3.1.9 \(\int (b \tan (c+d x))^{7/2} \, dx\) [9]

3.1.9.1 Optimal result

Integrand size = 12, antiderivative size = 232 \[ \int (b \tan (c+d x))^{7/2} \, dx=-\frac {b^{7/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} d}+\frac {b^{7/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {b \tan (c+d x)}}{\sqrt {b}}\right )}{\sqrt {2} d}-\frac {b^{7/2} \log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)-\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} d}+\frac {b^{7/2} \log \left (\sqrt {b}+\sqrt {b} \tan (c+d x)+\sqrt {2} \sqrt {b \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 b^3 \sqrt {b \tan (c+d x)}}{d}+\frac {2 b (b \tan (c+d x))^{5/2}}{5 d} \]

output

-1/2*b^(7/2)*arctan(1-2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))/d*2^(1/2)+1/2* 
b^(7/2)*arctan(1+2^(1/2)*(b*tan(d*x+c))^(1/2)/b^(1/2))/d*2^(1/2)-1/4*b^(7/ 
2)*ln(b^(1/2)-2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))/d*2^(1/2)+1 
/4*b^(7/2)*ln(b^(1/2)+2^(1/2)*(b*tan(d*x+c))^(1/2)+b^(1/2)*tan(d*x+c))/d*2 
^(1/2)-2*b^3*(b*tan(d*x+c))^(1/2)/d+2/5*b*(b*tan(d*x+c))^(5/2)/d
 

3.1.9.2 Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 174, normalized size of antiderivative = 0.75 \[ \int (b \tan (c+d x))^{7/2} \, dx=\frac {(b \tan (c+d x))^{7/2} \left (-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}+\frac {\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}-\frac {\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2}}+\frac {\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2}}-2 \sqrt {\tan (c+d x)}+\frac {2}{5} \tan ^{\frac {5}{2}}(c+d x)\right )}{d \tan ^{\frac {7}{2}}(c+d x)} \]

input

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

output

((b*Tan[c + d*x])^(7/2)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]) 
 + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2] - Log[1 - Sqrt[2]*Sqrt[T 
an[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2]) + Log[1 + Sqrt[2]*Sqrt[Tan[c + d* 
x]] + Tan[c + d*x]]/(2*Sqrt[2]) - 2*Sqrt[Tan[c + d*x]] + (2*Tan[c + d*x]^( 
5/2))/5))/(d*Tan[c + d*x]^(7/2))
 

3.1.9.3 Rubi [A] (warning: unable to verify)

Time = 0.55 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.98, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.250, Rules used = {3042, 3954, 3042, 3954, 3042, 3957, 266, 755, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3954

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3957

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 755

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

input

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

output

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

3.1.9.3.1 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 755

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2] 
], s = Denominator[Rt[a/b, 2]]}, Simp[1/(2*r)   Int[(r - s*x^2)/(a + b*x^4) 
, x], x] + Simp[1/(2*r)   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 3954

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

3.1.9.4 Maple [A] (verified)

Time = 0.20 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {2 b \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-b^{2} \sqrt {b \tan \left (d x +c \right )}+\frac {b^{2} \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{d}\)	\(169\)
default	\(\frac {2 b \left (\frac {\left (b \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-b^{2} \sqrt {b \tan \left (d x +c \right )}+\frac {b^{2} \left (b^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {b \tan \left (d x +c \right )+\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}{b \tan \left (d x +c \right )-\left (b^{2}\right )^{\frac {1}{4}} \sqrt {b \tan \left (d x +c \right )}\, \sqrt {2}+\sqrt {b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {b \tan \left (d x +c \right )}}{\left (b^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8}\right )}{d}\)	\(169\)

input

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

output

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

3.1.9.5 Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.23 (sec) , antiderivative size = 203, normalized size of antiderivative = 0.88 \[ \int (b \tan (c+d x))^{7/2} \, dx=\frac {5 \, \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d \log \left (\sqrt {b \tan \left (d x + c\right )} b^{3} + \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d\right ) + 5 i \, \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d \log \left (\sqrt {b \tan \left (d x + c\right )} b^{3} + i \, \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d\right ) - 5 i \, \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d \log \left (\sqrt {b \tan \left (d x + c\right )} b^{3} - i \, \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d\right ) - 5 \, \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d \log \left (\sqrt {b \tan \left (d x + c\right )} b^{3} - \left (-\frac {b^{14}}{d^{4}}\right )^{\frac {1}{4}} d\right ) + 4 \, {\left (b^{3} \tan \left (d x + c\right )^{2} - 5 \, b^{3}\right )} \sqrt {b \tan \left (d x + c\right )}}{10 \, d} \]

