3.2.52 \(\int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx\) [152]

3.2.52.1 Optimal result
3.2.52.2 Mathematica [A] (verified)
3.2.52.3 Rubi [A] (verified)
3.2.52.4 Maple [F]
3.2.52.5 Fricas [A] (verification not implemented)
3.2.52.6 Sympy [F(-2)]
3.2.52.7 Maxima [A] (verification not implemented)
3.2.52.8 Giac [F]
3.2.52.9 Mupad [F(-1)]

3.2.52.1 Optimal result

Integrand size = 23, antiderivative size = 410 \[ \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx=\frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {i 3^{-1-m} a^3 e^{3 e-\frac {3 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 f (c+d x)}{d}\right )}{8 f}-\frac {3\ 2^{-3-m} a^3 e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {15 i a^3 e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )}{8 f}+\frac {15 i a^3 e^{-e+\frac {c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )}{8 f}+\frac {3\ 2^{-3-m} a^3 e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{f}-\frac {i 3^{-1-m} a^3 e^{-3 e+\frac {3 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 f (c+d x)}{d}\right )}{8 f} \]

output
5/2*a^3*(d*x+c)^(1+m)/d/(1+m)-1/8*I*3^(-1-m)*a^3*exp(3*e-3*c*f/d)*(d*x+c)^ 
m*GAMMA(1+m,-3*f*(d*x+c)/d)/f/((-f*(d*x+c)/d)^m)-3*2^(-3-m)*a^3*exp(2*e-2* 
c*f/d)*(d*x+c)^m*GAMMA(1+m,-2*f*(d*x+c)/d)/f/((-f*(d*x+c)/d)^m)+15/8*I*a^3 
*exp(e-c*f/d)*(d*x+c)^m*GAMMA(1+m,-f*(d*x+c)/d)/f/((-f*(d*x+c)/d)^m)+15/8* 
I*a^3*exp(-e+c*f/d)*(d*x+c)^m*GAMMA(1+m,f*(d*x+c)/d)/f/((f*(d*x+c)/d)^m)+3 
*2^(-3-m)*a^3*exp(-2*e+2*c*f/d)*(d*x+c)^m*GAMMA(1+m,2*f*(d*x+c)/d)/f/((f*( 
d*x+c)/d)^m)-1/8*I*3^(-1-m)*a^3*exp(-3*e+3*c*f/d)*(d*x+c)^m*GAMMA(1+m,3*f* 
(d*x+c)/d)/f/((f*(d*x+c)/d)^m)
 
3.2.52.2 Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 410, normalized size of antiderivative = 1.00 \[ \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx=\frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {i 3^{-1-m} a^3 e^{3 e-\frac {3 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 f (c+d x)}{d}\right )}{8 f}-\frac {3\ 2^{-3-m} a^3 e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {15 i a^3 e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )}{8 f}+\frac {15 i a^3 e^{-e+\frac {c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )}{8 f}+\frac {3\ 2^{-3-m} a^3 e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{f}-\frac {i 3^{-1-m} a^3 e^{-3 e+\frac {3 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 f (c+d x)}{d}\right )}{8 f} \]

input
Integrate[(c + d*x)^m*(a + I*a*Sinh[e + f*x])^3,x]
 
output
(5*a^3*(c + d*x)^(1 + m))/(2*d*(1 + m)) - ((I/8)*3^(-1 - m)*a^3*E^(3*e - ( 
3*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (-3*f*(c + d*x))/d])/(f*(-((f*(c + d*x) 
)/d))^m) - (3*2^(-3 - m)*a^3*E^(2*e - (2*c*f)/d)*(c + d*x)^m*Gamma[1 + m, 
(-2*f*(c + d*x))/d])/(f*(-((f*(c + d*x))/d))^m) + (((15*I)/8)*a^3*E^(e - ( 
c*f)/d)*(c + d*x)^m*Gamma[1 + m, -((f*(c + d*x))/d)])/(f*(-((f*(c + d*x))/ 
d))^m) + (((15*I)/8)*a^3*E^(-e + (c*f)/d)*(c + d*x)^m*Gamma[1 + m, (f*(c + 
 d*x))/d])/(f*((f*(c + d*x))/d)^m) + (3*2^(-3 - m)*a^3*E^(-2*e + (2*c*f)/d 
)*(c + d*x)^m*Gamma[1 + m, (2*f*(c + d*x))/d])/(f*((f*(c + d*x))/d)^m) - ( 
(I/8)*3^(-1 - m)*a^3*E^(-3*e + (3*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (3*f*(c 
 + d*x))/d])/(f*((f*(c + d*x))/d)^m)
 
