3.1.84 \(\int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [84]

3.1.84.1 Optimal result

Integrand size = 28, antiderivative size = 141 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {4 \sqrt {2} a^{5/2} (i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {4 a^2 (i A+B) \sqrt {a+i a \tan (c+d x)}}{d}+\frac {2 a (i A+B) (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d} \]

output

-4*a^(5/2)*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2 
^(1/2)/d+4*a^2*(I*A+B)*(a+I*a*tan(d*x+c))^(1/2)/d+2/3*a*(I*A+B)*(a+I*a*tan 
(d*x+c))^(3/2)/d+2/5*B*(a+I*a*tan(d*x+c))^(5/2)/d
 

3.1.84.2 Mathematica [A] (verified)

Time = 0.87 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.82 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {-60 i \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 a^2 \sqrt {a+i a \tan (c+d x)} \left (35 i A+38 B+(-5 A+11 i B) \tan (c+d x)-3 B \tan ^2(c+d x)\right )}{15 d} \]

input

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

output

((-60*I)*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqr 
t[2]*Sqrt[a])] + 2*a^2*Sqrt[a + I*a*Tan[c + d*x]]*((35*I)*A + 38*B + (-5*A 
 + (11*I)*B)*Tan[c + d*x] - 3*B*Tan[c + d*x]^2))/(15*d)
 

3.1.84.3 Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.97, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4010, 3042, 3959, 3042, 3959, 3042, 3961, 219}

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.

\(\displaystyle \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))dx\)

\(\Big \downarrow \) 4010

\(\displaystyle (A-i B) \int (i \tan (c+d x) a+a)^{5/2}dx+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-i B) \int (i \tan (c+d x) a+a)^{5/2}dx+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3959

\(\displaystyle (A-i B) \left (2 a \int (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-i B) \left (2 a \int (i \tan (c+d x) a+a)^{3/2}dx+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3959

\(\displaystyle (A-i B) \left (2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle (A-i B) \left (2 a \left (2 a \int \sqrt {i \tan (c+d x) a+a}dx+\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 3961

\(\displaystyle (A-i B) \left (2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {4 i a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

\(\Big \downarrow \) 219

\(\displaystyle (A-i B) \left (2 a \left (\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\right )+\frac {2 i a (a+i a \tan (c+d x))^{3/2}}{3 d}\right )+\frac {2 B (a+i a \tan (c+d x))^{5/2}}{5 d}\)

input

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

output

(2*B*(a + I*a*Tan[c + d*x])^(5/2))/(5*d) + (A - I*B)*((((2*I)/3)*a*(a + I* 
a*Tan[c + d*x])^(3/2))/d + 2*a*(((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I 
*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((2*I)*a*Sqrt[a + I*a*Tan[c + d*x 
]])/d))
 

3.1.84.3.1 Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 3959

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[2*a   Int[(a + b*Tan[c + d* 
x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && GtQ[n 
, 1]
 

rule 3961

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) 
  Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a 
, b, c, d}, x] && EqQ[a^2 + b^2, 0]
 

rule 4010

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

3.1.84.4 Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.00

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(141\)
default	\(\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}-\frac {i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}+2 A \,a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(141\)
parts	\(\frac {2 i A a \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+2 a \sqrt {a +i a \tan \left (d x +c \right )}-2 a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}+\frac {B \left (\frac {2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a^{2} \sqrt {a +i a \tan \left (d x +c \right )}-4 a^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )\right )}{d}\)	\(164\)

input

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

output

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

3.1.84.5 Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 421 vs. \(2 (108) = 216\).

Time = 0.25 (sec) , antiderivative size = 421, normalized size of antiderivative = 2.99 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2 \, {\left (15 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 15 \, \sqrt {2} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a^{2}}\right ) - 2 \, \sqrt {2} {\left (2 \, {\left (-10 i \, A - 13 \, B\right )} a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} + 35 \, {\left (-i \, A - B\right )} a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, {\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

input

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

output

2/15*(15*sqrt(2)*sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I* 
c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) + 
sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e 
^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 15*sqrt(2)* 
sqrt(-(A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d 
*x + 2*I*c) + d)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) - sqrt(-(A^2 - 2*I* 
A*B - B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c 
) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 2*sqrt(2)*(2*(-10*I*A - 13*B 
)*a^2*e^(5*I*d*x + 5*I*c) + 35*(-I*A - B)*a^2*e^(3*I*d*x + 3*I*c) + 15*(-I 
*A - B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I* 
d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)
 

3.1.84.6 Sympy [F]

\[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (c + d x \right )}\right )\, dx \]

input

integrate((a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)
 

output

Integral((I*a*(tan(c + d*x) - I))**(5/2)*(A + B*tan(c + d*x)), x)
 

3.1.84.7 Maxima [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.95 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2 i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 3 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A - i \, B\right )} a^{2} + 30 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A - i \, B\right )} a^{3}\right )}}{15 \, a d} \]

input

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

output

2/15*I*(15*sqrt(2)*(A - I*B)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan( 
d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) - 3*I*(I*a* 
tan(d*x + c) + a)^(5/2)*B*a + 5*(I*a*tan(d*x + c) + a)^(3/2)*(A - I*B)*a^2 
 + 30*sqrt(I*a*tan(d*x + c) + a)*(A - I*B)*a^3)/(a*d)
 

3.1.84.8 Giac [F(-1)]

Timed out. \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

input

integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
 

output

Timed out
 

3.1.84.9 Mupad [B] (verification not implemented)

Time = 1.16 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.33 \[ \int (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{5\,d}+\frac {A\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {2\,B\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,d}+\frac {A\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{d}+\frac {4\,B\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{d}-\frac {\sqrt {2}\,A\,{\left (-a\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,4{}\mathrm {i}}{d}+\frac {\sqrt {2}\,B\,a^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,4{}\mathrm {i}}{d} \]

input

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

output

(2*B*(a + a*tan(c + d*x)*1i)^(5/2))/(5*d) + (A*a*(a + a*tan(c + d*x)*1i)^( 
3/2)*2i)/(3*d) + (2*B*a*(a + a*tan(c + d*x)*1i)^(3/2))/(3*d) + (A*a^2*(a + 
 a*tan(c + d*x)*1i)^(1/2)*4i)/d + (4*B*a^2*(a + a*tan(c + d*x)*1i)^(1/2))/ 
d - (2^(1/2)*A*(-a)^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2* 
(-a)^(1/2)))*4i)/d + (2^(1/2)*B*a^(5/2)*atan((2^(1/2)*(a + a*tan(c + d*x)* 
1i)^(1/2)*1i)/(2*a^(1/2)))*4i)/d