\(\int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\) [639]

Optimal result

Integrand size = 25, antiderivative size = 229 \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {i \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a-b} d}-\frac {i \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {i a+b} d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {8 b \sqrt {a+b \tan (c+d x)}}{15 a^2 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 a^3 d \sqrt {\tan (c+d x)}} \] 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)^(1 
/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)^(1/2)/d-2/5*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2)+8/15*b*(a+b*t 
an(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(3/2)+2/15*(15*a^2-8*b^2)*(a+b*tan(d*x+c 
))^(1/2)/a^3/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.86 (sec) , antiderivative size = 197, normalized size of antiderivative = 0.86 \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {-\frac {15 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {15 \sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-3 a^2+4 a b \tan (c+d x)+\left (15 a^2-8 b^2\right ) \tan ^2(c+d x)\right )}{a^3 \tan ^{\frac {5}{2}}(c+d x)}}{15 d} \] Input:

Integrate[1/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]
 

Output:

((-15*(-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqr 
t[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + (15*(-1)^(1/4)*ArcTan[((-1)^(1/4) 
*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b 
] + (2*Sqrt[a + b*Tan[c + d*x]]*(-3*a^2 + 4*a*b*Tan[c + d*x] + (15*a^2 - 8 
*b^2)*Tan[c + d*x]^2))/(a^3*Tan[c + d*x]^(5/2)))/(15*d)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.07, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {3042, 4052, 27, 3042, 4132, 27, 3042, 4133, 27, 3042, 4058, 613, 104, 218, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

\(\displaystyle -\frac {-\frac {2 \int -\frac {15 a^2-8 b^2-8 b^2 \tan ^2(c+d x)}{2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {\int \frac {15 a^2-8 b^2-8 b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\int \frac {15 a^2-8 b^2-8 b^2 \tan (c+d x)^2}{\tan (c+d x)^{3/2} \sqrt {a+b \tan (c+d x)}}dx}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4133

\(\displaystyle -\frac {\frac {-\frac {2 \int \frac {15 a^3 \sqrt {\tan (c+d x)}}{2 \sqrt {a+b \tan (c+d x)}}dx}{a}-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {\frac {-15 a^2 \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {-15 a^2 \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}dx-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 4058

\(\displaystyle -\frac {\frac {-\frac {15 a^2 \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{d}-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}}{3 a}-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}}{5 a}-\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}\)

\(\Big \downarrow \) 613

\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\frac {1}{2} \int \frac {1}{\sqrt {\tan (c+d x)} (\tan (c+d x)+i) \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)-\frac {1}{2} \int \frac {1}{(i-\tan (c+d x)) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}d\tan (c+d x)\right )}{d}}{3 a}}{5 a}\)

\(\Big \downarrow \) 104

\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}-\int \frac {1}{i-\frac {(a+i b) \tan (c+d x)}{a+b \tan (c+d x)}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}}{3 a}}{5 a}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\int \frac {1}{\frac {(a-i b) \tan (c+d x)}{a+b \tan (c+d x)}+i}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}+\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}\right )}{d}}{3 a}}{5 a}\)

\(\Big \downarrow \) 221

\(\displaystyle -\frac {2 \sqrt {a+b \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}-\frac {-\frac {8 b \sqrt {a+b \tan (c+d x)}}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {-\frac {2 \left (15 a^2-8 b^2\right ) \sqrt {a+b \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}}-\frac {15 a^2 \left (\frac {i \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-b+i a}}-\frac {i \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b+i a}}\right )}{d}}{3 a}}{5 a}\)

Input:

Int[1/(Tan[c + d*x]^(7/2)*Sqrt[a + b*Tan[c + d*x]]),x]
 

Output:

(-2*Sqrt[a + b*Tan[c + d*x]])/(5*a*d*Tan[c + d*x]^(5/2)) - ((-8*b*Sqrt[a + 
 b*Tan[c + d*x]])/(3*a*d*Tan[c + d*x]^(3/2)) + ((-15*a^2*((I*ArcTan[(Sqrt[ 
I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[I*a - b] - (I 
*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqr 
t[I*a + b]))/d - (2*(15*a^2 - 8*b^2)*Sqrt[a + b*Tan[c + d*x]])/(a*d*Sqrt[T 
an[c + d*x]]))/(3*a))/(5*a)
 

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 218

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

rule 221

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

rule 613

Int[Sqrt[(e_.)*(x_)]/(Sqrt[(c_) + (d_.)*(x_)]*((a_) + (b_.)*(x_)^2)), x_Sym 
bol] :> Simp[e/(2*b)   Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] + x)), x 
], x] - Simp[e/(2*b)   Int[1/(Sqrt[e*x]*Sqrt[c + d*x]*(Rt[-a/b, 2] - x)), x 
], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042

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

rule 4052

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

rule 4058

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

rule 4132

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*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[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] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4133

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

Maple [B] (warning: unable to verify)

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

Time = 0.76 (sec) , antiderivative size = 947049, normalized size of antiderivative = 4135.59

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

Input:

int(1/tan(d*x+c)^(7/2)/(a+b*tan(d*x+c))^(1/2),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. 4103 vs. \(2 (183) = 366\).

Time = 0.49 (sec) , antiderivative size = 4103, normalized size of antiderivative = 17.92 \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

Integral(1/(sqrt(a + b*tan(c + d*x))*tan(c + d*x)**(7/2)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx=\frac {-16 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{2} b^{2}+8 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right ) a b -6 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, a^{2}-15 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{3} b +\tan \left (d x +c \right )^{2} a}d x \right ) \tan \left (d x +c \right )^{3} a^{3} d}{15 \tan \left (d x +c \right )^{3} a^{3} d} \] Input:

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

Output:

( - 16*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2*b**2 + 
8*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)*a*b - 6*sqrt(ta 
n(c + d*x))*sqrt(tan(c + d*x)*b + a)*a**2 - 15*int((sqrt(tan(c + d*x))*sqr 
t(tan(c + d*x)*b + a))/(tan(c + d*x)**3*b + tan(c + d*x)**2*a),x)*tan(c + 
d*x)**3*a**3*d)/(15*tan(c + d*x)**3*a**3*d)