3.2.52.3 Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 394, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3799, 25, 25, 3042, 3793, 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.

\(\displaystyle \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^m (a+a \sin (i e+i f x))^3dx\)

\(\Big \downarrow \) 3799

\(\displaystyle 8 a^3 \int -(c+d x)^m \sinh ^6\left (\frac {e}{2}+\frac {f x}{2}-\frac {i \pi }{4}\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle -8 a^3 \int -(c+d x)^m \cosh ^6\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )dx\)

\(\Big \downarrow \) 25

\(\displaystyle 8 a^3 \int (c+d x)^m \cosh ^6\left (\frac {e}{2}+\frac {f x}{2}+\frac {i \pi }{4}\right )dx\)

\(\Big \downarrow \) 3042

\(\displaystyle 8 a^3 \int (c+d x)^m \sin \left (\frac {i e}{2}+\frac {i f x}{2}+\frac {\pi }{4}\right )^6dx\)

\(\Big \downarrow \) 3793

\(\displaystyle 8 a^3 \int \left (-\frac {3}{16} \cosh (2 e+2 f x) (c+d x)^m+\frac {15}{32} i \sinh (e+f x) (c+d x)^m-\frac {1}{32} i \sinh (3 e+3 f x) (c+d x)^m+\frac {5}{16} (c+d x)^m\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle 8 a^3 \left (-\frac {i 3^{-m-1} e^{3 e-\frac {3 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {3 f (c+d x)}{d}\right )}{64 f}-\frac {3\ 2^{-m-6} e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {15 i e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,-\frac {f (c+d x)}{d}\right )}{64 f}+\frac {15 i e^{\frac {c f}{d}-e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {f (c+d x)}{d}\right )}{64 f}+\frac {3\ 2^{-m-6} e^{\frac {2 c f}{d}-2 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {2 f (c+d x)}{d}\right )}{f}-\frac {i 3^{-m-1} e^{\frac {3 c f}{d}-3 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (m+1,\frac {3 f (c+d x)}{d}\right )}{64 f}+\frac {5 (c+d x)^{m+1}}{16 d (m+1)}\right )\)

input
Int[(c + d*x)^m*(a + I*a*Sinh[e + f*x])^3,x]
 
output
8*a^3*((5*(c + d*x)^(1 + m))/(16*d*(1 + m)) - ((I/64)*3^(-1 - m)*E^(3*e - 
(3*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (-3*f*(c + d*x))/d])/(f*(-((f*(c + d*x 
))/d))^m) - (3*2^(-6 - m)*E^(2*e - (2*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (-2 
*f*(c + d*x))/d])/(f*(-((f*(c + d*x))/d))^m) + (((15*I)/64)*E^(e - (c*f)/d 
)*(c + d*x)^m*Gamma[1 + m, -((f*(c + d*x))/d)])/(f*(-((f*(c + d*x))/d))^m) 
 + (((15*I)/64)*E^(-e + (c*f)/d)*(c + d*x)^m*Gamma[1 + m, (f*(c + d*x))/d] 
)/(f*((f*(c + d*x))/d)^m) + (3*2^(-6 - m)*E^(-2*e + (2*c*f)/d)*(c + d*x)^m 
*Gamma[1 + m, (2*f*(c + d*x))/d])/(f*((f*(c + d*x))/d)^m) - ((I/64)*3^(-1 
- m)*E^(-3*e + (3*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (3*f*(c + d*x))/d])/(f* 
((f*(c + d*x))/d)^m))
 

