Integrand size = 23, antiderivative size = 1093 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx =\text {Too large to display} \] Output:
-2*(-1)^(2/3)*a^(2/3)*b^(8/3)*arctan(((-1)^(1/3)*b^(1/3)-a^(1/3)*tan(1/2*d *x+1/2*c))/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/(a^(2/3)-(-1)^(2/3)*b^(2/3) )^(1/2)/(a^2-b^2)^2/d-2/3*b^2*(2*a^2+b^2)*arctan(((-1)^(1/3)*b^(1/3)-a^(1/ 3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-(-1)^(2/3)*b^(2/3))^(1/2))/a^(2/3)/(a^(2/3 )-(-1)^(2/3)*b^(2/3))^(1/2)/(a^2-b^2)^2/d+2*a^(2/3)*b^(8/3)*arctan((b^(1/3 )+a^(1/3)*tan(1/2*d*x+1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/(a^(2/3)-b^(2/3))^( 1/2)/(a^2-b^2)^2/d+2/3*b^2*(2*a^2+b^2)*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x +1/2*c))/(a^(2/3)-b^(2/3))^(1/2))/a^(2/3)/(a^(2/3)-b^(2/3))^(1/2)/(a^2-b^2 )^2/d+2/3*b^(4/3)*(a^2+2*b^2)*arctan((b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/ (a^(2/3)-b^(2/3))^(1/2))/(a^(2/3)-b^(2/3))^(1/2)/(a^2-b^2)^2/d-2*(-1)^(1/3 )*a^(2/3)*b^(8/3)*arctan(((-1)^(2/3)*b^(1/3)+a^(1/3)*tan(1/2*d*x+1/2*c))/( a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2))/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2)/(a^2 -b^2)^2/d+2/3*b^2*(2*a^2+b^2)*arctan(((-1)^(2/3)*b^(1/3)+a^(1/3)*tan(1/2*d *x+1/2*c))/(a^(2/3)+(-1)^(1/3)*b^(2/3))^(1/2))/a^(2/3)/(a^(2/3)+(-1)^(1/3) *b^(2/3))^(1/2)/(a^2-b^2)^2/d-2/3*b^(4/3)*(a^2+2*b^2)*arctanh((b^(1/3)-(-1 )^(1/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/(-(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2))/( -(-1)^(2/3)*a^(2/3)+b^(2/3))^(1/2)/(a^2-b^2)^2/d-2/3*b^(4/3)*(a^2+2*b^2)*a rctanh((b^(1/3)+(-1)^(2/3)*a^(1/3)*tan(1/2*d*x+1/2*c))/((-1)^(1/3)*a^(2/3) +b^(2/3))^(1/2))/((-1)^(1/3)*a^(2/3)+b^(2/3))^(1/2)/(a^2-b^2)^2/d+1/12*cos (d*x+c)/(a+b)/d/(1-sin(d*x+c))^2+1/12*cos(d*x+c)/(a+b)/d/(1-sin(d*x+c))...
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 2.28 (sec) , antiderivative size = 679, normalized size of antiderivative = 0.62 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\frac {4 i b^2 \text {RootSum}\left [-i b+3 i b \text {$\#$1}^2+8 a \text {$\#$1}^3-3 i b \text {$\#$1}^4+i b \text {$\#$1}^6\&,\frac {2 a^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )+4 b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right )-i a^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )-2 i b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right )+12 i a b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}+6 a b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}-20 a^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2-16 b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^2+10 i a^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2+8 i b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^2-12 i a b \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^3-6 a b \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^3+2 a^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4+4 b^2 \arctan \left (\frac {\sin (c+d x)}{\cos (c+d x)-\text {$\#$1}}\right ) \text {$\#$1}^4-i a^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4-2 i b^2 \log \left (1-2 \cos (c+d x) \text {$\#$1}+\text {$\#$1}^2\right ) \text {$\#$1}^4}{b \text {$\#$1}-4 i a \text {$\#$1}^2-2 b \text {$\#$1}^3+b \text {$\#$1}^5}\&\right ]+\sec ^3(c+d x) \left (4 a^2 b+32 b^3-3 b \left (5 a^2+13 b^2\right ) \cos (c+d x)+12 b \left (a^2+2 b^2\right ) \cos (2 (c+d x))-5 a^2 b \cos (3 (c+d x))-13 b^3 \cos (3 (c+d x))+12 a^3 \sin (c+d x)-30 a b^2 \sin (c+d x)+4 a^3 \sin (3 (c+d x))-22 a b^2 \sin (3 (c+d x))\right )}{24 (a-b)^2 (a+b)^2 d} \] Input:
Integrate[Sec[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]
Output:
((4*I)*b^2*RootSum[(-I)*b + (3*I)*b*#1^2 + 8*a*#1^3 - (3*I)*b*#1^4 + I*b*# 1^6 & , (2*a^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)] + 4*b^2*ArcTan[Sin [c + d*x]/(Cos[c + d*x] - #1)] - I*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] - (2*I)*b^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2] + (12*I)*a*b*ArcTan[Sin[c + d *x]/(Cos[c + d*x] - #1)]*#1 + 6*a*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1 - 20*a^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^2 - 16*b^2*ArcTan[Sin[ c + d*x]/(Cos[c + d*x] - #1)]*#1^2 + (10*I)*a^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 + (8*I)*b^2*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^2 - (12*I)*a *b*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^3 - 6*a*b*Log[1 - 2*Cos[c + d*x]*#1 + #1^2]*#1^3 + 2*a^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^ 4 + 4*b^2*ArcTan[Sin[c + d*x]/(Cos[c + d*x] - #1)]*#1^4 - I*a^2*Log[1 - 2* Cos[c + d*x]*#1 + #1^2]*#1^4 - (2*I)*b^2*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) & ] + Sec[c + d*x]^3*(4*a ^2*b + 32*b^3 - 3*b*(5*a^2 + 13*b^2)*Cos[c + d*x] + 12*b*(a^2 + 2*b^2)*Cos [2*(c + d*x)] - 5*a^2*b*Cos[3*(c + d*x)] - 13*b^3*Cos[3*(c + d*x)] + 12*a^ 3*Sin[c + d*x] - 30*a*b^2*Sin[c + d*x] + 4*a^3*Sin[3*(c + d*x)] - 22*a*b^2 *Sin[3*(c + d*x)]))/(24*(a - b)^2*(a + b)^2*d)
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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cos (c+d x)^4 \left (a+b \sin (c+d x)^3\right )}dx\) |
| \(\Big \downarrow \) 3707 |
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)}dx\) |
Input:
Int[Sec[c + d*x]^4/(a + b*Sin[c + d*x]^3),x]
Output:
$Aborted
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]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.72 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.27
| method | result | size |
| derivativedivides | \(\frac {\frac {b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\left (2 a^{2}+b^{2}\right ) \textit {\_R}^{4}-6 a b \,\textit {\_R}^{3}+2 \left (4 a^{2}+5 b^{2}\right ) \textit {\_R}^{2}-6 a \textit {\_R} b +2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{\left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5 b +2 a}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a -5 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(291\) |
| default | \(\frac {\frac {b^{2} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (\left (2 a^{2}+b^{2}\right ) \textit {\_R}^{4}-6 a b \,\textit {\_R}^{3}+2 \left (4 a^{2}+5 b^{2}\right ) \textit {\_R}^{2}-6 a \textit {\_R} b +2 a^{2}+b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +a \textit {\_R}}\right )}{3 \left (a -b \right )^{2} \left (a +b \right )^{2}}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (2 a +2 b \right )}-\frac {1}{\left (2 a +2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5 b +2 a}{2 \left (a +b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \left (2 a -2 b \right )}+\frac {1}{\left (2 a -2 b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {2 a -5 b}{2 \left (a -b \right )^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d}\) | \(291\) |
| risch | \(\text {Expression too large to display}\) | \(2867\) |
Input:
int(sec(d*x+c)^4/(a+b*sin(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(1/3*b^2/(a-b)^2/(a+b)^2*sum(((2*a^2+b^2)*_R^4-6*a*b*_R^3+2*(4*a^2+5*b ^2)*_R^2-6*a*_R*b+2*a^2+b^2)/(_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))-2/3/(tan(1/2* d*x+1/2*c)-1)^3/(2*a+2*b)-1/(2*a+2*b)/(tan(1/2*d*x+1/2*c)-1)^2-1/2*(5*b+2* a)/(a+b)^2/(tan(1/2*d*x+1/2*c)-1)-2/3/(tan(1/2*d*x+1/2*c)+1)^3/(2*a-2*b)+1 /(2*a-2*b)/(tan(1/2*d*x+1/2*c)+1)^2-1/2*(2*a-5*b)/(a-b)^2/(tan(1/2*d*x+1/2 *c)+1))
Result contains complex when optimal does not.