input

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

output

1/10*(5*(-b^14/d^4)^(1/4)*d*log(sqrt(b*tan(d*x + c))*b^3 + (-b^14/d^4)^(1/ 
4)*d) + 5*I*(-b^14/d^4)^(1/4)*d*log(sqrt(b*tan(d*x + c))*b^3 + I*(-b^14/d^ 
4)^(1/4)*d) - 5*I*(-b^14/d^4)^(1/4)*d*log(sqrt(b*tan(d*x + c))*b^3 - I*(-b 
^14/d^4)^(1/4)*d) - 5*(-b^14/d^4)^(1/4)*d*log(sqrt(b*tan(d*x + c))*b^3 - ( 
-b^14/d^4)^(1/4)*d) + 4*(b^3*tan(d*x + c)^2 - 5*b^3)*sqrt(b*tan(d*x + c))) 
/d
 

3.1.9.6 Sympy [F]

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

input

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

output

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

3.1.9.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.80 \[ \int (b \tan (c+d x))^{7/2} \, dx=\frac {10 \, \sqrt {2} b^{\frac {9}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} + 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right ) + 10 \, \sqrt {2} b^{\frac {9}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {b} - 2 \, \sqrt {b \tan \left (d x + c\right )}\right )}}{2 \, \sqrt {b}}\right ) + 5 \, \sqrt {2} b^{\frac {9}{2}} \log \left (b \tan \left (d x + c\right ) + \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right ) - 5 \, \sqrt {2} b^{\frac {9}{2}} \log \left (b \tan \left (d x + c\right ) - \sqrt {2} \sqrt {b \tan \left (d x + c\right )} \sqrt {b} + b\right ) + 8 \, \left (b \tan \left (d x + c\right )\right )^{\frac {5}{2}} b^{2} - 40 \, \sqrt {b \tan \left (d x + c\right )} b^{4}}{20 \, b d} \]

input

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

output

1/20*(10*sqrt(2)*b^(9/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(b) + 2*sqrt(b*ta 
n(d*x + c)))/sqrt(b)) + 10*sqrt(2)*b^(9/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sq 
rt(b) - 2*sqrt(b*tan(d*x + c)))/sqrt(b)) + 5*sqrt(2)*b^(9/2)*log(b*tan(d*x 
 + c) + sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b) - 5*sqrt(2)*b^(9/2)*log( 
b*tan(d*x + c) - sqrt(2)*sqrt(b*tan(d*x + c))*sqrt(b) + b) + 8*(b*tan(d*x 
+ c))^(5/2)*b^2 - 40*sqrt(b*tan(d*x + c))*b^4)/(b*d)
 

3.1.9.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.1.9.9 Mupad [B] (verification not implemented)

Time = 3.38 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.40 \[ \int (b \tan (c+d x))^{7/2} \, dx=\frac {2\,b\,{\left (b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,d}-\frac {2\,b^3\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{d}-\frac {{\left (-1\right )}^{1/4}\,b^{7/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}}{\sqrt {b}}\right )\,1{}\mathrm {i}}{d}-\frac {{\left (-1\right )}^{1/4}\,b^{7/2}\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}\,\sqrt {b\,\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}}{\sqrt {b}}\right )}{d} \]

input

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

output

(2*b*(b*tan(c + d*x))^(5/2))/(5*d) - (2*b^3*(b*tan(c + d*x))^(1/2))/d - (( 
-1)^(1/4)*b^(7/2)*atan(((-1)^(1/4)*(b*tan(c + d*x))^(1/2))/b^(1/2))*1i)/d 
- ((-1)^(1/4)*b^(7/2)*atan(((-1)^(1/4)*(b*tan(c + d*x))^(1/2)*1i)/b^(1/2)) 
)/d