Integrand size = 20, antiderivative size = 1523 \[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx =\text {Too large to display} \] Output:
-6*I*b^2*d^2*(d*x+c)*polylog(2,-a*exp(I*(f*x+e))/(b+I*(a^2-b^2)^(1/2)))/a^ 2/(a^2-b^2)/f^3-6*I*b^2*d^2*(d*x+c)*polylog(2,-a*exp(I*(f*x+e))/(b-I*(a^2- b^2)^(1/2)))/a^2/(a^2-b^2)/f^3-6*I*b^3*d^2*(d*x+c)*polylog(3,-a*exp(I*(f*x +e))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/f^3-12*I*b*d^2*(d*x+c)*pol ylog(3,-a*exp(I*(f*x+e))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/f^3+6* I*b^3*d^2*(d*x+c)*polylog(3,-a*exp(I*(f*x+e))/(b+(-a^2+b^2)^(1/2)))/a^2/(- a^2+b^2)^(3/2)/f^3+12*I*b*d^2*(d*x+c)*polylog(3,-a*exp(I*(f*x+e))/(b-(-a^2 +b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/f^3+6*b^3*d^3*polylog(4,-a*exp(I*(f*x+e ))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(3/2)/f^4-12*b*d^3*polylog(4,-a*ex p(I*(f*x+e))/(b-(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/f^4+12*b*d^3*polyl og(4,-a*exp(I*(f*x+e))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^(1/2)/f^4+6*b^ 2*d^3*polylog(3,-a*exp(I*(f*x+e))/(b+I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/f^4 +6*b^2*d^3*polylog(3,-a*exp(I*(f*x+e))/(b-I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2 )/f^4-6*b^3*d^3*polylog(4,-a*exp(I*(f*x+e))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^ 2+b^2)^(3/2)/f^4-I*b^2*(d*x+c)^3/a^2/(a^2-b^2)/f+b^2*(d*x+c)^3*sin(f*x+e)/ a/(a^2-b^2)/f/(b+a*cos(f*x+e))+1/4*(d*x+c)^4/a^2/d+3*b^2*d*(d*x+c)^2*ln(1+ a*exp(I*(f*x+e))/(b+I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/f^2+3*b^2*d*(d*x+c)^ 2*ln(1+a*exp(I*(f*x+e))/(b-I*(a^2-b^2)^(1/2)))/a^2/(a^2-b^2)/f^2+3*b^3*d*( d*x+c)^2*polylog(2,-a*exp(I*(f*x+e))/(b+(-a^2+b^2)^(1/2)))/a^2/(-a^2+b^2)^ (3/2)/f^2-3*b^3*d*(d*x+c)^2*polylog(2,-a*exp(I*(f*x+e))/(b-(-a^2+b^2)^(...
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(20116\) vs. \(2(1523)=3046\).
Time = 19.83 (sec) , antiderivative size = 20116, normalized size of antiderivative = 13.21 \[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx=\text {Result too large to show} \] Input:
Integrate[(c + d*x)^3/(a + b*Sec[e + f*x])^2,x]
Output:
Result too large to show
Time = 3.30 (sec) , antiderivative size = 1523, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4679, 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.
