\(\int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [613]

Optimal result

Integrand size = 25, antiderivative size = 181 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {i \sqrt {i a-b} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i \sqrt {i a+b} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 b \sqrt {a+b \tan (c+d x)}}{3 a d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.89 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {-3 (-1)^{3/4} \sqrt {-a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 (-1)^{3/4} \sqrt {a+i b} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {2 (a+b \tan (c+d x))^{3/2}}{a \tan ^{\frac {3}{2}}(c+d x)}}{3 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4051, 27, 3042, 4132, 27, 2011, 3042, 4058, 610, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4051

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4058

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

\(\Big \downarrow \) 610

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 610

Int[(((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_))/((a_) + (b_.)*(x_)^2), x_S 
ymbol] :> Simp[e^(m + 1/2)   Int[ExpandIntegrand[1/(Sqrt[e*x]*Sqrt[c + d*x] 
), x^(m + 1/2)*((c + d*x)^(n + 1/2)/(a + b*x^2)), x], x], x] /; FreeQ[{a, b 
, c, d, e}, x] && IGtQ[n + 1/2, 0] && ILtQ[m - 1/2, 0]
 

rule 2009

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

rule 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

rule 3042

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

rule 4051

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

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])))
 

Maple [B] (warning: unable to verify)

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

Time = 0.62 (sec) , antiderivative size = 1089917, normalized size of antiderivative = 6021.64

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

Input:

int((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),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. 2743 vs. \(2 (141) = 282\).

Time = 0.31 (sec) , antiderivative size = 2743, normalized size of antiderivative = 15.15 \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

1/24*(3*a*d*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^ 
3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*ta 
n(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x 
 + c) + a)*sqrt(tan(d*x + c)) + ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2* 
(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x + c) - (3*a^4*b + 4*a^2*b^3)*d + (a^ 
4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan(d*x + c)^2 - 4*(a^3*b + 2*a*b^3) 
*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2))/(t 
an(d*x + c)^2 + 1))*tan(d*x + c)^2 + 3*a*d*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/ 
d^2)*log(-(2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 3*a^3*b^2 + 4*a*b^4)*tan(d*x + 
c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) - (a^4 + 3*a^2*b^2 + 4*b^4)*d^2 
)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)*sqrt(tan(d*x + c)) + ((a^4*b + 
4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^3*b^2 + 4*a*b^4)*d*tan(d*x + c) 
 - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - (a^4 + 6*a^2*b^2 + 8*b^4)*d^3*tan( 
d*x + c)^2 - 4*(a^3*b + 2*a*b^3)*d^3*tan(d*x + c))*sqrt(-a^2/d^4))*sqrt(-( 
d^2*sqrt(-a^2/d^4) + b)/d^2))/(tan(d*x + c)^2 + 1))*tan(d*x + c)^2 - 3*a*d 
*sqrt(-(d^2*sqrt(-a^2/d^4) + b)/d^2)*log((2*(2*a^4*b + 4*a^2*b^3 + (a^5 + 
3*a^3*b^2 + 4*a*b^4)*tan(d*x + c) + (2*(a^3*b + 2*a*b^3)*d^2*tan(d*x + c) 
- (a^4 + 3*a^2*b^2 + 4*b^4)*d^2)*sqrt(-a^2/d^4))*sqrt(b*tan(d*x + c) + a)* 
sqrt(tan(d*x + c)) - ((a^4*b + 4*a^2*b^3)*d*tan(d*x + c)^2 - 2*(a^5 + 3*a^ 
3*b^2 + 4*a*b^4)*d*tan(d*x + c) - (3*a^4*b + 4*a^2*b^3)*d + (a^4*d^3 - ...
 

Sympy [F]

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

integrate((a+b*tan(d*x+c))**(1/2)/tan(d*x+c)**(5/2),x)
 

Output:

Integral(sqrt(a + b*tan(c + d*x))/tan(c + d*x)**(5/2), x)
 

Maxima [F]

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

integrate((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),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. \[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int((a + b*tan(c + d*x))^(1/2)/tan(c + d*x)^(5/2),x)
 

Output:

int((a + b*tan(c + d*x))^(1/2)/tan(c + d*x)^(5/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {a+b \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{3}}d x \] Input:

int((a+b*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),x)
 

Output:

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