3.13.26 \(\int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx\) [1226]

3.13.26.1 Optimal result

Integrand size = 25, antiderivative size = 67 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2 i \sqrt [4]{a-i a x}}{5 a^2 (a+i a x)^{5/4}}+\frac {4 i \sqrt [4]{a-i a x}}{5 a^3 \sqrt [4]{a+i a x}} \]

output

2/5*I*(a-I*a*x)^(1/4)/a^2/(a+I*a*x)^(5/4)+4/5*I*(a-I*a*x)^(1/4)/a^3/(a+I*a 
*x)^(1/4)
 

3.13.26.2 Mathematica [A] (verified)

Time = 2.33 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2 (3+2 i x) \sqrt [4]{a-i a x}}{5 a^3 (-i+x) \sqrt [4]{a+i a x}} \]

input

Integrate[1/((a - I*a*x)^(3/4)*(a + I*a*x)^(9/4)),x]
 

output

(2*(3 + (2*I)*x)*(a - I*a*x)^(1/4))/(5*a^3*(-I + x)*(a + I*a*x)^(1/4))
 

3.13.26.3 Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {55, 48}

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.

\(\displaystyle \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {2 \int \frac {1}{(a-i a x)^{3/4} (i x a+a)^{5/4}}dx}{5 a}+\frac {2 i \sqrt [4]{a-i a x}}{5 a^2 (a+i a x)^{5/4}}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {4 i \sqrt [4]{a-i a x}}{5 a^3 \sqrt [4]{a+i a x}}+\frac {2 i \sqrt [4]{a-i a x}}{5 a^2 (a+i a x)^{5/4}}\)

input

Int[1/((a - I*a*x)^(3/4)*(a + I*a*x)^(9/4)),x]
 

output

(((2*I)/5)*(a - I*a*x)^(1/4))/(a^2*(a + I*a*x)^(5/4)) + (((4*I)/5)*(a - I* 
a*x)^(1/4))/(a^3*(a + I*a*x)^(1/4))
 

3.13.26.3.1 Defintions of rubi rules used

rule 48

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

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

3.13.26.4 Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.55

method	result	size
gosper	\(-\frac {2 \left (x +i\right ) \left (-x +i\right ) \left (-2 x +3 i\right )}{5 \left (-i a x +a \right )^{\frac {3}{4}} \left (i a x +a \right )^{\frac {9}{4}}}\)	\(37\)
risch	\(\frac {\frac {4}{5} x^{2}-\frac {2}{5} i x +\frac {6}{5}}{a^{2} \left (-a \left (i x -1\right )\right )^{\frac {3}{4}} \left (a \left (i x +1\right )\right )^{\frac {1}{4}} \left (x -i\right )}\)	\(44\)

input

int(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x,method=_RETURNVERBOSE)
 

output

-2/5*(x+I)*(-x+I)*(-2*x+3*I)/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4)
 

3.13.26.5 Fricas [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2 \, {\left (i \, a x + a\right )}^{\frac {3}{4}} {\left (-i \, a x + a\right )}^{\frac {1}{4}} {\left (2 \, x - 3 i\right )}}{5 \, {\left (a^{4} x^{2} - 2 i \, a^{4} x - a^{4}\right )}} \]

input

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x, algorithm="fricas")
 

output

2/5*(I*a*x + a)^(3/4)*(-I*a*x + a)^(1/4)*(2*x - 3*I)/(a^4*x^2 - 2*I*a^4*x 
- a^4)
 

3.13.26.6 Sympy [F]

\[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\int \frac {1}{\left (i a \left (x - i\right )\right )^{\frac {9}{4}} \left (- i a \left (x + i\right )\right )^{\frac {3}{4}}}\, dx \]

input

integrate(1/(a-I*a*x)**(3/4)/(a+I*a*x)**(9/4),x)
 

output

Integral(1/((I*a*(x - I))**(9/4)*(-I*a*(x + I))**(3/4)), x)
 

3.13.26.7 Maxima [F]

\[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\int { \frac {1}{{\left (i \, a x + a\right )}^{\frac {9}{4}} {\left (-i \, a x + a\right )}^{\frac {3}{4}}} \,d x } \]

input

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x, algorithm="maxima")
 

output

integrate(1/((I*a*x + a)^(9/4)*(-I*a*x + a)^(3/4)), x)
 

3.13.26.8 Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\text {Exception raised: TypeError} \]

input

integrate(1/(a-I*a*x)^(3/4)/(a+I*a*x)^(9/4),x, algorithm="giac")
 

output

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 
]ext_reduce Error: Bad Argument TypeDone
 

3.13.26.9 Mupad [B] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.57 \[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {2\,\left (3+x\,2{}\mathrm {i}\right )\,{\left (-a\,\left (-1+x\,1{}\mathrm {i}\right )\right )}^{1/4}}{5\,a^3\,\left (x-\mathrm {i}\right )\,{\left (a\,\left (1+x\,1{}\mathrm {i}\right )\right )}^{1/4}} \]

input

int(1/((a - a*x*1i)^(3/4)*(a + a*x*1i)^(9/4)),x)
 

output

(2*(x*2i + 3)*(-a*(x*1i - 1))^(1/4))/(5*a^3*(x - 1i)*(a*(x*1i + 1))^(1/4))
 

3.13.26.10 Reduce [F]

\[ \int \frac {1}{(a-i a x)^{3/4} (a+i a x)^{9/4}} \, dx=\frac {\int \frac {1}{2 \left (i x +1\right )^{\frac {1}{4}} \left (-i x +1\right )^{\frac {3}{4}} i x -\left (i x +1\right )^{\frac {1}{4}} \left (-i x +1\right )^{\frac {3}{4}} x^{2}+\left (i x +1\right )^{\frac {1}{4}} \left (-i x +1\right )^{\frac {3}{4}}}d x}{a^{3}} \]

input

int(1/((a*i*x + a)**(1/4)*( - a*i*x + a)**(3/4)*a**2*(2*i*x - x**2 + 1)),x 
)
 

output

int(1/(2*(i*x + 1)**(1/4)*( - i*x + 1)**(3/4)*i*x - (i*x + 1)**(1/4)*( - i 
*x + 1)**(3/4)*x**2 + (i*x + 1)**(1/4)*( - i*x + 1)**(3/4)),x)/a**3