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

Optimal. Leaf size=410 \[ \frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {i 3^{-1-m} a^3 e^{3 e-\frac {3 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {3 f (c+d x)}{d}\right )}{8 f}-\frac {3\ 2^{-3-m} a^3 e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )}{f}+\frac {15 i a^3 e^{e-\frac {c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,-\frac {f (c+d x)}{d}\right )}{8 f}+\frac {15 i a^3 e^{-e+\frac {c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {f (c+d x)}{d}\right )}{8 f}+\frac {3\ 2^{-3-m} a^3 e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{f}-\frac {i 3^{-1-m} a^3 e^{-3 e+\frac {3 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {3 f (c+d x)}{d}\right )}{8 f} \]

[Out]

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*GAMM
A(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)

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Rubi [A]
time = 0.42, antiderivative size = 410, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3399, 3393, 3388, 2212, 3389} \begin {gather*} -\frac {i a^3 3^{-m-1} e^{3 e-\frac {3 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\frac {3 f (c+d x)}{d}\right )}{8 f}-\frac {3 a^3 2^{-m-3} e^{2 e-\frac {2 c f}{d}} (c+d x)^m \left (-\frac {f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,-\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} \text {Gamma}\left (m+1,-\frac {f (c+d x)}{d}\right )}{8 f}+\frac {15 i a^3 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 )}{8 f}+\frac {3 a^3 2^{-m-3} e^{\frac {2 c f}{d}-2 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {2 f (c+d x)}{d}\right )}{f}-\frac {i a^3 3^{-m-1} e^{\frac {3 c f}{d}-3 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {3 f (c+d x)}{d}\right )}{8 f}+\frac {5 a^3 (c+d x)^{m+1}}{2 d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m*(a + I*a*Sinh[e + f*x])^3,x]

[Out]

(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)

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 3388

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

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 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[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 3399

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(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])

Rubi steps

\begin {align*} \int (c+d x)^m (a+i a \sinh (e+f x))^3 \, dx &=\left (8 a^3\right ) \int (c+d x)^m \sin ^6\left (\frac {1}{2} \left (i e+\frac {\pi }{2}\right )+\frac {i f x}{2}\right ) \, dx\\ &=\left (8 a^3\right ) \int \left (\frac {5}{16} (c+d x)^m-\frac {3}{16} (c+d x)^m \cosh (2 e+2 f x)+\frac {15}{32} i (c+d x)^m \sinh (e+f x)-\frac {1}{32} i (c+d x)^m \sinh (3 e+3 f x)\right ) \, dx\\ &=\frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{4} \left (i a^3\right ) \int (c+d x)^m \sinh (3 e+3 f x) \, dx+\frac {1}{4} \left (15 i a^3\right ) \int (c+d x)^m \sinh (e+f x) \, dx-\frac {1}{2} \left (3 a^3\right ) \int (c+d x)^m \cosh (2 e+2 f x) \, dx\\ &=\frac {5 a^3 (c+d x)^{1+m}}{2 d (1+m)}-\frac {1}{8} \left (i a^3\right ) \int e^{-i (3 i e+3 i f x)} (c+d x)^m \, dx+\frac {1}{8} \left (i a^3\right ) \int e^{i (3 i e+3 i f x)} (c+d x)^m \, dx+\frac {1}{8} \left (15 i a^3\right ) \int e^{-i (i e+i f x)} (c+d x)^m \, dx-\frac {1}{8} \left (15 i a^3\right ) \int e^{i (i e+i f x)} (c+d x)^m \, dx-\frac {1}{4} \left (3 a^3\right ) \int e^{-i (2 i e+2 i f x)} (c+d x)^m \, dx-\frac {1}{4} \left (3 a^3\right ) \int e^{i (2 i e+2 i f x)} (c+d x)^m \, 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}\\ \end {align*}

