Integrand size = 23, antiderivative size = 79 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {a-b} d}+\frac {\sin (c+d x)}{2 a d \left (a-(a-b) \sin ^2(c+d x)\right )} \] Output:
1/2*arctanh((a-b)^(1/2)*sin(d*x+c)/a^(1/2))/a^(3/2)/(a-b)^(1/2)/d+1/2*sin( d*x+c)/a/d/(a-(a-b)*sin(d*x+c)^2)
Time = 0.17 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {\text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{\sqrt {a-b}}+\frac {\sqrt {a} \sin (c+d x)}{a+(-a+b) \sin ^2(c+d x)}}{2 a^{3/2} d} \] Input:
Integrate[Sec[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]
Output:
(ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]]/Sqrt[a - b] + (Sqrt[a]*Sin[c + d*x])/(a + (-a + b)*Sin[c + d*x]^2))/(2*a^(3/2)*d)
Time = 0.25 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4159, 215, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^3}{\left (a+b \tan (c+d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4159 |
\(\displaystyle \frac {\int \frac {1}{\left (a-(a-b) \sin ^2(c+d x)\right )^2}d\sin (c+d x)}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {\int \frac {1}{a-(a-b) \sin ^2(c+d x)}d\sin (c+d x)}{2 a}+\frac {\sin (c+d x)}{2 a \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {\text {arctanh}\left (\frac {\sqrt {a-b} \sin (c+d x)}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {a-b}}+\frac {\sin (c+d x)}{2 a \left (a-(a-b) \sin ^2(c+d x)\right )}}{d}\) |
Input:
Int[Sec[c + d*x]^3/(a + b*Tan[c + d*x]^2)^2,x]
Output:
(ArcTanh[(Sqrt[a - b]*Sin[c + d*x])/Sqrt[a]]/(2*a^(3/2)*Sqrt[a - b]) + Sin [c + d*x]/(2*a*(a - (a - b)*Sin[c + d*x]^2)))/d
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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]
Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ ))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 *x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} , x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
Time = 10.13 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.01
method | result | size |
derivativedivides | \(\frac {-\frac {\sin \left (d x +c \right )}{2 a \left (a \sin \left (d x +c \right )^{2}-b \sin \left (d x +c \right )^{2}-a \right )}+\frac {\operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}}{d}\) | \(80\) |
default | \(\frac {-\frac {\sin \left (d x +c \right )}{2 a \left (a \sin \left (d x +c \right )^{2}-b \sin \left (d x +c \right )^{2}-a \right )}+\frac {\operatorname {arctanh}\left (\frac {\left (a -b \right ) \sin \left (d x +c \right )}{\sqrt {a \left (a -b \right )}}\right )}{2 a \sqrt {a \left (a -b \right )}}}{d}\) | \(80\) |
risch | \(-\frac {i \left ({\mathrm e}^{3 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}\right )}{a d \left (a \,{\mathrm e}^{4 i \left (d x +c \right )}-b \,{\mathrm e}^{4 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{2 i \left (d x +c \right )}+a -b \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{4 \sqrt {a^{2}-a b}\, d a}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{\sqrt {a^{2}-a b}}-1\right )}{4 \sqrt {a^{2}-a b}\, d a}\) | \(192\) |
Input:
int(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/d*(-1/2*sin(d*x+c)/a/(a*sin(d*x+c)^2-b*sin(d*x+c)^2-a)+1/2/a/(a*(a-b))^( 1/2)*arctanh((a-b)*sin(d*x+c)/(a*(a-b))^(1/2)))
Time = 0.13 (sec) , antiderivative size = 266, normalized size of antiderivative = 3.37 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\left [\frac {{\left ({\left (a - b\right )} \cos \left (d x + c\right )^{2} + b\right )} \sqrt {a^{2} - a b} \log \left (-\frac {{\left (a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - a b} \sin \left (d x + c\right ) - 2 \, a + b}{{\left (a - b\right )} \cos \left (d x + c\right )^{2} + b}\right ) + 2 \, {\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a^{2} b^{2}\right )} d\right )}}, -\frac {{\left ({\left (a - b\right )} \cos \left (d x + c\right )^{2} + b\right )} \sqrt {-a^{2} + a b} \arctan \left (\frac {\sqrt {-a^{2} + a b} \sin \left (d x + c\right )}{a}\right ) - {\left (a^{2} - a b\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} d \cos \left (d x + c\right )^{2} + {\left (a^{3} b - a^{2} b^{2}\right )} d\right )}}\right ] \] Input:
integrate(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="fricas")
Output:
