3.196 \(\int \frac {\tan ^3(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=54 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{a^{5/2} d}-\frac {4}{a^2 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

2*arctanh((a+a*sec(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d-4/a^2/d/(a+a*sec(d*x+c))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3880, 78, 63, 207} \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a \sec (c+d x)+a}}{\sqrt {a}}\right )}{a^{5/2} d}-\frac {4}{a^2 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(2*ArcTanh[Sqrt[a + a*Sec[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) - 4/(a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 207

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

Rule 3880

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

Rubi steps

\begin {align*} \int \frac {\tan ^3(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {-a+a x}{x (a+a x)^{3/2}} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac {4}{a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {\operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+a x}} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac {4}{a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {2 \operatorname {Subst}\left (\int \frac {1}{-1+\frac {x^2}{a}} \, dx,x,\sqrt {a+a \sec (c+d x)}\right )}{a^3 d}\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+a \sec (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}-\frac {4}{a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 50, normalized size = 0.93 \[ \frac {2 \left (\sqrt {\sec (c+d x)+1} \tanh ^{-1}\left (\sqrt {\sec (c+d x)+1}\right )-2\right )}{a^2 d \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.67, size = 245, normalized size = 4.54 \[ \left [\frac {\sqrt {a} {\left (\cos \left (d x + c\right ) + 1\right )} \log \left (-8 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} - 8 \, a \cos \left (d x + c\right ) - a\right ) - 8 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, {\left (a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {-a} {\left (\cos \left (d x + c\right ) + 1\right )} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + a}\right ) + 4 \, \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{a^{3} d \cos \left (d x + c\right ) + a^{3} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 4.92, size = 104, normalized size = 1.93 \[ \frac {2 \, {\left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {\sqrt {2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )}}{a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 1.14, size = 154, normalized size = 2.85 \[ -\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (\cos \left (d x +c \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )+\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}+4 \cos \left (d x +c \right )\right )}{d \left (1+\cos \left (d x +c \right )\right ) a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x)

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.45, size = 125, normalized size = 2.31 \[ -\frac {\frac {3 \, \log \left (\frac {\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {2 \, {\left (4 \, a + \frac {3 \, a}{\cos \left (d x + c\right )}\right )}}{{\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a^{2}} + \frac {6}{\sqrt {a + \frac {a}{\cos \left (d x + c\right )}} a^{2}} - \frac {2}{{\left (a + \frac {a}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/3*(3*log((sqrt(a + a/cos(d*x + c)) - sqrt(a))/(sqrt(a + a/cos(d*x + c)) + sqrt(a)))/a^(5/2) + 2*(4*a + 3*a/
cos(d*x + c))/((a + a/cos(d*x + c))^(3/2)*a^2) + 6/(sqrt(a + a/cos(d*x + c))*a^2) - 2/((a + a/cos(d*x + c))^(3
/2)*a))/d

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3/(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(tan(c + d*x)^3/(a + a/cos(c + d*x))^(5/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral(tan(c + d*x)**3/(a*(sec(c + d*x) + 1))**(5/2), x)

________________________________________________________________________________________