\(\int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) [451]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{(a+b)^{5/2} d}-\frac {a \tan (c+d x)}{(a+b)^2 d}+\frac {\tan ^3(c+d x)}{3 (a+b) d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3274, 308, 211} \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )}{d (a+b)^{5/2}}+\frac {\tan ^3(c+d x)}{3 d (a+b)}-\frac {a \tan (c+d x)}{d (a+b)^2} \]

[In]

Int[Tan[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

(a^(3/2)*ArcTan[(Sqrt[a + b]*Tan[c + d*x])/Sqrt[a]])/((a + b)^(5/2)*d) - (a*Tan[c + d*x])/((a + b)^2*d) + Tan[
c + d*x]^3/(3*(a + b)*d)

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 3274

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]}, Dist[ff/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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{d} \\ & = \frac {\text {Subst}\left (\int \left (-\frac {a}{(a+b)^2}+\frac {x^2}{a+b}+\frac {a^2}{(a+b)^2 \left (a+(a+b) x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{d} \\ & = -\frac {a \tan (c+d x)}{(a+b)^2 d}+\frac {\tan ^3(c+d x)}{3 (a+b) d}+\frac {a^2 \text {Subst}\left (\int \frac {1}{a+(a+b) x^2} \, dx,x,\tan (c+d x)\right )}{(a+b)^2 d} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {3 a^{3/2} \arctan \left (\frac {\sqrt {a+b} \tan (c+d x)}{\sqrt {a}}\right )+\sqrt {a+b} \left (-4 a-b+(a+b) \sec ^2(c+d x)\right ) \tan (c+d x)}{3 (a+b)^{5/2} d} \]

[In]

Integrate[Tan[c + d*x]^4/(a + b*Sin[c + d*x]^2),x]

[Out]

(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)

Maple [A] (verified)

Time = 1.74 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(\frac {\frac {\frac {a \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\tan ^{3}\left (d x +c \right )\right ) b}{3}-\tan \left (d x +c \right ) a}{\left (a +b \right )^{2}}+\frac {a^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{2} \sqrt {a \left (a +b \right )}}}{d}\)	\(78\)
default	\(\frac {\frac {\frac {a \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {\left (\tan ^{3}\left (d x +c \right )\right ) b}{3}-\tan \left (d x +c \right ) a}{\left (a +b \right )^{2}}+\frac {a^{2} \arctan \left (\frac {\left (a +b \right ) \tan \left (d x +c \right )}{\sqrt {a \left (a +b \right )}}\right )}{\left (a +b \right )^{2} \sqrt {a \left (a +b \right )}}}{d}\)	\(78\)
risch	\(-\frac {2 i \left (6 a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+6 a \,{\mathrm e}^{2 i \left (d x +c \right )}+4 a +b \right )}{3 d \left (a +b \right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i \sqrt {-a \left (a +b \right )}-2 a -b}{b}\right )}{2 \left (a +b \right )^{3} d}-\frac {\sqrt {-a \left (a +b \right )}\, a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {2 i \sqrt {-a \left (a +b \right )}+2 a +b}{b}\right )}{2 \left (a +b \right )^{3} d}\)	\(170\)

[In]

int(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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)*t
an(d*x+c)/(a*(a+b))^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 366, normalized size of antiderivative = 4.95 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\left [\frac {3 \, a \sqrt {-\frac {a}{a + b}} \cos \left (d x + c\right )^{3} \log \left (\frac {{\left (8 \, a^{2} + 8 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (4 \, a^{2} + 5 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left ({\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{3} - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {-\frac {a}{a + b}} \sin \left (d x + c\right ) + a^{2} + 2 \, a b + b^{2}}{b^{2} \cos \left (d x + c\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}}\right ) - 4 \, {\left ({\left (4 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{3}}, -\frac {3 \, a \sqrt {\frac {a}{a + b}} \arctan \left (\frac {{\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sqrt {\frac {a}{a + b}}}{2 \, a \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ) \cos \left (d x + c\right )^{3} + 2 \, {\left ({\left (4 \, a + b\right )} \cos \left (d x + c\right )^{2} - a - b\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} d \cos \left (d x + c\right )^{3}}\right ] \]

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

[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))
*sin(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)]

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \]

[In]

integrate(tan(d*x+c)**4/(a+b*sin(d*x+c)**2),x)

[Out]

Integral(tan(c + d*x)**4/(a + b*sin(c + d*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.15 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {3 \, a^{2} \arctan \left (\frac {{\left (a + b\right )} \tan \left (d x + c\right )}{\sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} {\left (a^{2} + 2 \, a b + b^{2}\right )}} + \frac {{\left (a + b\right )} \tan \left (d x + c\right )^{3} - 3 \, a \tan \left (d x + c\right )}{a^{2} + 2 \, a b + b^{2}}}{3 \, d} \]

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (64) = 128\).

Time = 0.91 (sec) , antiderivative size = 164, normalized size of antiderivative = 2.22 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {d x + c}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a + 2 \, b\right ) + \arctan \left (\frac {a \tan \left (d x + c\right ) + b \tan \left (d x + c\right )}{\sqrt {a^{2} + a b}}\right )\right )} a^{2}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a^{2} + a b}} + \frac {a^{2} \tan \left (d x + c\right )^{3} + 2 \, a b \tan \left (d x + c\right )^{3} + b^{2} \tan \left (d x + c\right )^{3} - 3 \, a^{2} \tan \left (d x + c\right ) - 3 \, a b \tan \left (d x + c\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}}}{3 \, d} \]

[In]

integrate(tan(d*x+c)^4/(a+b*sin(d*x+c)^2),x, algorithm="giac")

[Out]

1/3*(3*(pi*floor((d*x + c)/pi + 1/2)*sgn(2*a + 2*b) + arctan((a*tan(d*x + c) + b*tan(d*x + c))/sqrt(a^2 + a*b)
))*a^2/((a^2 + 2*a*b + b^2)*sqrt(a^2 + a*b)) + (a^2*tan(d*x + c)^3 + 2*a*b*tan(d*x + c)^3 + b^2*tan(d*x + c)^3
- 3*a^2*tan(d*x + c) - 3*a*b*tan(d*x + c))/(a^3 + 3*a^2*b + 3*a*b^2 + b^3))/d

Mupad [B] (verification not implemented)

Time = 13.95 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^4(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d\,\left (a+b\right )}-\frac {a\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\left (a+b\right )}^2}+\frac {a^{3/2}\,\mathrm {atan}\left (\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (2\,a+2\,b\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,\sqrt {a}\,{\left (a+b\right )}^{5/2}}\right )}{d\,{\left (a+b\right )}^{5/2}} \]

[In]

int(tan(c + d*x)^4/(a + b*sin(c + d*x)^2),x)

[Out]

tan(c + d*x)^3/(3*d*(a + b)) - (a*tan(c + d*x))/(d*(a + b)^2) + (a^(3/2)*atan((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))