3.5.2 \(\int \frac {\sec ^2(c+d x)}{(a+b \sin ^3(c+d x))^2} \, dx\) [402]

3.5.2.1 Optimal result
3.5.2.2 Mathematica [C] (warning: unable to verify)
3.5.2.3 Rubi [N/A]
3.5.2.4 Maple [N/A] (verified)
3.5.2.5 Fricas [C] (verification not implemented)
3.5.2.6 Sympy [F(-1)]
3.5.2.7 Maxima [F(-2)]
3.5.2.8 Giac [N/A]
3.5.2.9 Mupad [B] (verification not implemented)

3.5.2.1 Optimal result

Integrand size = 23, antiderivative size = 23 \[ \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Int}\left (\frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2},x\right ) \]

output
Unintegrable(sec(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x)
 
3.5.2.2 Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 2.21 (sec) , antiderivative size = 845, normalized size of antiderivative = 36.74 \[ \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\frac {-\frac {i b \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {16 a^2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+2 b^3 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-8 i a^2 b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-i b^3 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+20 i a^3 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}+16 i a b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}+10 a^3 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}+8 a b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}-120 a^2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+12 b^3 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+60 i a^2 b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-6 i b^3 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-20 i a^3 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^3-16 i a b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^3-10 a^3 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3-8 a b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3+16 a^2 b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+2 b^3 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-8 i a^2 b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4-i b^3 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]}{a \left (a^2-b^2\right )^2}+\frac {18 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {18 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {12 b \cos (c+d x) \left (-2 a^3-7 a b^2+3 a b^2 \cos (2 (c+d x))+2 b \left (2 a^2+b^2\right ) \sin (c+d x)\right )}{a (a-b)^2 (a+b)^2 (4 a+3 b \sin (c+d x)-b \sin (3 (c+d x)))}}{18 d} \]

input
Integrate[Sec[c + d*x]^2/(a + b*Sin[c + d*x]^3)^2,x]
 
output
(((-I)*b*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*#1^ 
6 & , (16*a^2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] + 2*b^3*ArcTan[Si 
n[c + d*x]/(Cos[c + d*x] - #1)] - (8*I)*a^2*b*Log[1 - 2*Cos[c + d*x]*#1 + 
#1^2] - I*b^3*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + (20*I)*a^3*ArcTan[Sin[c 
+ d*x]/(Cos[c + d*x] - #1)]*#1 + (16*I)*a*b^2*ArcTan[Sin[c + d*x]/(Cos[c + 
 d*x] - #1)]*#1 + 10*a^3*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 + 8*a*b^2*Lo 
g[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 - 120*a^2*b*ArcTan[Sin[c + d*x]/(Cos[c 
+ d*x] - #1)]*#1^2 + 12*b^3*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 
+ (60*I)*a^2*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (6*I)*b^3*Log[1 - 
2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (20*I)*a^3*ArcTan[Sin[c + d*x]/(Cos[c + d 
*x] - #1)]*#1^3 - (16*I)*a*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1 
^3 - 10*a^3*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^3 - 8*a*b^2*Log[1 - 2*Cos 
[c + d*x]*#1 + #1^2]*#1^3 + 16*a^2*b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - # 
1)]*#1^4 + 2*b^3*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - (8*I)*a^2 
*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^4 - I*b^3*Log[1 - 2*Cos[c + d*x]*# 
1 + #1^2]*#1^4)/(b*#1 - (4*I)*a*#1^2 - 2*b*#1^3 + b*#1^5) & ])/(a*(a^2 - b 
^2)^2) + (18*Sin[(c + d*x)/2])/((a + b)^2*(Cos[(c + d*x)/2] - Sin[(c + d*x 
)/2])) + (18*Sin[(c + d*x)/2])/((a - b)^2*(Cos[(c + d*x)/2] + Sin[(c + d*x 
)/2])) + (12*b*Cos[c + d*x]*(-2*a^3 - 7*a*b^2 + 3*a*b^2*Cos[2*(c + d*x)] + 
 2*b*(2*a^2 + b^2)*Sin[c + d*x]))/(a*(a - b)^2*(a + b)^2*(4*a + 3*b*Sin...
 
3.5.2.3 Rubi [N/A]

Not integrable

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {3042, 3707}

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.

