\(\int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [640]

Optimal result

Integrand size = 25, antiderivative size = 250 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b^{5/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2 a^2 \tan ^{\frac {3}{2}}(c+d x)}{b \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}}+\frac {\left (3 a^2+b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{b^2 \left (a^2+b^2\right ) d} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.02 (sec) , antiderivative size = 270, normalized size of antiderivative = 1.08 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {5 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}-\frac {5 (-1)^{3/4} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{3/2}}-\frac {5 \sqrt {\tan (c+d x)}}{(a-i b) \sqrt {a+b \tan (c+d x)}}-\frac {5 \sqrt {\tan (c+d x)}}{(a+i b) \sqrt {a+b \tan (c+d x)}}+\frac {2 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},-\frac {b \tan (c+d x)}{a}\right ) \tan ^{\frac {5}{2}}(c+d x) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{a \sqrt {a+b \tan (c+d x)}}}{5 d} \] Input:

Integrate[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

((5*(-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)^(3/2) - (5*(-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)^(3/ 
2) - (5*Sqrt[Tan[c + d*x]])/((a - I*b)*Sqrt[a + b*Tan[c + d*x]]) - (5*Sqrt 
[Tan[c + d*x]])/((a + I*b)*Sqrt[a + b*Tan[c + d*x]]) + (2*Hypergeometric2F 
1[3/2, 5/2, 7/2, -((b*Tan[c + d*x])/a)]*Tan[c + d*x]^(5/2)*Sqrt[1 + (b*Tan 
[c + d*x])/a])/(a*Sqrt[a + b*Tan[c + d*x]]))/(5*d)
 

Rubi [A] (verified)

Time = 1.34 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.11, 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, 4130, 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.

\(\displaystyle \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (c+d x)^{7/2}}{(a+b \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

Input:

Int[Tan[c + d*x]^(7/2)/(a + b*Tan[c + d*x])^(3/2),x]
 

Output:

(-2*a^2*Tan[c + d*x]^(3/2))/(b*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]) + ( 
-((-(((a - I*b)*b^2*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*T 
an[c + d*x]]])/Sqrt[I*a - b]) + (3*a*(a^2 + b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan 
[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] - ((a + I*b)*b^2*ArcTanh[(S 
qrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a + b]) 
/(b*d)) + ((3*a^2 + b^2)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/(b*d 
))/(b*(a^2 + b^2))
 

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 4130

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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, 
b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && 
 NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

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]
 

Maple [B] (warning: unable to verify)

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

Time = 1.14 (sec) , antiderivative size = 764550, normalized size of antiderivative = 3058.20

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

Input:

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7819 vs. \(2 (206) = 412\).

Time = 2.10 (sec) , antiderivative size = 15640, normalized size of antiderivative = 62.56 \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(tan(d*x+c)**(7/2)/(a+b*tan(d*x+c))**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int { \frac {\tan \left (d x + c\right )^{\frac {7}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(tan(d*x + c)^(7/2)/(b*tan(d*x + c) + a)^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")
 

Output:

Exception raised: RuntimeError >> an error occurred running a Giac command 
:INPUT:sage2OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const ve 
cteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{7/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^(3/2),x)
 

Output:

int(tan(c + d*x)^(7/2)/(a + b*tan(c + d*x))^(3/2), x)
 

Reduce [F]

\[ \int \frac {\tan ^{\frac {7}{2}}(c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{3}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \] Input:

int(tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(3/2),x)
 

Output:

int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**3)/(tan(c + 
 d*x)**2*b**2 + 2*tan(c + d*x)*a*b + a**2),x)