Integrand size = 23, antiderivative size = 74 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{5/2} d}-\frac {a \tan (c+d x)}{(a+b)^2 d}+\frac {\tan ^3(c+d x)}{3 (a+b) d} \] Output:
a^(3/2)*arctan((a+b)^(1/2)*tan(d*x+c)/a^(1/2))/(a+b)^(5/2)/d-a*tan(d*x+c)/ (a+b)^2/d+1/3*tan(d*x+c)^3/(a+b)/d
Time = 0.60 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {3 a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a+b} \left (-4 a-b+(a+b) \sec ^2(c+d x)\right ) \tan (c+d x)}{3 (a+b)^{5/2} d} \] Input:
Integrate[Tan[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]
Output:
(3*a^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]] + Sqrt[a + b]*(-4*a - b + (a + b)*Sec[c + d*x]^2)*Tan[c + d*x])/(3*(a + b)^(5/2)*d)
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3674, 254, 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.
\(\displaystyle \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x)^4}{a+b \sin (c+d x)^2}dx\) |
\(\Big \downarrow \) 3674 |
\(\displaystyle \frac {\int \frac {\tan ^4(c+d x)}{(a+b) \tan ^2(c+d x)+a}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\int \left (\frac {a^2}{(a+b)^2 \left ((a+b) \tan ^2(c+d x)+a\right )}-\frac {a}{(a+b)^2}+\frac {\tan ^2(c+d x)}{a+b}\right )d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{5/2}}+\frac {\tan ^3(c+d x)}{3 (a+b)}-\frac {a \tan (c+d x)}{(a+b)^2}}{d}\) |
Input:
Int[Tan[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]
Output:
((a^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/(a + b)^(5/2) - (a*T an[c + d*x])/(a + b)^2 + Tan[c + d*x]^3/(3*(a + b)))/d
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*((d_.)*tan[(e_.) + (f_.) *(x_)])^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[f f/f Subst[Int[(d*ff*x)^m*((a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(p + 1) ), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m}, x] && IntegerQ [p]
Time = 1.73 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05
method | result | size |
derivativedivides | \(\frac {\frac {\frac {a \tan \left (d x +c \right )^{3}}{3}+\frac {\tan \left (d x +c \right )^{3} b}{3}-\tan \left (d x +c \right ) a}{\left (a +b \right )^{2}}+\frac {a^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{2} \sqrt {a \left (a +b \right )}}}{d}\) | \(78\) |
default | \(\frac {\frac {\frac {a \tan \left (d x +c \right )^{3}}{3}+\frac {\tan \left (d x +c \right )^{3} b}{3}-\tan \left (d x +c \right ) a}{\left (a +b \right )^{2}}+\frac {a^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{2} \sqrt {a \left (a +b \right )}}}{d}\) | \(78\) |
risch | \(-\frac {2 i \left (6 \,{\mathrm e}^{4 i \left (d x +c \right )} a +3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a +b \right )}{3 d \left (a +b \right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{3} d}-\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{3} d}\) | \(170\) |
Input:
int(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(1/(a+b)^2*(1/3*a*tan(d*x+c)^3+1/3*tan(d*x+c)^3*b-tan(d*x+c)*a)+a^2/(a +b)^2/(a*(a+b))^(1/2)*arctan((a+b)*tan(d*x+c)/(a*(a+b))^(1/2)))
Time = 0.11 (sec) , antiderivative size = 366, normalized size of antiderivative = 4.95 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [\frac {3 \, a \sqrt {-\frac {a}{a + b}} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \, {\left ({\left (4 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, a \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (4 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{3}}\right ] \] Input:
integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="fricas")
Output:
[1/12*(3*a*sqrt(-a/(a + b))*cos(d*x + c)^3*log(((8*a^2 + 8*a*b + b^2)*cos( d*x + c)^4 - 2*(4*a^2 + 5*a*b + b^2)*cos(d*x + c)^2 - 4*((2*a^2 + 3*a*b + b^2)*cos(d*x + c)^3 - (a^2 + 2*a*b + b^2)*cos(d*x + c))*sqrt(-a/(a + b))*s in(d*x + c) + a^2 + 2*a*b + b^2)/(b^2*cos(d*x + c)^4 - 2*(a*b + b^2)*cos(d *x + c)^2 + a^2 + 2*a*b + b^2)) - 4*((4*a + b)*cos(d*x + c)^2 - a - b)*sin (d*x + c))/((a^2 + 2*a*b + b^2)*d*cos(d*x + c)^3), -1/6*(3*a*sqrt(a/(a + b ))*arctan(1/2*((2*a + b)*cos(d*x + c)^2 - a - b)*sqrt(a/(a + b))/(a*cos(d* x + c)*sin(d*x + c)))*cos(d*x + c)^3 + 2*((4*a + b)*cos(d*x + c)^2 - a - b )*sin(d*x + c))/((a^2 + 2*a*b + b^2)*d*cos(d*x + c)^3)]
\[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(tan(d*x+c)**4/(a+b*sin(d*x+c)**2),x)
Output:
Integral(tan(c + d*x)**4/(a + b*sin(c + d*x)**2), x)
Time = 0.11 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {3 \, a^{2} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {{\left (a + b\right )} \tan \left (d x + c\right )^{3} - 3 \, a \tan \left (d x + c\right )}{a^{2} + 2 \, a b + b^{2}}}{3 \, d} \] Input:
integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="maxima")
Output:
1/3*(3*a^2*arctan((a + b)*tan(d*x + c)/sqrt((a + b)*a))/(sqrt((a + b)*a)*( a^2 + 2*a*b + b^2)) + ((a + b)*tan(d*x + c)^3 - 3*a*tan(d*x + c))/(a^2 + 2 *a*b + b^2))/d
Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (64) = 128\).