\(\displaystyle \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cos (c+d x)^2 \left (a+b \sin (c+d x)^3\right )^2}dx\)

\(\Big \downarrow \) 3707

\(\displaystyle \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2}dx\)

input
Int[Sec[c + d*x]^2/(a + b*Sin[c + d*x]^3)^2,x]
 
output
$Aborted
 

3.5.2.3.1 Defintions of rubi rules used

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3707
Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_.)*((a_) + (b_.)*((c_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> Unintegrable[(d*Cos[e + f*x])^m*(a + 
 b*(c*Sin[e + f*x])^n)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x]
 
3.5.2.4 Maple [N/A] (verified)

Time = 6.54 (sec) , antiderivative size = 398, normalized size of antiderivative = 17.30

method result size
derivativedivides \(\frac {\frac {2 b \left (\frac {-\frac {\left (2 a^{2}+b^{2}\right ) b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a}+\left (-\frac {a^{2}}{3}+\frac {4 b^{2}}{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 b \left (a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a}+\left (-\frac {2 a^{2}}{3}-\frac {10 b^{2}}{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (2 a^{2}+b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a}-\frac {a^{2}}{3}-\frac {2 b^{2}}{3}}{a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (b \left (-11 a^{2}+2 b^{2}\right ) \textit {\_R}^{4}+2 a \left (5 a^{2}+4 b^{2}\right ) \textit {\_R}^{3}-54 a^{2} b \,\textit {\_R}^{2}+2 a \left (5 a^{2}+4 b^{2}\right ) \textit {\_R} -11 a^{2} b +2 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}}{18 a}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) \(398\)
default \(\frac {\frac {2 b \left (\frac {-\frac {\left (2 a^{2}+b^{2}\right ) b \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a}+\left (-\frac {a^{2}}{3}+\frac {4 b^{2}}{3}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 b \left (a^{2}+2 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 a}+\left (-\frac {2 a^{2}}{3}-\frac {10 b^{2}}{3}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {\left (2 a^{2}+b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{3 a}-\frac {a^{2}}{3}-\frac {2 b^{2}}{3}}{a \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +3 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +a}+\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (b \left (-11 a^{2}+2 b^{2}\right ) \textit {\_R}^{4}+2 a \left (5 a^{2}+4 b^{2}\right ) \textit {\_R}^{3}-54 a^{2} b \,\textit {\_R}^{2}+2 a \left (5 a^{2}+4 b^{2}\right ) \textit {\_R} -11 a^{2} b +2 b^{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}}{18 a}\right )}{\left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {1}{\left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {1}{\left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) \(398\)
risch \(\text {Expression too large to display}\) \(5261\)

input
int(sec(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x,method=_RETURNVERBOSE)
 
output
1/d*(2*b/(a-b)^2/(a+b)^2*((-1/3*(2*a^2+b^2)*b/a*tan(1/2*d*x+1/2*c)^5+(-1/3 
*a^2+4/3*b^2)*tan(1/2*d*x+1/2*c)^4-4/3*b*(a^2+2*b^2)/a*tan(1/2*d*x+1/2*c)^ 
3+(-2/3*a^2-10/3*b^2)*tan(1/2*d*x+1/2*c)^2+1/3*(2*a^2+b^2)*b/a*tan(1/2*d*x 
+1/2*c)-1/3*a^2-2/3*b^2)/(a*tan(1/2*d*x+1/2*c)^6+3*tan(1/2*d*x+1/2*c)^4*a+ 
8*tan(1/2*d*x+1/2*c)^3*b+3*tan(1/2*d*x+1/2*c)^2*a+a)+1/18/a*sum((b*(-11*a^ 
2+2*b^2)*_R^4+2*a*(5*a^2+4*b^2)*_R^3-54*a^2*b*_R^2+2*a*(5*a^2+4*b^2)*_R-11 
*a^2*b+2*b^3)/(_R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tan(1/2*d*x+1/2*c)-_R),_R 
=RootOf(_Z^6*a+3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a)))-1/(a+b)^2/(tan(1/2*d*x+1/2* 
c)-1)-1/(a-b)^2/(tan(1/2*d*x+1/2*c)+1))
 