Time = 11.05 (sec) , antiderivative size = 85064, normalized size of antiderivative = 77.83 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sec(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{a + b \sin ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(sec(d*x+c)**4/(a+b*sin(d*x+c)**3),x)
Output:
Integral(sec(c + d*x)**4/(a + b*sin(c + d*x)**3), x)
Exception generated. \[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(sec(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
\[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{b \sin \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sec(d*x+c)^4/(a+b*sin(d*x+c)^3),x, algorithm="giac")
Output:
sage0*x
Time = 47.76 (sec) , antiderivative size = 323390, normalized size of antiderivative = 295.87 \[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
int(1/(cos(c + d*x)^4*(a + b*sin(c + d*x)^3)),x)
Output:
(14*b^3*cos(c/2 + (d*x)/2)^6)/(3*(a^4*d*cos(c/2 + (d*x)/2)^6 + b^4*d*cos(c /2 + (d*x)/2)^6 - a^4*d*sin(c/2 + (d*x)/2)^6 - b^4*d*sin(c/2 + (d*x)/2)^6 - 2*a^2*b^2*d*cos(c/2 + (d*x)/2)^6 + 2*a^2*b^2*d*sin(c/2 + (d*x)/2)^6 + 3* a^4*d*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4 - 3*a^4*d*cos(c/2 + (d*x)/ 2)^4*sin(c/2 + (d*x)/2)^2 + 3*b^4*d*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2 )^4 - 3*b^4*d*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^2 - 6*a^2*b^2*d*cos( c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4 + 6*a^2*b^2*d*cos(c/2 + (d*x)/2)^4*s in(c/2 + (d*x)/2)^2)) - (4*a^3*cos(c/2 + (d*x)/2)^3*sin(c/2 + (d*x)/2)^3)/ (3*(a^4*d*cos(c/2 + (d*x)/2)^6 + b^4*d*cos(c/2 + (d*x)/2)^6 - a^4*d*sin(c/ 2 + (d*x)/2)^6 - b^4*d*sin(c/2 + (d*x)/2)^6 - 2*a^2*b^2*d*cos(c/2 + (d*x)/ 2)^6 + 2*a^2*b^2*d*sin(c/2 + (d*x)/2)^6 + 3*a^4*d*cos(c/2 + (d*x)/2)^2*sin (c/2 + (d*x)/2)^4 - 3*a^4*d*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^2 + 3* b^4*d*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4 - 3*b^4*d*cos(c/2 + (d*x)/ 2)^4*sin(c/2 + (d*x)/2)^2 - 6*a^2*b^2*d*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d* x)/2)^4 + 6*a^2*b^2*d*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^2)) + (6*b^3 *cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4)/(a^4*d*cos(c/2 + (d*x)/2)^6 + b^4*d*cos(c/2 + (d*x)/2)^6 - a^4*d*sin(c/2 + (d*x)/2)^6 - b^4*d*sin(c/2 + (d*x)/2)^6 - 2*a^2*b^2*d*cos(c/2 + (d*x)/2)^6 + 2*a^2*b^2*d*sin(c/2 + (d*x )/2)^6 + 3*a^4*d*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2)^4 - 3*a^4*d*cos(c /2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)^2 + 3*b^4*d*cos(c/2 + (d*x)/2)^2*sin...
\[ \int \frac {\sec ^4(c+d x)}{a+b \sin ^3(c+d x)} \, dx=\text {too large to display} \] Input:
int(sec(d*x+c)^4/(a+b*sin(d*x+c)^3),x)
Output:
(4416*cos(c + d*x)*int(tan((c + d*x)/2)**4/(tan((c + d*x)/2)**14*a - tan(( c + d*x)/2)**12*a + 8*tan((c + d*x)/2)**11*b - 3*tan((c + d*x)/2)**10*a - 32*tan((c + d*x)/2)**9*b + 3*tan((c + d*x)/2)**8*a + 48*tan((c + d*x)/2)** 7*b + 3*tan((c + d*x)/2)**6*a - 32*tan((c + d*x)/2)**5*b - 3*tan((c + d*x) /2)**4*a + 8*tan((c + d*x)/2)**3*b - tan((c + d*x)/2)**2*a + a),x)*sin(c + d*x)**2*a**3*b**2*d + 172032*cos(c + d*x)*int(tan((c + d*x)/2)**4/(tan((c + d*x)/2)**14*a - tan((c + d*x)/2)**12*a + 8*tan((c + d*x)/2)**11*b - 3*t an((c + d*x)/2)**10*a - 32*tan((c + d*x)/2)**9*b + 3*tan((c + d*x)/2)**8*a + 48*tan((c + d*x)/2)**7*b + 3*tan((c + d*x)/2)**6*a - 32*tan((c + d*x)/2 )**5*b - 3*tan((c + d*x)/2)**4*a + 8*tan((c + d*x)/2)**3*b - tan((c + d*x) /2)**2*a + a),x)*sin(c + d*x)**2*a*b**4*d - 4416*cos(c + d*x)*int(tan((c + d*x)/2)**4/(tan((c + d*x)/2)**14*a - tan((c + d*x)/2)**12*a + 8*tan((c + d*x)/2)**11*b - 3*tan((c + d*x)/2)**10*a - 32*tan((c + d*x)/2)**9*b + 3*ta n((c + d*x)/2)**8*a + 48*tan((c + d*x)/2)**7*b + 3*tan((c + d*x)/2)**6*a - 32*tan((c + d*x)/2)**5*b - 3*tan((c + d*x)/2)**4*a + 8*tan((c + d*x)/2)** 3*b - tan((c + d*x)/2)**2*a + a),x)*a**3*b**2*d - 172032*cos(c + d*x)*int( tan((c + d*x)/2)**4/(tan((c + d*x)/2)**14*a - tan((c + d*x)/2)**12*a + 8*t an((c + d*x)/2)**11*b - 3*tan((c + d*x)/2)**10*a - 32*tan((c + d*x)/2)**9* b + 3*tan((c + d*x)/2)**8*a + 48*tan((c + d*x)/2)**7*b + 3*tan((c + d*x)/2 )**6*a - 32*tan((c + d*x)/2)**5*b - 3*tan((c + d*x)/2)**4*a + 8*tan((c ...