3.2.52.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3793
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In 
t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f 
, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
 

rule 3799
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Simp[(2*a)^n   Int[(c + d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) 
+ f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^ 
2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])
 
3.2.52.4 Maple [F]

\[\int \left (d x +c \right )^{m} \left (a +i a \sinh \left (f x +e \right )\right )^{3}d x\]

input
int((d*x+c)^m*(a+I*a*sinh(f*x+e))^3,x)
 
output
int((d*x+c)^m*(a+I*a*sinh(f*x+e))^3,x)
 
3.2.52.5 Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 374, normalized size of antiderivative = 0.91 \[ \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx=\frac {{\left (-i \, a^{3} d m - i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {3 \, f}{d}\right ) + 3 \, d e - 3 \, c f}{d}\right )} \Gamma \left (m + 1, \frac {3 \, {\left (d f x + c f\right )}}{d}\right ) + 9 \, {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right )} \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - 45 \, {\left (-i \, a^{3} d m - i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (\frac {f}{d}\right ) + d e - c f}{d}\right )} \Gamma \left (m + 1, \frac {d f x + c f}{d}\right ) - 45 \, {\left (-i \, a^{3} d m - i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {f}{d}\right ) - d e + c f}{d}\right )} \Gamma \left (m + 1, -\frac {d f x + c f}{d}\right ) - 9 \, {\left (a^{3} d m + a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {2 \, f}{d}\right ) - 2 \, d e + 2 \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {2 \, {\left (d f x + c f\right )}}{d}\right ) + {\left (-i \, a^{3} d m - i \, a^{3} d\right )} e^{\left (-\frac {d m \log \left (-\frac {3 \, f}{d}\right ) - 3 \, d e + 3 \, c f}{d}\right )} \Gamma \left (m + 1, -\frac {3 \, {\left (d f x + c f\right )}}{d}\right ) + 60 \, {\left (a^{3} d f x + a^{3} c f\right )} {\left (d x + c\right )}^{m}}{24 \, {\left (d f m + d f\right )}} \]

input
integrate((d*x+c)^m*(a+I*a*sinh(f*x+e))^3,x, algorithm="fricas")
 
output
1/24*((-I*a^3*d*m - I*a^3*d)*e^(-(d*m*log(3*f/d) + 3*d*e - 3*c*f)/d)*gamma 
(m + 1, 3*(d*f*x + c*f)/d) + 9*(a^3*d*m + a^3*d)*e^(-(d*m*log(2*f/d) + 2*d 
*e - 2*c*f)/d)*gamma(m + 1, 2*(d*f*x + c*f)/d) - 45*(-I*a^3*d*m - I*a^3*d) 
*e^(-(d*m*log(f/d) + d*e - c*f)/d)*gamma(m + 1, (d*f*x + c*f)/d) - 45*(-I* 
a^3*d*m - I*a^3*d)*e^(-(d*m*log(-f/d) - d*e + c*f)/d)*gamma(m + 1, -(d*f*x 
 + c*f)/d) - 9*(a^3*d*m + a^3*d)*e^(-(d*m*log(-2*f/d) - 2*d*e + 2*c*f)/d)* 
gamma(m + 1, -2*(d*f*x + c*f)/d) + (-I*a^3*d*m - I*a^3*d)*e^(-(d*m*log(-3* 
f/d) - 3*d*e + 3*c*f)/d)*gamma(m + 1, -3*(d*f*x + c*f)/d) + 60*(a^3*d*f*x 
+ a^3*c*f)*(d*x + c)^m)/(d*f*m + d*f)
 
