\(\int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [627]

Optimal result

Integrand size = 25, antiderivative size = 183 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {2 b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {(i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.33 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\frac {2 \sqrt {a} b^{5/2} \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}-\frac {\sqrt [4]{-1} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}+\sqrt [4]{-1} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\tan (c+d x)}+2 a^2 \sqrt {a+b \tan (c+d x)}}{\sqrt {\tan (c+d x)}}}{d} \] Input:

Integrate[(a + b*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 4048, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4048

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}+\frac {2 \left (\frac {1}{2} (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+b^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-\frac {1}{2} (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )\right )}{d}\)

Input:

Int[(a + b*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(3/2),x]
 

Output:

(2*(((I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* 
Tan[c + d*x]]])/2 + b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
b*Tan[c + d*x]]] - ((I*a + b)^(5/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d* 
x]])/Sqrt[a + b*Tan[c + d*x]]])/2))/d - (2*a^2*Sqrt[a + b*Tan[c + d*x]])/( 
d*Sqrt[Tan[c + d*x]])
 

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 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)

Leaf count of result is larger than twice the leaf count of optimal. \(2432\) vs. \(2(151)=302\).

Time = 2.64 (sec) , antiderivative size = 2433, normalized size of antiderivative = 13.30

method	result	size
default	\(\text {Expression too large to display}\)	\(2433\)

Input:

int((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/2/d*2^(1/2)/(b+(a^2+b^2)^(1/2))^(1/2)*((2*csc(d*x+c)-2*cot(d*x+c))*(b+( 
a^2+b^2)^(1/2))^(1/2)*b*(a^2+b^2)^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*ln(1/(- 
cos(d*x+c)+1)*(a*(-cos(d*x+c)+1)^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(-cos(d*x+ 
c)+1)-2*(-2*(a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2) 
*(-b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*b*(-cos(d*x+c)+1)-a*sin(d*x+c)))+ 
(-2*csc(d*x+c)+2*cot(d*x+c))*(b+(a^2+b^2)^(1/2))^(1/2)*b*(a^2+b^2)^(1/2)*( 
-b+(a^2+b^2)^(1/2))^(1/2)*ln(1/(-cos(d*x+c)+1)*(a*(-cos(d*x+c)+1)^2*csc(d* 
x+c)+2*(a^2+b^2)^(1/2)*(-cos(d*x+c)+1)+2*(-2*(a*cos(d*x+c)+b*sin(d*x+c))*s 
in(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2* 
b*(-cos(d*x+c)+1)-a*sin(d*x+c)))+(-2*csc(d*x+c)+2*cot(d*x+c))*b^3*arctan(1 
/(b+(a^2+b^2)^(1/2))^(1/2)*(-(-b+(a^2+b^2)^(1/2))^(1/2)*(csc(d*x+c)-cot(d* 
x+c))+(-2*(a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))/ 
(-cos(d*x+c)+1)*sin(d*x+c))+(-2*csc(d*x+c)+2*cot(d*x+c))*b^3*arctan(1/(b+( 
a^2+b^2)^(1/2))^(1/2)*((-b+(a^2+b^2)^(1/2))^(1/2)*(csc(d*x+c)-cot(d*x+c))+ 
(-2*(a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c)+1)^2)^(1/2))/(-cos( 
d*x+c)+1)*sin(d*x+c))+4*b^(5/2)*2^(1/2)*arctan(1/2*2^(1/2)/b^(1/2)/(-cos(d 
*x+c)+1)*sin(d*x+c)*(-2*(a*cos(d*x+c)+b*sin(d*x+c))*sin(d*x+c)/(cos(d*x+c) 
+1)^2)^(1/2))*(b+(a^2+b^2)^(1/2))^(1/2)*(csc(d*x+c)-cot(d*x+c))+(2*csc(d*x 
+c)-2*cot(d*x+c))*a^2*(a^2+b^2)^(1/2)*arctan(1/(b+(a^2+b^2)^(1/2))^(1/2)*( 
-(-b+(a^2+b^2)^(1/2))^(1/2)*(csc(d*x+c)-cot(d*x+c))+(-2*(a*cos(d*x+c)+b...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4698 vs. \(2 (147) = 294\).

Time = 1.14 (sec) , antiderivative size = 9395, normalized size of antiderivative = 51.34 \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

integrate((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate((a+b*tan(d*x+c))**(5/2)/tan(d*x+c)**(3/2),x)
 

Output:

Integral((a + b*tan(c + d*x))**(5/2)/tan(c + d*x)**(3/2), x)
 

Maxima [F]

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

integrate((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(3/2),x, algorithm="maxima")
 

Output:

integrate((b*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(3/2), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:

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

int((a + b*tan(c + d*x))^(5/2)/tan(c + d*x)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

int((a + b*tan(c + d*x))^(5/2)/tan(c + d*x)^(3/2), x)
 

Reduce [F]

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

int((a+b*tan(d*x+c))^(5/2)/tan(d*x+c)^(3/2),x)
 

Output:

(2*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*a*b - 2*int((sqrt(tan(c + d 
*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**2)/(tan(c + d*x)*b + a),x)*a*b 
**2*d - int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x))/(ta 
n(c + d*x)*b + a),x)*a**2*b*d + int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)* 
b + a))/tan(c + d*x)**2,x)*a**2*d + int((sqrt(tan(c + d*x))*sqrt(tan(c + d 
*x)*b + a))/(tan(c + d*x)**2*b + tan(c + d*x)*a),x)*a**2*b*d + int(sqrt(ta 
n(c + d*x))*sqrt(tan(c + d*x)*b + a),x)*b**2*d)/d