3.7.48 \(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [648]

3.7.48.1 Optimal result

Integrand size = 25, antiderivative size = 251 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{5/2} d}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{5/2} d}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{3 b \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}-\frac {2 a^2 \left (a^2+3 b^2\right ) \sqrt {\tan (c+d x)}}{b^2 \left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \]

output

-arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a-b)^(5/ 
2)/d+2*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(5/2)/d- 
arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/(I*a+b)^(5/ 
2)/d-2*a^2*(a^2+3*b^2)*tan(d*x+c)^(1/2)/b^2/(a^2+b^2)^2/d/(a+b*tan(d*x+c)) 
^(1/2)-2/3*a^2*tan(d*x+c)^(3/2)/b/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)
 

3.7.48.2 Mathematica [A] (verified)

Time = 6.25 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.69 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {(-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2} d}-\frac {(-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2} d}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{3 (i a-b) d (a+b \tan (c+d x))^{3/2}}-\frac {2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d (a+b \tan (c+d x))^{3/2}}+\frac {\tan ^{\frac {3}{2}}(c+d x)}{3 (i a+b) d (a+b \tan (c+d x))^{3/2}}-\frac {\sqrt {\tan (c+d x)}}{(a-i b)^2 d \sqrt {a+b \tan (c+d x)}}-\frac {\sqrt {\tan (c+d x)}}{(a+i b)^2 d \sqrt {a+b \tan (c+d x)}}-\frac {2 \sqrt {\tan (c+d x)}}{b^2 d \sqrt {a+b \tan (c+d x)}}+\frac {2 \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {a+b \tan (c+d x)}}{\sqrt {a} b^{5/2} d \sqrt {1+\frac {b \tan (c+d x)}{a}}} \]

input

Integrate[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^(5/2),x]
 

output

-(((-1)^(3/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
 + b*Tan[c + d*x]]])/((-a + I*b)^(5/2)*d)) - ((-1)^(3/4)*ArcTan[((-1)^(1/4 
)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/((a + I*b)^ 
(5/2)*d) - Tan[c + d*x]^(3/2)/(3*(I*a - b)*d*(a + b*Tan[c + d*x])^(3/2)) - 
 (2*Tan[c + d*x]^(3/2))/(3*b*d*(a + b*Tan[c + d*x])^(3/2)) + Tan[c + d*x]^ 
(3/2)/(3*(I*a + b)*d*(a + b*Tan[c + d*x])^(3/2)) - Sqrt[Tan[c + d*x]]/((a 
- I*b)^2*d*Sqrt[a + b*Tan[c + d*x]]) - Sqrt[Tan[c + d*x]]/((a + I*b)^2*d*S 
qrt[a + b*Tan[c + d*x]]) - (2*Sqrt[Tan[c + d*x]])/(b^2*d*Sqrt[a + b*Tan[c 
+ d*x]]) + (2*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[a + b*Tan 
[c + d*x]])/(Sqrt[a]*b^(5/2)*d*Sqrt[1 + (b*Tan[c + d*x])/a])
 

3.7.48.3 Rubi [A] (verified)

Time = 1.43 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.24, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4048, 27, 3042, 4128, 27, 3042, 4138, 2035, 2257, 2009}

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.

\(\displaystyle \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{7/2}}{(a+b \tan (c+d x))^{5/2}}dx\)

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4128

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{3 b d \left (a^2+b^2\right ) (a+b \tan (c+d x))^{3/2}}+\frac {-\frac {2 a^2 \left (a^2+3 b^2\right ) \sqrt {\tan (c+d x)}}{b d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {2 \left (\frac {\left (a^2+b^2\right )^2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}+\frac {b^2 (a-i b)^2 \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {-b+i a}}+\frac {b^2 (a+i b)^2 \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{2 \sqrt {b+i a}}\right )}{b d \left (a^2+b^2\right )}}{b \left (a^2+b^2\right )}\)

input

Int[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^(5/2),x]
 

output

(-2*a^2*Tan[c + d*x]^(3/2))/(3*b*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) 
 + ((2*(((a - I*b)^2*b^2*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a 
+ b*Tan[c + d*x]]])/(2*Sqrt[I*a - b]) + ((a^2 + b^2)^2*ArcTanh[(Sqrt[b]*Sq 
rt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] + ((a + I*b)^2*b^2*Ar 
cTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*Sqr 
t[I*a + b])))/(b*(a^2 + b^2)*d) - (2*a^2*(a^2 + 3*b^2)*Sqrt[Tan[c + d*x]]) 
/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))/(b*(a^2 + b^2))
 

3.7.48.3.1 Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

rule 4048

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m 
 - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Simp[1 
/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + 
f*x])^(n + 1)*Simp[a^2*d*(b*d*(m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c 
*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3*a*b^2*d) 
*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*( 
n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 2] && LtQ 
[n, -1] && IntegerQ[2*m]
 

rule 4128

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Tan[e + 
 f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Sim 
p[1/(d*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e 
 + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c*m + a*d* 
(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b 
*(d*(B*c - A*d)*(m + n + 1) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], 
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ 
[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
 

rule 4138

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

3.7.48.4 Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.20 (sec) , antiderivative size = 1490358, normalized size of antiderivative = 5937.68

\[\text {output too large to display}\]

input

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(5/2),x)
 

output

result too large to display
 

3.7.48.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 11532 vs. \(2 (209) = 418\).

Time = 4.33 (sec) , antiderivative size = 23066, normalized size of antiderivative = 91.90 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

output

Too large to include
 

3.7.48.6 Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)**(7/2)/(a+b*tan(d*x+c))**(5/2),x)
 

output

Timed out
 

3.7.48.7 Maxima [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

input

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

output

integrate(tan(d*x + c)^(7/2)/(b*tan(d*x + c) + a)^(5/2), x)
 

3.7.48.8 Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

input

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
 

output

Timed out
 

3.7.48.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]

input

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^(5/2),x)
 

output

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^(5/2), x)