[1/4*(((a - b)*cos(d*x + c)^2 + b)*sqrt(a^2 - a*b)*log(-((a - b)*cos(d*x + c)^2 - 2*sqrt(a^2 - a*b)*sin(d*x + c) - 2*a + b)/((a - b)*cos(d*x + c)^2 + b)) + 2*(a^2 - a*b)*sin(d*x + c))/((a^4 - 2*a^3*b + a^2*b^2)*d*cos(d*x + c)^2 + (a^3*b - a^2*b^2)*d), -1/2*(((a - b)*cos(d*x + c)^2 + b)*sqrt(-a^2 + a*b)*arctan(sqrt(-a^2 + a*b)*sin(d*x + c)/a) - (a^2 - a*b)*sin(d*x + c) )/((a^4 - 2*a^3*b + a^2*b^2)*d*cos(d*x + c)^2 + (a^3*b - a^2*b^2)*d)]
\[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \tan ^{2}{\left (c + d x \right )}\right )^{2}}\, dx \] Input:
integrate(sec(d*x+c)**3/(a+b*tan(d*x+c)**2)**2,x)
Output:
Integral(sec(c + d*x)**3/(a + b*tan(c + d*x)**2)**2, x)
Exception generated. \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^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(b-a>0)', see `assume?` for more details)Is
Time = 0.71 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.15 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {\arctan \left (-\frac {a \sin \left (d x + c\right ) - b \sin \left (d x + c\right )}{\sqrt {-a^{2} + a b}}\right )}{\sqrt {-a^{2} + a b} a} - \frac {\sin \left (d x + c\right )}{{\left (a \sin \left (d x + c\right )^{2} - b \sin \left (d x + c\right )^{2} - a\right )} a}}{2 \, d} \] Input:
integrate(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x, algorithm="giac")
Output:
1/2*(arctan(-(a*sin(d*x + c) - b*sin(d*x + c))/sqrt(-a^2 + a*b))/(sqrt(-a^ 2 + a*b)*a) - sin(d*x + c)/((a*sin(d*x + c)^2 - b*sin(d*x + c)^2 - a)*a))/ d
Time = 10.29 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.37 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a}}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (4\,b-2\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {\mathrm {atanh}\left (\frac {4\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^{3/2}\,\sqrt {a-b}\,\left (\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a-b}-\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a}+\frac {2}{a}-\frac {2}{a-b}+\frac {4\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{a\,b-a^2}\right )}\right )}{2\,a^{3/2}\,d\,\sqrt {a-b}} \] Input:
int(1/(cos(c + d*x)^3*(a + b*tan(c + d*x)^2)^2),x)
Output:
(tan(c/2 + (d*x)/2)^3/a + tan(c/2 + (d*x)/2)/a)/(d*(a - tan(c/2 + (d*x)/2) ^2*(2*a - 4*b) + a*tan(c/2 + (d*x)/2)^4)) - atanh((4*b*tan(c/2 + (d*x)/2)) /(a^(3/2)*(a - b)^(1/2)*((2*tan(c/2 + (d*x)/2)^2)/(a - b) - (2*tan(c/2 + ( d*x)/2)^2)/a + 2/a - 2/(a - b) + (4*b*tan(c/2 + (d*x)/2)^2)/(a*b - a^2)))) /(2*a^(3/2)*d*(a - b)^(1/2))
Time = 0.22 (sec) , antiderivative size = 382, normalized size of antiderivative = 4.84 \[ \int \frac {\sec ^3(c+d x)}{\left (a+b \tan ^2(c+d x)\right )^2} \, dx=\frac {-\sqrt {a}\, \sqrt {a -b}\, \mathrm {log}\left (-2 \sqrt {a -b}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {a}\right ) \sin \left (d x +c \right )^{2} a +\sqrt {a}\, \sqrt {a -b}\, \mathrm {log}\left (-2 \sqrt {a -b}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {a}\right ) \sin \left (d x +c \right )^{2} b +\sqrt {a}\, \sqrt {a -b}\, \mathrm {log}\left (-2 \sqrt {a -b}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {a}\right ) a +\sqrt {a}\, \sqrt {a -b}\, \mathrm {log}\left (2 \sqrt {a -b}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {a}\right ) \sin \left (d x +c \right )^{2} a -\sqrt {a}\, \sqrt {a -b}\, \mathrm {log}\left (2 \sqrt {a -b}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {a}\right ) \sin \left (d x +c \right )^{2} b -\sqrt {a}\, \sqrt {a -b}\, \mathrm {log}\left (2 \sqrt {a -b}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {a}\right ) a -2 \sin \left (d x +c \right ) a^{2}+2 \sin \left (d x +c \right ) a b}{4 a^{2} d \left (\sin \left (d x +c \right )^{2} a^{2}-2 \sin \left (d x +c \right )^{2} a b +\sin \left (d x +c \right )^{2} b^{2}-a^{2}+a b \right )} \] Input:
int(sec(d*x+c)^3/(a+b*tan(d*x+c)^2)^2,x)
Output:
( - sqrt(a)*sqrt(a - b)*log( - 2*sqrt(a - b)*tan((c + d*x)/2) + sqrt(a)*ta n((c + d*x)/2)**2 + sqrt(a))*sin(c + d*x)**2*a + sqrt(a)*sqrt(a - b)*log( - 2*sqrt(a - b)*tan((c + d*x)/2) + sqrt(a)*tan((c + d*x)/2)**2 + sqrt(a))* sin(c + d*x)**2*b + sqrt(a)*sqrt(a - b)*log( - 2*sqrt(a - b)*tan((c + d*x) /2) + sqrt(a)*tan((c + d*x)/2)**2 + sqrt(a))*a + sqrt(a)*sqrt(a - b)*log(2 *sqrt(a - b)*tan((c + d*x)/2) + sqrt(a)*tan((c + d*x)/2)**2 + sqrt(a))*sin (c + d*x)**2*a - sqrt(a)*sqrt(a - b)*log(2*sqrt(a - b)*tan((c + d*x)/2) + sqrt(a)*tan((c + d*x)/2)**2 + sqrt(a))*sin(c + d*x)**2*b - sqrt(a)*sqrt(a - b)*log(2*sqrt(a - b)*tan((c + d*x)/2) + sqrt(a)*tan((c + d*x)/2)**2 + sq rt(a))*a - 2*sin(c + d*x)*a**2 + 2*sin(c + d*x)*a*b)/(4*a**2*d*(sin(c + d* x)**2*a**2 - 2*sin(c + d*x)**2*a*b + sin(c + d*x)**2*b**2 - a**2 + a*b))