\(\int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx\) [16]

Optimal result

Integrand size = 12, antiderivative size = 179 \[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{7/2} d}+\frac {\arctan \left (1+\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} b^{7/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}+\sqrt {b} \tan (c+d x)}\right )}{\sqrt {2} b^{7/2} d}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}+\frac {2}{b^3 d \sqrt {b \tan (c+d x)}} \] Output:

-1/2*arctan(1-2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))*2^(1/2)/b^(7/2)/d+1/2* 
arctan(1+2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))*2^(1/2)/b^(7/2)/d-1/2*arcta 
nh(2^(1/2)*(b*tan(d*x+c))^(1/2)/(b^(1/2)+b^(1/2)*tan(d*x+c)))*2^(1/2)/b^(7 
/2)/d-2/5/b/d/(b*tan(d*x+c))^(5/2)+2/b^3/d/(b*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=\frac {10-2 \cot ^2(c+d x)+5 \arctan \left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}-5 \text {arctanh}\left (\sqrt [4]{-\tan ^2(c+d x)}\right ) \sqrt [4]{-\tan ^2(c+d x)}}{5 b^3 d \sqrt {b \tan (c+d x)}} \] Input:

Integrate[(b*Tan[c + d*x])^(-7/2),x]
 

Output:

(10 - 2*Cot[c + d*x]^2 + 5*ArcTan[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d*x]^ 
2)^(1/4) - 5*ArcTanh[(-Tan[c + d*x]^2)^(1/4)]*(-Tan[c + d*x]^2)^(1/4))/(5* 
b^3*d*Sqrt[b*Tan[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.26, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {3042, 3955, 3042, 3955, 3042, 3957, 266, 826, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

\(\displaystyle \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(b \tan (c+d x))^{7/2}}dx\)

\(\Big \downarrow \) 3955

\(\displaystyle -\frac {\int \frac {1}{(b \tan (c+d x))^{3/2}}dx}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \frac {1}{(b \tan (c+d x))^{3/2}}dx}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3955

\(\displaystyle -\frac {-\frac {\int \sqrt {b \tan (c+d x)}dx}{b^2}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {-\frac {\int \sqrt {b \tan (c+d x)}dx}{b^2}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3957

\(\displaystyle -\frac {-\frac {\int \frac {\sqrt {b \tan (c+d x)}}{\tan ^2(c+d x) b^2+b^2}d(b \tan (c+d x))}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 266

\(\displaystyle -\frac {-\frac {2 \int \frac {b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 826

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \int \frac {b^2 \tan ^2(c+d x)+b}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 1476

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}+\frac {1}{2} \int \frac {1}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 1082

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \left (\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-1}d\left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}-\frac {\int \frac {1}{-b^2 \tan ^2(c+d x)-1}d\left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 217

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )-\frac {1}{2} \int \frac {b-b^2 \tan ^2(c+d x)}{b^4 \tan ^4(c+d x)+b^2}d\sqrt {b \tan (c+d x)}\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 1479

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \left (\frac {\int -\frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}+\frac {\int -\frac {\sqrt {2} \left (\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}-\frac {\int \frac {\sqrt {2} \left (\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \left (-\frac {\int \frac {\sqrt {2} \sqrt {b}-2 \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)-\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {2} \sqrt {b}}-\frac {\int \frac {\sqrt {b}+\sqrt {2} \sqrt {b \tan (c+d x)}}{b^2 \tan ^2(c+d x)+\sqrt {2} b^{3/2} \tan (c+d x)+b}d\sqrt {b \tan (c+d x)}}{2 \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {-\frac {2 \left (\frac {1}{2} \left (\frac {\arctan \left (\sqrt {2} \sqrt {b} \tan (c+d x)+1\right )}{\sqrt {2} \sqrt {b}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {b} \tan (c+d x)\right )}{\sqrt {2} \sqrt {b}}\right )+\frac {1}{2} \left (\frac {\log \left (-\sqrt {2} b^{3/2} \tan (c+d x)+b^2 \tan ^2(c+d x)+b\right )}{2 \sqrt {2} \sqrt {b}}-\frac {\log \left (\sqrt {2} b^{3/2} \tan (c+d x)+b^2 \tan ^2(c+d x)+b\right )}{2 \sqrt {2} \sqrt {b}}\right )\right )}{b d}-\frac {2}{b d \sqrt {b \tan (c+d x)}}}{b^2}-\frac {2}{5 b d (b \tan (c+d x))^{5/2}}\)