\(\displaystyle \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c+d x)^3}{\left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4679 |
\(\displaystyle \int \left (\frac {b^2 (c+d x)^3}{a^2 (a \cos (e+f x)+b)^2}-\frac {2 b (c+d x)^3}{a^2 (a \cos (e+f x)+b)}+\frac {(c+d x)^3}{a^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(c+d x)^4}{4 a^2 d}+\frac {2 i b \log \left (\frac {e^{i (e+f x)} a}{b-\sqrt {b^2-a^2}}+1\right ) (c+d x)^3}{a^2 \sqrt {b^2-a^2} f}-\frac {i b^3 \log \left (\frac {e^{i (e+f x)} a}{b-\sqrt {b^2-a^2}}+1\right ) (c+d x)^3}{a^2 \left (b^2-a^2\right )^{3/2} f}-\frac {2 i b \log \left (\frac {e^{i (e+f x)} a}{b+\sqrt {b^2-a^2}}+1\right ) (c+d x)^3}{a^2 \sqrt {b^2-a^2} f}+\frac {i b^3 \log \left (\frac {e^{i (e+f x)} a}{b+\sqrt {b^2-a^2}}+1\right ) (c+d x)^3}{a^2 \left (b^2-a^2\right )^{3/2} f}+\frac {b^2 \sin (e+f x) (c+d x)^3}{a \left (a^2-b^2\right ) f (b+a \cos (e+f x))}-\frac {i b^2 (c+d x)^3}{a^2 \left (a^2-b^2\right ) f}+\frac {3 b^2 d \log \left (\frac {e^{i (e+f x)} a}{b-i \sqrt {a^2-b^2}}+1\right ) (c+d x)^2}{a^2 \left (a^2-b^2\right ) f^2}+\frac {3 b^2 d \log \left (\frac {e^{i (e+f x)} a}{b+i \sqrt {a^2-b^2}}+1\right ) (c+d x)^2}{a^2 \left (a^2-b^2\right ) f^2}+\frac {6 b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right ) (c+d x)^2}{a^2 \sqrt {b^2-a^2} f^2}-\frac {3 b^3 d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right ) (c+d x)^2}{a^2 \left (b^2-a^2\right )^{3/2} f^2}-\frac {6 b d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right ) (c+d x)^2}{a^2 \sqrt {b^2-a^2} f^2}+\frac {3 b^3 d \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right ) (c+d x)^2}{a^2 \left (b^2-a^2\right )^{3/2} f^2}-\frac {6 i b^2 d^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b-i \sqrt {a^2-b^2}}\right ) (c+d x)}{a^2 \left (a^2-b^2\right ) f^3}-\frac {6 i b^2 d^2 \operatorname {PolyLog}\left (2,-\frac {a e^{i (e+f x)}}{b+i \sqrt {a^2-b^2}}\right ) (c+d x)}{a^2 \left (a^2-b^2\right ) f^3}+\frac {12 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right ) (c+d x)}{a^2 \sqrt {b^2-a^2} f^3}-\frac {6 i b^3 d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right ) (c+d x)}{a^2 \left (b^2-a^2\right )^{3/2} f^3}-\frac {12 i b d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right ) (c+d x)}{a^2 \sqrt {b^2-a^2} f^3}+\frac {6 i b^3 d^2 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right ) (c+d x)}{a^2 \left (b^2-a^2\right )^{3/2} f^3}+\frac {6 b^2 d^3 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b-i \sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right ) f^4}+\frac {6 b^2 d^3 \operatorname {PolyLog}\left (3,-\frac {a e^{i (e+f x)}}{b+i \sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right ) f^4}-\frac {12 b d^3 \operatorname {PolyLog}\left (4,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a^2 \sqrt {b^2-a^2} f^4}+\frac {6 b^3 d^3 \operatorname {PolyLog}\left (4,-\frac {a e^{i (e+f x)}}{b-\sqrt {b^2-a^2}}\right )}{a^2 \left (b^2-a^2\right )^{3/2} f^4}+\frac {12 b d^3 \operatorname {PolyLog}\left (4,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a^2 \sqrt {b^2-a^2} f^4}-\frac {6 b^3 d^3 \operatorname {PolyLog}\left (4,-\frac {a e^{i (e+f x)}}{b+\sqrt {b^2-a^2}}\right )}{a^2 \left (b^2-a^2\right )^{3/2} f^4}\) |
Input:
Int[(c + d*x)^3/(a + b*Sec[e + f*x])^2,x]
Output:
((-I)*b^2*(c + d*x)^3)/(a^2*(a^2 - b^2)*f) + (c + d*x)^4/(4*a^2*d) + (3*b^ 2*d*(c + d*x)^2*Log[1 + (a*E^(I*(e + f*x)))/(b - I*Sqrt[a^2 - b^2])])/(a^2 *(a^2 - b^2)*f^2) + (3*b^2*d*(c + d*x)^2*Log[1 + (a*E^(I*(e + f*x)))/(b + I*Sqrt[a^2 - b^2])])/(a^2*(a^2 - b^2)*f^2) - (I*b^3*(c + d*x)^3*Log[1 + (a *E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*f) + (( 2*I)*b*(c + d*x)^3*Log[1 + (a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2])])/(a ^2*Sqrt[-a^2 + b^2]*f) + (I*b^3*(c + d*x)^3*Log[1 + (a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2])])/(a^2*(-a^2 + b^2)^(3/2)*f) - ((2*I)*b*(c + d*x)^3*L og[1 + (a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2])])/(a^2*Sqrt[-a^2 + b^2]* f) - ((6*I)*b^2*d^2*(c + d*x)*PolyLog[2, -((a*E^(I*(e + f*x)))/(b - I*Sqrt [a^2 - b^2]))])/(a^2*(a^2 - b^2)*f^3) - ((6*I)*b^2*d^2*(c + d*x)*PolyLog[2 , -((a*E^(I*(e + f*x)))/(b + I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*f^3) - (3*b^3*d*(c + d*x)^2*PolyLog[2, -((a*E^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^ 2]))])/(a^2*(-a^2 + b^2)^(3/2)*f^2) + (6*b*d*(c + d*x)^2*PolyLog[2, -((a*E ^(I*(e + f*x)))/(b - Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*f^2) + (3* b^3*d*(c + d*x)^2*PolyLog[2, -((a*E^(I*(e + f*x)))/(b + Sqrt[-a^2 + b^2])) ])/(a^2*(-a^2 + b^2)^(3/2)*f^2) - (6*b*d*(c + d*x)^2*PolyLog[2, -((a*E^(I* (e + f*x)))/(b + Sqrt[-a^2 + b^2]))])/(a^2*Sqrt[-a^2 + b^2]*f^2) + (6*b^2* d^3*PolyLog[3, -((a*E^(I*(e + f*x)))/(b - I*Sqrt[a^2 - b^2]))])/(a^2*(a^2 - b^2)*f^4) + (6*b^2*d^3*PolyLog[3, -((a*E^(I*(e + f*x)))/(b + I*Sqrt[a...
