3.2.54 \(\int (c+d x)^m (a+i a \sinh (e+f x)) \, dx\) [154]

Optimal. Leaf size=135 \[ \frac {a (c+d x)^{1+m}}{d (1+m)}+\frac {i a e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )}{2 f}+\frac {i a e^{-e+\frac {c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )}{2 f} \]

[Out]

a*(d*x+c)^(1+m)/d/(1+m)+1/2*I*a*exp(e-c*f/d)*(d*x+c)^m*GAMMA(1+m,-f*(d*x+c)/d)/f/((-f*(d*x+c)/d)^m)+1/2*I*a*ex
p(-e+c*f/d)*(d*x+c)^m*GAMMA(1+m,f*(d*x+c)/d)/f/((f*(d*x+c)/d)^m)

________________________________________________________________________________________

Rubi [A]
time = 0.10, antiderivative size = 135, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3398, 3389, 2212} \begin {gather*} \frac {i a e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {f (c+d x)}{d}\right )}{2 f}+\frac {i a e^{\frac {c f}{d}-e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {f (c+d x)}{d}\right )}{2 f}+\frac {a (c+d x)^{m+1}}{d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*(a + I*a*Sinh[e + f*x]),x]

[Out]

(a*(c + d*x)^(1 + m))/(d*(1 + m)) + ((I/2)*a*E^(e - (c*f)/d)*(c + d*x)^m*Gamma[1 + m, -((f*(c + d*x))/d)])/(f*
(-((f*(c + d*x))/d))^m) + ((I/2)*a*E^(-e + (c*f)/d)*(c + d*x)^m*Gamma[1 + m, (f*(c + d*x))/d])/(f*((f*(c + d*x
))/d)^m)

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3389

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/E^(I*(e + f*x))
, x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d, e, f, m}, x]

Rule 3398

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

Rubi steps

\begin {align*} \int (c+d x)^m (a+i a \sinh (e+f x)) \, dx &=\int \left (a (c+d x)^m+i a (c+d x)^m \sinh (e+f x)\right ) \, dx\\ &=\frac {a (c+d x)^{1+m}}{d (1+m)}+(i a) \int (c+d x)^m \sinh (e+f x) \, dx\\ &=\frac {a (c+d x)^{1+m}}{d (1+m)}+\frac {1}{2} (i a) \int e^{-i (i e+i f x)} (c+d x)^m \, dx-\frac {1}{2} (i a) \int e^{i (i e+i f x)} (c+d x)^m \, dx\\ &=\frac {a (c+d x)^{1+m}}{d (1+m)}+\frac {i a 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 )}{2 f}+\frac {i a 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 )}{2 f}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 4.67, size = 166, normalized size = 1.23 \begin {gather*} \frac {a (c+d x)^m \left (2 f (c+d x)+i d e^{-e+\frac {c f}{d}} (1+m) \left (f \left (\frac {c}{d}+x\right )\right )^{-m} \Gamma \left (1+m,f \left (\frac {c}{d}+x\right )\right )+i d e^{e-\frac {c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )\right ) (1+i \sinh (e+f x))}{2 d f (1+m) \left (\cosh \left (\frac {1}{2} (e+f x)\right )+i \sinh \left (\frac {1}{2} (e+f x)\right )\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m*(a + I*a*Sinh[e + f*x]),x]

[Out]

(a*(c + d*x)^m*(2*f*(c + d*x) + (I*d*E^(-e + (c*f)/d)*(1 + m)*Gamma[1 + m, f*(c/d + x)])/(f*(c/d + x))^m + (I*
d*E^(e - (c*f)/d)*(1 + m)*Gamma[1 + m, -((f*(c + d*x))/d)])/(-((f*(c + d*x))/d))^m)*(1 + I*Sinh[e + f*x]))/(2*
d*f*(1 + m)*(Cosh[(e + f*x)/2] + I*Sinh[(e + f*x)/2])^2)

________________________________________________________________________________________

Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{m} \left (a +i a \sinh \left (f x +e \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m*(a+I*a*sinh(f*x+e)),x)

[Out]

int((d*x+c)^m*(a+I*a*sinh(f*x+e)),x)

________________________________________________________________________________________

Maxima [A]
time = 0.06, size = 103, normalized size = 0.76 \begin {gather*} \frac {1}{2} i \, {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (\frac {c f}{d} - e\right )} E_{-m}\left (\frac {{\left (d x + c\right )} f}{d}\right )}{d} - \frac {{\left (d x + c\right )}^{m + 1} e^{\left (-\frac {c f}{d} + e\right )} E_{-m}\left (-\frac {{\left (d x + c\right )} f}{d}\right )}{d}\right )} a + \frac {{\left (d x + c\right )}^{m + 1} a}{d {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*(a+I*a*sinh(f*x+e)),x, algorithm="maxima")

[Out]

1/2*I*((d*x + c)^(m + 1)*e^(c*f/d - e)*exp_integral_e(-m, (d*x + c)*f/d)/d - (d*x + c)^(m + 1)*e^(-c*f/d + e)*
exp_integral_e(-m, -(d*x + c)*f/d)/d)*a + (d*x + c)^(m + 1)*a/(d*(m + 1))

________________________________________________________________________________________

Fricas [A]
time = 0.10, size = 136, normalized size = 1.01 \begin {gather*} \frac {{\left (i \, a d m + i \, a d\right )} e^{\left (-\frac {d m \log \left (\frac {f}{d}\right ) - c f + d e}{d}\right )} \Gamma \left (m + 1, \frac {d f x + c f}{d}\right ) + {\left (i \, a d m + i \, a d\right )} e^{\left (-\frac {d m \log \left (-\frac {f}{d}\right ) + c f - d e}{d}\right )} \Gamma \left (m + 1, -\frac {d f x + c f}{d}\right ) + 2 \, {\left (a d f x + a c f\right )} {\left (d x + c\right )}^{m}}{2 \, {\left (d f m + d f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*(a+I*a*sinh(f*x+e)),x, algorithm="fricas")

[Out]

1/2*((I*a*d*m + I*a*d)*e^(-(d*m*log(f/d) - c*f + d*e)/d)*gamma(m + 1, (d*f*x + c*f)/d) + (I*a*d*m + I*a*d)*e^(
-(d*m*log(-f/d) + c*f - d*e)/d)*gamma(m + 1, -(d*f*x + c*f)/d) + 2*(a*d*f*x + a*c*f)*(d*x + c)^m)/(d*f*m + d*f
)

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**m*(a+I*a*sinh(f*x+e)),x)

[Out]

Exception raised: TypeError >> cannot determine truth value of Relational

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m*(a+I*a*sinh(f*x+e)),x, algorithm="giac")

[Out]

integrate((I*a*sinh(f*x + e) + a)*(d*x + c)^m, x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,{\left (c+d\,x\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________