Input:

Int[(b*Tan[c + d*x])^(-7/2),x]
 

Output:

-2/(5*b*d*(b*Tan[c + d*x])^(5/2)) - ((-2*((-(ArcTan[1 - Sqrt[2]*Sqrt[b]*Ta 
n[c + d*x]]/(Sqrt[2]*Sqrt[b])) + ArcTan[1 + Sqrt[2]*Sqrt[b]*Tan[c + d*x]]/ 
(Sqrt[2]*Sqrt[b]))/2 + (Log[b - Sqrt[2]*b^(3/2)*Tan[c + d*x] + b^2*Tan[c + 
 d*x]^2]/(2*Sqrt[2]*Sqrt[b]) - Log[b + Sqrt[2]*b^(3/2)*Tan[c + d*x] + b^2* 
Tan[c + d*x]^2]/(2*Sqrt[2]*Sqrt[b]))/2))/(b*d) - 2/(b*d*Sqrt[b*Tan[c + d*x 
]]))/b^2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

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

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 826

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 
2]], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*s)   Int[(r + s*x^2)/(a + b*x^ 
4), x], x] - Simp[1/(2*s)   Int[(r - s*x^2)/(a + b*x^4), x], x]] /; FreeQ[{ 
a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] 
 && AtomQ[SplitProduct[SumBaseQ, b]]))
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3955

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Tan[c + d*x] 
)^(n + 1)/(b*d*(n + 1)), x] - Simp[1/b^2   Int[(b*Tan[c + d*x])^(n + 2), x] 
, x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {2 b \left (-\frac {1}{5 b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{b^{4} \sqrt {b \tan \left (d x +c \right )}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 b^{4} \left (b^{2}\right )^{\frac {1}{4}}}\right )}{d}\)	\(171\)
default	\(\frac {2 b \left (-\frac {1}{5 b^{2} \left (b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}+\frac {1}{b^{4} \sqrt {b \tan \left (d x +c \right )}}+\frac {\sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 b^{4} \left (b^{2}\right )^{\frac {1}{4}}}\right )}{d}\)	\(171\)

Input:

int(1/(b*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
 

Output:

2/d*b*(-1/5/b^2/(b*tan(d*x+c))^(5/2)+1/b^4/(b*tan(d*x+c))^(1/2)+1/8/b^4/(b 
^2)^(1/4)*2^(1/2)*(ln((b*tan(d*x+c)-(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2^(1/ 
2)+(b^2)^(1/2))/(b*tan(d*x+c)+(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)*2^(1/2)+(b^ 
2)^(1/2)))+2*arctan(2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1)-2*arctan(- 
2^(1/2)/(b^2)^(1/4)*(b*tan(d*x+c))^(1/2)+1)))
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.11 \[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=\frac {10 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )}}{\sqrt {b}} + 1\right ) \tan \left (d x + c\right )^{3} + 10 \, \sqrt {2} \sqrt {b} \arctan \left (\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )}}{\sqrt {b}} - 1\right ) \tan \left (d x + c\right )^{3} - 5 \, \sqrt {2} \sqrt {b} \log \left (\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )}}{\sqrt {b}} + \tan \left (d x + c\right ) + 1\right ) \tan \left (d x + c\right )^{3} + 5 \, \sqrt {2} \sqrt {b} \log \left (-\frac {\sqrt {2} \sqrt {b \tan \left (d x + c\right )}}{\sqrt {b}} + \tan \left (d x + c\right ) + 1\right ) \tan \left (d x + c\right )^{3} + 8 \, \sqrt {b \tan \left (d x + c\right )} {\left (5 \, \tan \left (d x + c\right )^{2} - 1\right )}}{20 \, b^{4} d \tan \left (d x + c\right )^{3}} \] Input:

integrate(1/(b*tan(d*x+c))^(7/2),x, algorithm="fricas")
 