Time = 0.77 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.30 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a^{2} \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )}{{\left (a^{2} d + 2 \, a b d + b^{2} d\right )} \sqrt {a^{2} + a b}} + \frac {a^{2} d^{2} \tan \left (d x + c\right )^{3} + 2 \, a b d^{2} \tan \left (d x + c\right )^{3} + b^{2} d^{2} \tan \left (d x + c\right )^{3} - 3 \, a^{2} d^{2} \tan \left (d x + c\right ) - 3 \, a b d^{2} \tan \left (d x + c\right )}{3 \, {\left (a^{3} d^{3} + 3 \, a^{2} b d^{3} + 3 \, a b^{2} d^{3} + b^{3} d^{3}\right )}} \] Input:
integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="giac")
Output:
a^2*arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b))/((a^2*d + 2* a*b*d + b^2*d)*sqrt(a^2 + a*b)) + 1/3*(a^2*d^2*tan(d*x + c)^3 + 2*a*b*d^2* tan(d*x + c)^3 + b^2*d^2*tan(d*x + c)^3 - 3*a^2*d^2*tan(d*x + c) - 3*a*b*d ^2*tan(d*x + c))/(a^3*d^3 + 3*a^2*b*d^3 + 3*a*b^2*d^3 + b^3*d^3)
Time = 34.69 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d\,\left (a+b\right )}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\left (a+b\right )}^2}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )}{d\,{\left (a+b\right )}^{5/2}} \] Input:
int(tan(c + d*x)^4/(a + b*sin(c + d*x)^2),x)
Output:
tan(c + d*x)^3/(3*d*(a + b)) - (a*tan(c + d*x))/(d*(a + b)^2) + (a^(3/2)*a tan((tan(c + d*x)*(2*a + 2*b)*(2*a*b + a^2 + b^2))/(2*a^(1/2)*(a + b)^(5/2 ))))/(d*(a + b)^(5/2))
Time = 0.41 (sec) , antiderivative size = 1290, normalized size of antiderivative = 17.43 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x)
Output:
( - 6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*at an((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*c os(c + d*x)*sin(c + d*x)**2 + 6*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b) *sqrt(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)* sqrt(a + b) + a + 2*b)))*cos(c + d*x) + 6*sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b ) + a + 2*b)))*cos(c + d*x)*sin(c + d*x)**2*a + 6*sqrt(a)*sqrt(2*sqrt(b)*s qrt(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sq rt(a + b) + a + 2*b)))*cos(c + d*x)*sin(c + d*x)**2*b - 6*sqrt(a)*sqrt(2*s qrt(b)*sqrt(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sq rt(b)*sqrt(a + b) + a + 2*b)))*cos(c + d*x)*a - 6*sqrt(a)*sqrt(2*sqrt(b)*s qrt(a + b) + a + 2*b)*atan((tan((c + d*x)/2)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sq rt(a + b) + a + 2*b)))*cos(c + d*x)*b - 3*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt (2*sqrt(b)*sqrt(a + b) - a - 2*b)*cos(c + d*x)*log( - sqrt(2*sqrt(b)*sqrt( a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**2 + 3*sqrt(b)* sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b)*cos(c + d*x)*log ( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2)) + 3* sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b)*cos(c + d*x)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2)) *sin(c + d*x)**2 - 3*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a ...