3.2.52.6 Sympy [F(-2)]

Exception generated. \[ \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx=\text {Exception raised: TypeError} \]

input
integrate((d*x+c)**m*(a+I*a*sinh(f*x+e))**3,x)
 
output
Exception raised: TypeError >> cannot determine truth value of Relational
 
3.2.52.7 Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 375, normalized size of antiderivative = 0.91 \[ \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx=-\frac {1}{8} i \, {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-3 \, e + \frac {3 \, c f}{d}\right )} E_{-m}\left (\frac {3 \, {\left (d x + c\right )} f}{d}\right )}{d} - \frac {3 \, {\left (d x + c\right )}^{m + 1} e^{\left (-e + \frac {c f}{d}\right )} E_{-m}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} + \frac {3 \, {\left (d x + c\right )}^{m + 1} e^{\left (e - \frac {c f}{d}\right )} E_{-m}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d} - \frac {{\left (d x + c\right )}^{m + 1} e^{\left (3 \, e - \frac {3 \, c f}{d}\right )} E_{-m}\left (-\frac {3 \, {\left (d x + c\right )} f}{d}\right )}{d}\right )} a^{3} + \frac {3}{4} \, {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-2 \, e + \frac {2 \, c f}{d}\right )} E_{-m}\left (\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {{\left (d x + c\right )}^{m + 1} e^{\left (2 \, e - \frac {2 \, c f}{d}\right )} E_{-m}\left (-\frac {2 \, {\left (d x + c\right )} f}{d}\right )}{d} + \frac {2 \, {\left (d x + c\right )}^{m + 1}}{d {\left (m + 1\right )}}\right )} a^{3} + \frac {3}{2} i \, {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (-e + \frac {c f}{d}\right )} E_{-m}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} - \frac {{\left (d x + c\right )}^{m + 1} e^{\left (e - \frac {c f}{d}\right )} E_{-m}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} a^{3} + \frac {{\left (d x + c\right )}^{m + 1} a^{3}}{d {\left (m + 1\right )}} \]

input
integrate((d*x+c)^m*(a+I*a*sinh(f*x+e))^3,x, algorithm="maxima")
 
output
-1/8*I*((d*x + c)^(m + 1)*e^(-3*e + 3*c*f/d)*exp_integral_e(-m, 3*(d*x + c 
)*f/d)/d - 3*(d*x + c)^(m + 1)*e^(-e + c*f/d)*exp_integral_e(-m, (d*x + c) 
*f/d)/d + 3*(d*x + c)^(m + 1)*e^(e - c*f/d)*exp_integral_e(-m, -(d*x + c)* 
f/d)/d - (d*x + c)^(m + 1)*e^(3*e - 3*c*f/d)*exp_integral_e(-m, -3*(d*x + 
c)*f/d)/d)*a^3 + 3/4*((d*x + c)^(m + 1)*e^(-2*e + 2*c*f/d)*exp_integral_e( 
-m, 2*(d*x + c)*f/d)/d + (d*x + c)^(m + 1)*e^(2*e - 2*c*f/d)*exp_integral_ 
e(-m, -2*(d*x + c)*f/d)/d + 2*(d*x + c)^(m + 1)/(d*(m + 1)))*a^3 + 3/2*I*( 
(d*x + c)^(m + 1)*e^(-e + c*f/d)*exp_integral_e(-m, (d*x + c)*f/d)/d - (d* 
x + c)^(m + 1)*e^(e - c*f/d)*exp_integral_e(-m, -(d*x + c)*f/d)/d)*a^3 + ( 
d*x + c)^(m + 1)*a^3/(d*(m + 1))
 
3.2.52.8 Giac [F]

\[ \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx=\int { {\left (i \, a \sinh \left (f x + e\right ) + a\right )}^{3} {\left (d x + c\right )}^{m} \,d x } \]

input
integrate((d*x+c)^m*(a+I*a*sinh(f*x+e))^3,x, algorithm="giac")
 
output
integrate((I*a*sinh(f*x + e) + a)^3*(d*x + c)^m, x)
 
3.2.52.9 Mupad [F(-1)]

Timed out. \[ \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx=\int {\left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,{\left (c+d\,x\right )}^m \,d x \]

input
int((a + a*sinh(e + f*x)*1i)^3*(c + d*x)^m,x)
 
output
int((a + a*sinh(e + f*x)*1i)^3*(c + d*x)^m, x)