Output:

1/20*(10*sqrt(2)*sqrt(b)*arctan(sqrt(2)*sqrt(b*tan(d*x + c))/sqrt(b) + 1)* 
tan(d*x + c)^3 + 10*sqrt(2)*sqrt(b)*arctan(sqrt(2)*sqrt(b*tan(d*x + c))/sq 
rt(b) - 1)*tan(d*x + c)^3 - 5*sqrt(2)*sqrt(b)*log(sqrt(2)*sqrt(b*tan(d*x + 
 c))/sqrt(b) + tan(d*x + c) + 1)*tan(d*x + c)^3 + 5*sqrt(2)*sqrt(b)*log(-s 
qrt(2)*sqrt(b*tan(d*x + c))/sqrt(b) + tan(d*x + c) + 1)*tan(d*x + c)^3 + 8 
*sqrt(b*tan(d*x + c))*(5*tan(d*x + c)^2 - 1))/(b^4*d*tan(d*x + c)^3)
 

Sympy [F]

\[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=\int \frac {1}{\left (b \tan {\left (c + d x \right )}\right )^{\frac {7}{2}}}\, dx \] Input:

integrate(1/(b*tan(d*x+c))**(7/2),x)
 

Output:

Integral((b*tan(c + d*x))**(-7/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.09 \[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=\frac {\frac {5 \, {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right )}{\sqrt {b}} - \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}} + \frac {\sqrt {2} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right )}{\sqrt {b}}\right )}}{b^{2}} + \frac {8 \, {\left (5 \, b^{2} \tan \left (d x + c\right )^{2} - b^{2}\right )}}{\left (b \tan \left (d x + c\right )\right )^{\frac {5}{2}} b^{2}}}{20 \, b d} \] Input:

integrate(1/(b*tan(d*x+c))^(7/2),x, algorithm="maxima")
 

Output:

1/20*(5*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*tan(d*x 
+ c)))/sqrt(b))/sqrt(b) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(b) - 
 2*sqrt(b*tan(d*x + c)))/sqrt(b))/sqrt(b) - sqrt(2)*log(b*tan(d*x + c) + s 
qrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/sqrt(b) + sqrt(2)*log(b*tan(d*x + 
 c) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b)/sqrt(b))/b^2 + 8*(5*b^2*ta 
n(d*x + c)^2 - b^2)/((b*tan(d*x + c))^(5/2)*b^2))/(b*d)
 

Giac [F(-1)]

Timed out. \[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=\text {Timed out} \] Input:

integrate(1/(b*tan(d*x+c))^(7/2),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 0.38 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.51 \[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=\frac {{\left (-1\right )}^{1/4}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{7/2}\,d}-\frac {\frac {2}{5\,b}-\frac {2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{b}}{d\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}-\frac {{\left (-1\right )}^{1/4}\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )}{b^{7/2}\,d} \] Input:

int(1/(b*tan(c + d*x))^(7/2),x)
 

Output:

((-1)^(1/4)*atan(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2)))/(b^(7/2)*d) 
 - (2/(5*b) - (2*tan(c + d*x)^2)/b)/(d*(b*tan(c + d*x))^(5/2)) - ((-1)^(1/ 
4)*atanh(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2)))/(b^(7/2)*d)
 

Reduce [F]

\[ \int \frac {1}{(b \tan (c+d x))^{7/2}} \, dx=\frac {\sqrt {b}\, \left (\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{4}}d x \right )}{b^{4}} \] Input:

int(1/(b*tan(d*x+c))^(7/2),x)
 

Output:

(sqrt(b)*int(sqrt(tan(c + d*x))/tan(c + d*x)**4,x))/b**4