\(\int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [336]

Optimal result

Integrand size = 23, antiderivative size = 88 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {a+b \sec (c+d x)}}{b^2 d} \]

[Out]

2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+2*(a^2-b^2)/a/b^2/d/(a+b*sec(d*x+c))^(1/2)+2*(a+b*sec(d*x+
c))^(1/2)/b^2/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3970, 912, 1275, 212} \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {a+b \sec (c+d x)}}{b^2 d} \]

[In]

Int[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (2*(a^2 - b^2))/(a*b^2*d*Sqrt[a + b*Sec[c + d*x]])
+ (2*Sqrt[a + b*Sec[c + d*x]])/(b^2*d)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 912

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 + a*e^2)/e^2 - 2*c*
d*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]

Rule 1275

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

Rule 3970

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[(b^2 - x^2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.26 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.75 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (-b^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1+\frac {b \sec (c+d x)}{a}\right )+a (2 a+b \sec (c+d x))\right )}{a b^2 d \sqrt {a+b \sec (c+d x)}} \]

[In]

Integrate[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*(-(b^2*Hypergeometric2F1[-1/2, 1, 1/2, 1 + (b*Sec[c + d*x])/a]) + a*(2*a + b*Sec[c + d*x])))/(a*b^2*d*Sqrt[
a + b*Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(730\) vs. \(2(78)=156\).

Time = 9.56 (sec) , antiderivative size = 731, normalized size of antiderivative = 8.31

method	result	size
default	\(\frac {\left (\cos \left (d x +c \right )^{3} a^{\frac {5}{2}} \ln \left (4 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a}+4 a \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 b \right ) b^{2}+2 \cos \left (d x +c \right )^{2} a^{\frac {3}{2}} \ln \left (4 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a}+4 a \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 b \right ) b^{3}+2 \sin \left (d x +c \right )^{2} \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{2} b^{2}+4 \cos \left (d x +c \right )^{3} \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{4}+2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{2} b^{2} \sin \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )^{2} \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{4}+6 \cos \left (d x +c \right )^{2} \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{3} b -2 \cos \left (d x +c \right )^{2} \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a \,b^{3}+\cos \left (d x +c \right ) \ln \left (4 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a}+4 a \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 b \right ) \sqrt {a}\, b^{4}+6 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{3} b -2 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a \,b^{3}\right ) \sqrt {a +b \sec \left (d x +c \right )}}{d \,a^{2} b^{2} \left (b +a \cos \left (d x +c \right )\right )^{2} \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \left (\cos \left (d x +c \right )+1\right )}\)	\(731\)

[In]

int(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/d/a^2/b^2*(cos(d*x+c)^3*a^(5/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)
+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^2+2*cos(d*x+c)^2*a^(3/2)
*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*c
os(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^3+2*sin(d*x+c)^2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/
(cos(d*x+c)+1)^2)^(1/2)*a^2*b^2+4*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^4+2*((b+
a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^2*sin(d*x+c)^2+4*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x
+c)/(cos(d*x+c)+1)^2)^(1/2)*a^4+6*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^3*b-2*co
s(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b^3+cos(d*x+c)*ln(4*cos(d*x+c)*((b+a*cos(d*x
+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x
+c)+1)^2)^(1/2)+2*b)*a^(1/2)*b^4+6*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^3*b-2*cos
(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b^3)*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))^2/
((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)/(cos(d*x+c)+1)

Fricas [A] (verification not implemented)

none

Time = 0.44 (sec) , antiderivative size = 317, normalized size of antiderivative = 3.60 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\left [\frac {{\left (a b^{2} \cos \left (d x + c\right ) + b^{3}\right )} \sqrt {a} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} - 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \, {\left (a^{2} b + {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{2 \, {\left (a^{3} b^{2} d \cos \left (d x + c\right ) + a^{2} b^{3} d\right )}}, -\frac {{\left (a b^{2} \cos \left (d x + c\right ) + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) - 2 \, {\left (a^{2} b + {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{a^{3} b^{2} d \cos \left (d x + c\right ) + a^{2} b^{3} d}\right ] \]

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/2*((a*b^2*cos(d*x + c) + b^3)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 - 4*(2*a*cos(d*x
+ c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + 4*(a^2*b + (2*a^3 - a*b^2)*cos(d*
x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^2*d*cos(d*x + c) + a^2*b^3*d), -((a*b^2*cos(d*x + c) +
b^3)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b))
- 2*(a^2*b + (2*a^3 - a*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^2*d*cos(d*x + c) +
a^2*b^3*d)]

Sympy [F]

\[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(tan(d*x+c)**3/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(tan(c + d*x)**3/(a + b*sec(c + d*x))**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.25 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {\frac {\log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a} - \frac {2 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}}}{b^{2}} - \frac {2 \, a}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} b^{2}}}{d} \]

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-(log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a)))/a^(3/2) + 2/(sqrt(a + b/cos(d
*x + c))*a) - 2*sqrt(a + b/cos(d*x + c))/b^2 - 2*a/(sqrt(a + b/cos(d*x + c))*b^2))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 258 vs. \(2 (78) = 156\).

Time = 1.10 (sec) , antiderivative size = 258, normalized size of antiderivative = 2.93 \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \, {\left (\frac {\frac {{\left (2 \, a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - a^{2} b \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - a b^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} b^{2}} - \frac {2 \, a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + a^{2} b \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - a b^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2} b^{2}}}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}} + \frac {\arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} + \sqrt {a - b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )}}{d} \]

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-2*(((2*a^3*sgn(cos(d*x + c)) - a^2*b*sgn(cos(d*x + c)) - a*b^2*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/(a^2
*b^2) - (2*a^3*sgn(cos(d*x + c)) + a^2*b*sgn(cos(d*x + c)) - a*b^2*sgn(cos(d*x + c)))/(a^2*b^2))/sqrt(a*tan(1/
2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) + arctan(-1/2*(sqrt(a - b)*t
an(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2
+ a + b) + sqrt(a - b))/sqrt(-a))/(sqrt(-a)*a*sgn(cos(d*x + c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(tan(c + d*x)^3/(a + b/cos(c + d*x))^(3/2),x)

[Out]

int(tan(c + d*x)^3/(a + b/cos(c + d*x))^(3/2), x)