\(\int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx\) [550]

Optimal result

Integrand size = 21, antiderivative size = 155 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{5/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{5/2} d}+\frac {2 a}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}}+\frac {2 \left (a^2-b^2\right )}{\left (a^2+b^2\right )^2 d \sqrt {a+b \tan (c+d x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {(a+i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a-i b}\right )+(a-i b) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \tan (c+d x)}{a+i b}\right )}{3 \left (a^2+b^2\right ) d (a+b \tan (c+d x))^{3/2}} \] Input:

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

Output:

((a + I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a - I*b) 
] + (a - I*b)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Tan[c + d*x])/(a + I 
*b)])/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.83 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.18, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 4012, 3042, 4012, 3042, 4022, 3042, 4020, 25, 73, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(2*a)/(3*(a^2 + b^2)*d*(a + b*Tan[c + d*x])^(3/2)) + ((((-I)*(a + I*b)^2*A 
rcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (I*(a - I*b)^2*ArcT 
an[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d))/(a^2 + b^2) + (2*(a^2 - 
 b^2))/((a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]]))/(a^2 + b^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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 3042

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

rule 4012

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

rule 4020

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

rule 4022

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2156\) vs. \(2(133)=266\).

Time = 0.26 (sec) , antiderivative size = 2157, normalized size of antiderivative = 13.92

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2157\)
default	\(\text {Expression too large to display}\)	\(2157\)

Input:

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

Output:

2/3*a/(a^2+b^2)/d/(a+b*tan(d*x+c))^(3/2)+2/d*a^2/(a^2+b^2)^2/(a+b*tan(d*x+ 
c))^(1/2)-2/d/(a^2+b^2)^2/(a+b*tan(d*x+c))^(1/2)*b^2+2/d/(a^2+b^2)^(5/2)/( 
2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b* 
tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a*b^2+1/2/d/(a^2+b^2)^(7 
/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a 
-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2*b^2-6/d/(a^2+b^2)^(7/2 
)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a 
+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3*b^2-5/d/(a^2+b^2) 
^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2) 
-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a*b^4-1/2/d/(a^2 
+b^2)^3*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+ 
c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*b^2-1/2/d*b^2/(a^2+b 
^2)^(7/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a) 
^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+6/d*b^2/(a^2+b^2 
)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2* 
(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+5/d*b^4/(a^ 
2+b^2)^(7/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2 
)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a-2/d*b^2/ 
(a^2+b^2)^(5/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^( 
1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3289 vs. \(2 (129) = 258\).

Time = 0.15 (sec) , antiderivative size = 3289, normalized size of antiderivative = 21.22 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more 
details)Is
 

Giac [F(-2)]

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

integrate(tan(d*x+c)/(a+b*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:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[6,14,5]%%%}+%%%{6,[6,12,5]%%%}+%%%{15,[6,10,5]%%%}+ 
%%%{20,[6
 

Mupad [B] (verification not implemented)

Time = 7.14 (sec) , antiderivative size = 4630, normalized size of antiderivative = 29.87 \[ \int \frac {\tan (c+d x)}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

atan(((((1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^ 
2*10i - 10*a^3*b^2*d^2))^(1/2)*(48*a*b^20*d^4 - ((1/(a^5*d^2 - b^5*d^2*1i 
+ 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(a 
 + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 + 2880*a^5*b^18 
*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11*b^12*d^5 + 1344 
0*a^13*b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 640*a^19*b^4*d^5 
 + 64*a^21*b^2*d^5))/4 + 368*a^3*b^18*d^4 + 1216*a^5*b^16*d^4 + 2240*a^7*b 
^14*d^4 + 2464*a^9*b^12*d^4 + 1568*a^11*b^10*d^4 + 448*a^13*b^8*d^4 - 64*a 
^15*b^6*d^4 - 80*a^17*b^4*d^4 - 16*a^19*b^2*d^4))/2 - ((a + b*tan(c + d*x) 
)^(1/2)*(320*a^4*b^14*d^3 - 16*b^18*d^3 + 1024*a^6*b^12*d^3 + 1440*a^8*b^1 
0*d^3 + 1024*a^10*b^8*d^3 + 320*a^12*b^6*d^3 - 16*a^16*b^2*d^3))/2)*(1/(a^ 
5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3 
*b^2*d^2))^(1/2)*1i - (((1/(a^5*d^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2 
*5i + a^2*b^3*d^2*10i - 10*a^3*b^2*d^2))^(1/2)*(48*a*b^20*d^4 + ((1/(a^5*d 
^2 - b^5*d^2*1i + 5*a*b^4*d^2 - a^4*b*d^2*5i + a^2*b^3*d^2*10i - 10*a^3*b^ 
2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*(64*a*b^22*d^5 + 640*a^3*b^20*d^5 
 + 2880*a^5*b^18*d^5 + 7680*a^7*b^16*d^5 + 13440*a^9*b^14*d^5 + 16128*a^11 
*b^12*d^5 + 13440*a^13*b^10*d^5 + 7680*a^15*b^8*d^5 + 2880*a^17*b^6*d^5 + 
640*a^19*b^4*d^5 + 64*a^21*b^2*d^5))/4 + 368*a^3*b^18*d^4 + 1216*a^5*b^16* 
d^4 + 2240*a^7*b^14*d^4 + 2464*a^9*b^12*d^4 + 1568*a^11*b^10*d^4 + 448*...
 

Reduce [F]

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

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

Output:

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