3.5.2.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 43.88 (sec) , antiderivative size = 102913, normalized size of antiderivative = 4474.48 \[ \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Too large to display} \]

input
integrate(sec(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x, algorithm="fricas")
 
output
Too large to include
 
3.5.2.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Timed out} \]

input
integrate(sec(d*x+c)**2/(a+b*sin(d*x+c)**3)**2,x)
 
output
Timed out
 
3.5.2.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Exception raised: RuntimeError} \]

input
integrate(sec(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x, algorithm="maxima")
 
output
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 
3.5.2.8 Giac [N/A]

Not integrable

Time = 4.34 (sec) , antiderivative size = 3, normalized size of antiderivative = 0.13 \[ \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\int { \frac {\sec \left (d x + c\right )^{2}}{{\left (b \sin \left (d x + c\right )^{3} + a\right )}^{2}} \,d x } \]

input
integrate(sec(d*x+c)^2/(a+b*sin(d*x+c)^3)^2,x, algorithm="giac")
 
output
sage0*x
 
3.5.2.9 Mupad [B] (verification not implemented)

Time = 19.15 (sec) , antiderivative size = 3148, normalized size of antiderivative = 136.87 \[ \int \frac {\sec ^2(c+d x)}{\left (a+b \sin ^3(c+d x)\right )^2} \, dx=\text {Too large to display} \]

input
int(1/(cos(c + d*x)^2*(a + b*sin(c + d*x)^3)^2),x)
 
output
symsum(log(5479612416*a^8*b^36 - 180486144*a^6*b^38 - root(5314410*a^16*b^ 
4*d^6 - 5314410*a^14*b^6*d^6 - 2657205*a^18*b^2*d^6 + 2657205*a^12*b^8*d^6 
 - 531441*a^10*b^10*d^6 + 531441*a^20*d^6 + 11514555*a^12*b^4*d^4 + 206671 
5*a^14*b^2*d^4 + 1062882*a^10*b^6*d^4 - 295245*a^8*b^8*d^4 + 984150*a^8*b^ 
4*d^2 - 98415*a^6*b^6*d^2 + 15625*a^4*b^4 - 2000*a^2*b^6 + 64*b^8, d, k)*( 
tan(c/2 + (d*x)/2)*(764411904*a^6*b^40 - 27805483008*a^8*b^38 + 4372973568 
00*a^10*b^36 - 3672461721600*a^12*b^34 + 19250011791360*a^14*b^32 - 691506 
35753472*a^16*b^30 + 180165872001024*a^18*b^28 - 352655758540800*a^20*b^26 
 + 529923028377600*a^22*b^24 - 618699706859520*a^24*b^22 + 563713761042432 
*a^26*b^20 - 399760062234624*a^28*b^18 + 218398602240000*a^30*b^16 - 90108 
039168000*a^32*b^14 + 27130620764160*a^34*b^12 - 5617221156864*a^36*b^10 + 
 713536708608*a^38*b^8 - 41803776000*a^40*b^6) - root(5314410*a^16*b^4*d^6 
 - 5314410*a^14*b^6*d^6 - 2657205*a^18*b^2*d^6 + 2657205*a^12*b^8*d^6 - 53 
1441*a^10*b^10*d^6 + 531441*a^20*d^6 + 11514555*a^12*b^4*d^4 + 2066715*a^1 
4*b^2*d^4 + 1062882*a^10*b^6*d^4 - 295245*a^8*b^8*d^4 + 984150*a^8*b^4*d^2 
 - 98415*a^6*b^6*d^2 + 15625*a^4*b^4 - 2000*a^2*b^6 + 64*b^8, d, k)*(root( 
5314410*a^16*b^4*d^6 - 5314410*a^14*b^6*d^6 - 2657205*a^18*b^2*d^6 + 26572 
05*a^12*b^8*d^6 - 531441*a^10*b^10*d^6 + 531441*a^20*d^6 + 11514555*a^12*b 
^4*d^4 + 2066715*a^14*b^2*d^4 + 1062882*a^10*b^6*d^4 - 295245*a^8*b^8*d^4 
+ 984150*a^8*b^4*d^2 - 98415*a^6*b^6*d^2 + 15625*a^4*b^4 - 2000*a^2*b^6...