqrt((a + b)*b)))/((a*b^2 + b^3)*sqrt((a + b)*b)) + 4*cos(d*x + c)/ 
b^2)/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 342 vs. \(2 (93) = 186\).

Time = 0.48 (sec) , antiderivative size = 342, normalized size of antiderivative = 3.35 \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {{\left (3 \, a^{2} + 4 \, a b\right )} \arctan \left (\frac {b \cos \left (d x + c\right ) + a + b}{\sqrt {-a b - b^{2}} \cos \left (d x + c\right ) + \sqrt {-a b - b^{2}}}\right )}{{\left (a b^{2} + b^{3}\right )} \sqrt {-a b - b^{2}}} + \frac {2 \, {\left (3 \, a^{2} + 2 \, a b - \frac {6 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {14 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {8 \, b^{2} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, a b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}}{{\left (a b^{2} + b^{3}\right )} {\left (a - \frac {3 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {4 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}}{2 \, d} \] Input:

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

Output:

-1/2*((3*a^2 + 4*a*b)*arctan((b*cos(d*x + c) + a + b)/(sqrt(-a*b - b^2)*co 
s(d*x + c) + sqrt(-a*b - b^2)))/((a*b^2 + b^3)*sqrt(-a*b - b^2)) + 2*(3*a^ 
2 + 2*a*b - 6*a^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 14*a*b*(cos(d*x 
+ c) - 1)/(cos(d*x + c) + 1) - 8*b^2*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) 
 + 3*a^2*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 4*a*b*(cos(d*x + c) - 
 1)^2/(cos(d*x + c) + 1)^2)/((a*b^2 + b^3)*(a - 3*a*(cos(d*x + c) - 1)/(co 
s(d*x + c) + 1) - 4*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a*(cos(d*x 
 + c) - 1)^2/(cos(d*x + c) + 1)^2 + 4*b*(cos(d*x + c) - 1)^2/(cos(d*x + c) 
 + 1)^2 - a*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3)))/d
 

Mupad [B] (verification not implemented)

Time = 36.40 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx=\frac {a\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\cos \left (c+d\,x\right )}{\sqrt {a+b}}\right )\,\left (3\,a+4\,b\right )}{2\,b^{5/2}\,d\,{\left (a+b\right )}^{3/2}}-\frac {a^2\,\cos \left (c+d\,x\right )}{2\,d\,\left (a+b\right )\,\left (-b^3\,{\cos \left (c+d\,x\right )}^2+b^3+a\,b^2\right )}-\frac {\cos \left (c+d\,x\right )}{b^2\,d} \] Input:

int(sin(c + d*x)^5/(a + b*sin(c + d*x)^2)^2,x)
 

Output:

(a*atanh((b^(1/2)*cos(c + d*x))/(a + b)^(1/2))*(3*a + 4*b))/(2*b^(5/2)*d*( 
a + b)^(3/2)) - (a^2*cos(c + d*x))/(2*d*(a + b)*(a*b^2 + b^3 - b^3*cos(c + 
 d*x)^2)) - cos(c + d*x)/(b^2*d)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 772, normalized size of antiderivative = 7.57 \[ \int \frac {\sin ^5(c+d x)}{\left (a+b \sin ^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

int(sin(d*x+c)^5/(a+b*sin(d*x+c)^2)^2,x)
 

Output:

( - 4*cos(c + d*x)*sin(c + d*x)**2*a**2*b**2 - 8*cos(c + d*x)*sin(c + d*x) 
**2*a*b**3 - 4*cos(c + d*x)*sin(c + d*x)**2*b**4 - 6*cos(c + d*x)*a**3*b - 
 10*cos(c + d*x)*a**2*b**2 - 4*cos(c + d*x)*a*b**3 - 3*sqrt(b)*sqrt(a + b) 
*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))* 
sin(c + d*x)**2*a**2*b - 4*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt( 
a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2*a*b**2 - 3*s 
qrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)* 
tan((c + d*x)/2))*a**3 - 4*sqrt(b)*sqrt(a + b)*log( - sqrt(2*sqrt(b)*sqrt( 
a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*a**2*b - 3*sqrt(b)*sqrt(a + 
b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*s 
in(c + d*x)**2*a**2*b - 4*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + 
b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2*a*b**2 - 3*sqrt( 
b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c 
+ d*x)/2))*a**3 - 4*sqrt(b)*sqrt(a + b)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a 
 - 2*b) + sqrt(a)*tan((c + d*x)/2))*a**2*b + 3*sqrt(b)*sqrt(a + b)*log(2*s 
qrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**2*a**2 
*b + 4*sqrt(b)*sqrt(a + b)*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2 
*a + a + 2*b)*sin(c + d*x)**2*a*b**2 + 3*sqrt(b)*sqrt(a + b)*log(2*sqrt(b) 
*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*a**3 + 4*sqrt(b)*sqrt(a + 
b)*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*a**2*b ...