\(\int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\) [327]

Optimal result

Integrand size = 23, antiderivative size = 79 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {2 a \sqrt {a+b \sec (c+d x)}}{b^2 d}+\frac {2 (a+b \sec (c+d x))^{3/2}}{3 b^2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 79, 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, 1167, 213} \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {2 (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac {2 a \sqrt {a+b \sec (c+d x)}}{b^2 d} \]

[In]

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

[Out]

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

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 1167

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 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 \sqrt {a+x}} \, 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}{-a+x^2} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^2 d} \\ & = -\frac {2 \text {Subst}\left (\int \left (a-x^2+\frac {b^2}{-a+x^2}\right ) \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^2 d} \\ & = -\frac {2 a \sqrt {a+b \sec (c+d x)}}{b^2 d}+\frac {2 (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac {2 \text {Subst}\left (\int \frac {1}{-a+x^2} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{d} \\ & = \frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {2 a \sqrt {a+b \sec (c+d x)}}{b^2 d}+\frac {2 (a+b \sec (c+d x))^{3/2}}{3 b^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {(-2 a+b \sec (c+d x)) \sqrt {a+b \sec (c+d x)}}{b^2}\right )}{3 d} \]

[In]

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

[Out]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(311\) vs. \(2(67)=134\).

Time = 11.29 (sec) , antiderivative size = 312, normalized size of antiderivative = 3.95

method	result	size
default	\(-\frac {\sqrt {a +b \sec \left (d x +c \right )}\, \left (-3 \sqrt {a}\, \cos \left (d x +c \right ) \ln \left (4 \cos \left (d x +c \right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, \sqrt {a}+4 a \cos \left (d x +c \right )+4 \sqrt {a}\, \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}+2 b \right ) b^{2}+4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{2} \cos \left (d x +c \right )+4 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a^{2}-2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a b -2 \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}\, a b \sec \left (d x +c \right )\right )}{3 d a \,b^{2} \left (\cos \left (d x +c \right )+1\right ) \sqrt {\frac {\left (b +a \cos \left (d x +c \right )\right ) \cos \left (d x +c \right )}{\left (\cos \left (d x +c \right )+1\right )^{2}}}}\)	\(312\)

[In]

int(tan(d*x+c)^3/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/d/a/b^2*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)/((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(-3*a^
(1/2)*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)*b^2+4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x
+c)+1)^2)^(1/2)*a^2*cos(d*x+c)+4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2-2*((b+a*cos(d*x+c))*
cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b-2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a*b*sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 273, normalized size of antiderivative = 3.46 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=\left [\frac {3 \, \sqrt {a} b^{2} \cos \left (d x + c\right ) \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} - 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) - 4 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) - a b\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{6 \, a b^{2} d \cos \left (d x + c\right )}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right ) + 2 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) - a b\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{3 \, a b^{2} d \cos \left (d x + c\right )}\right ] \]

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(a)*b^2*cos(d*x + c)*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*(2*a^2*cos(d*x + c) - a*b)*sqrt((a*cos(
d*x + c) + b)/cos(d*x + c)))/(a*b^2*d*cos(d*x + c)), -1/3*(3*sqrt(-a)*b^2*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))*cos(d*x + c) + 2*(2*a^2*cos(d*x + c) - a*b)*sqrt((a
*cos(d*x + c) + b)/cos(d*x + c)))/(a*b^2*d*cos(d*x + c))]

Sympy [F]

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

[In]

integrate(tan(d*x+c)**3/(a+b*sec(d*x+c))**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.16 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {\frac {3 \, \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{b^{2}} + \frac {6 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a}{b^{2}}}{3 \, d} \]

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/3*(3*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a)))/sqrt(a) - 2*(a + b/cos(
d*x + c))^(3/2)/b^2 + 6*sqrt(a + b/cos(d*x + c))*a/b^2)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (67) = 134\).

Time = 0.82 (sec) , antiderivative size = 265, normalized size of antiderivative = 3.35 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx=-\frac {2 \, {\left (\frac {3 \, \arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} + \sqrt {a - b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {2 \, {\left (3 \, {\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )}^{2} - 3 \, a - b\right )}}{{\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} - \sqrt {a - b}\right )}^{3}}\right )}}{3 \, d \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} \]

[In]

integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int(tan(c + d*x)^3/(a + b/cos(c + d*x))^(1/2),x)

[Out]

int(tan(c + d*x)^3/(a + b/cos(c + d*x))^(1/2), x)