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt Q[m, 0]
\[\int \frac {\left (d x +c \right )^{3}}{\left (a +b \sec \left (f x +e \right )\right )^{2}}d x\]
Input:
int((d*x+c)^3/(a+b*sec(f*x+e))^2,x)
Output:
int((d*x+c)^3/(a+b*sec(f*x+e))^2,x)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7008 vs. \(2 (1361) = 2722\).
Time = 0.49 (sec) , antiderivative size = 7008, normalized size of antiderivative = 4.60 \[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx=\text {Too large to display} \] Input:
integrate((d*x+c)^3/(a+b*sec(f*x+e))^2,x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx=\int \frac {\left (c + d x\right )^{3}}{\left (a + b \sec {\left (e + f x \right )}\right )^{2}}\, dx \] Input:
integrate((d*x+c)**3/(a+b*sec(f*x+e))**2,x)
Output:
Integral((c + d*x)**3/(a + b*sec(e + f*x))**2, x)
Exception generated. \[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((d*x+c)^3/(a+b*sec(f*x+e))^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(4*a^2-4*b^2>0)', see `assume?` f or more de
\[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx=\int { \frac {{\left (d x + c\right )}^{3}}{{\left (b \sec \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:
integrate((d*x+c)^3/(a+b*sec(f*x+e))^2,x, algorithm="giac")
Output:
integrate((d*x + c)^3/(b*sec(f*x + e) + a)^2, x)
Timed out. \[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx=\text {Hanged} \] Input:
int((c + d*x)^3/(a + b/cos(e + f*x))^2,x)
Output:
\text{Hanged}
\[ \int \frac {(c+d x)^3}{(a+b \sec (e+f x))^2} \, dx=\text {too large to display} \] Input:
int((d*x+c)^3/(a+b*sec(f*x+e))^2,x)
Output:
( - 64*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b) /sqrt( - a**2 + b**2))*cos(e + f*x)*a**5*b*c**3*f**3 - 192*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2)) *cos(e + f*x)*a**5*b*c*d**2*f - 64*sqrt( - a**2 + b**2)*atan((tan((e + f*x )/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2))*cos(e + f*x)*a**4*b**2* c**3*f**3 - 96*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f* x)/2)*b)/sqrt( - a**2 + b**2))*cos(e + f*x)*a**4*b**2*c*d**2*f + 16*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2))*cos(e + f*x)*a**3*b**3*c**3*f**3 + 192*sqrt( - a**2 + b**2)*atan ((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2))*cos(e + f *x)*a**3*b**3*c*d**2*f + 32*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2))*cos(e + f*x)*a**2*b**4*c**3*f* *3 + 96*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b )/sqrt( - a**2 + b**2))*cos(e + f*x)*a**2*b**4*c*d**2*f + 8*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2) )*cos(e + f*x)*a*b**5*c**3*f**3 - 64*sqrt( - a**2 + b**2)*atan((tan((e + f *x)/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2))*a**4*b**2*c**3*f**3 - 192*sqrt( - a**2 + b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/s qrt( - a**2 + b**2))*a**4*b**2*c*d**2*f - 64*sqrt( - a**2 + b**2)*atan((ta n((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt( - a**2 + b**2))*a**3*b**3*...