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Mathematica [A]
time = 4.37, size = 452, normalized size = 1.10 \begin {gather*} \frac {2^{-3-m} 3^{-1-m} a^3 e^{-3 \left (e+\frac {c f}{d}\right )} (c+d x)^m \left (-\frac {f^2 (c+d x)^2}{d^2}\right )^{-m} \left (5\ 2^m 3^{2+m} d e^{2 e+\frac {4 c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^m \Gamma \left (1+m,f \left (\frac {c}{d}+x\right )\right )-2^m d e^{6 e} (1+m) \left (\frac {f (c+d x)}{d}\right )^m \Gamma \left (1+m,-\frac {3 f (c+d x)}{d}\right )+i 3^{2+m} d e^{5 e+\frac {c f}{d}} (1+m) \left (f \left (\frac {c}{d}+x\right )\right )^m \Gamma \left (1+m,-\frac {2 f (c+d x)}{d}\right )+5\ 2^m 3^{2+m} d e^{4 e+\frac {2 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 )-i 3^{2+m} d e^{e+\frac {5 c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )-2^m e^{\frac {3 c f}{d}} \left (20 i 3^{1+m} e^{3 e} f (c+d x) \left (-\frac {f^2 (c+d x)^2}{d^2}\right )^m+d e^{\frac {3 c f}{d}} (1+m) \left (-\frac {f (c+d x)}{d}\right )^m \Gamma \left (1+m,\frac {3 f (c+d x)}{d}\right )\right )\right ) (-i+\sinh (e+f x))^3}{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 )^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m*(a + I*a*Sinh[e + f*x])^3,x]

[Out]

(2^(-3 - m)*3^(-1 - m)*a^3*(c + d*x)^m*(5*2^m*3^(2 + m)*d*E^(2*e + (4*c*f)/d)*(1 + m)*(-((f*(c + d*x))/d))^m*G
amma[1 + m, f*(c/d + x)] - 2^m*d*E^(6*e)*(1 + m)*((f*(c + d*x))/d)^m*Gamma[1 + m, (-3*f*(c + d*x))/d] + I*3^(2
+ m)*d*E^(5*e + (c*f)/d)*(1 + m)*(f*(c/d + x))^m*Gamma[1 + m, (-2*f*(c + d*x))/d] + 5*2^m*3^(2 + m)*d*E^(4*e
+ (2*c*f)/d)*(1 + m)*((f*(c + d*x))/d)^m*Gamma[1 + m, -((f*(c + d*x))/d)] - I*3^(2 + m)*d*E^(e + (5*c*f)/d)*(1
+ m)*(-((f*(c + d*x))/d))^m*Gamma[1 + m, (2*f*(c + d*x))/d] - 2^m*E^((3*c*f)/d)*((20*I)*3^(1 + m)*E^(3*e)*f*(
c + d*x)*(-((f^2*(c + d*x)^2)/d^2))^m + d*E^((3*c*f)/d)*(1 + m)*(-((f*(c + d*x))/d))^m*Gamma[1 + m, (3*f*(c +
d*x))/d]))*(-I + Sinh[e + f*x])^3)/(d*E^(3*(e + (c*f)/d))*f*(1 + m)*(-((f^2*(c + d*x)^2)/d^2))^m*(Cosh[(e + f*
x)/2] + I*Sinh[(e + f*x)/2])^6)

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{m} \left (a +i a \sinh \left (f x +e \right )\right )^{3}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m*(a+I*a*sinh(f*x+e))^3,x)

[Out]

int((d*x+c)^m*(a+I*a*sinh(f*x+e))^3,x)

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Maxima [A]
time = 0.12, size = 383, normalized size = 0.93 \begin {gather*} -\frac {1}{8} i \, {\left (\frac {{\left (d x + c\right )}^{m + 1} e^{\left (\frac {3 \, c f}{d} - 3 \, e\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 (\frac {c f}{d} - e\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 (-\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 {3 \, c f}{d} + 3 \, e\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 (\frac {2 \, c f}{d} - 2 \, e\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 (-\frac {2 \, c f}{d} + 2 \, e\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 (\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^{3} + \frac {{\left (d x + c\right )}^{m + 1} a^{3}}{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))^3,x, algorithm="maxima")

[Out]

-1/8*I*((d*x + c)^(m + 1)*e^(3*c*f/d - 3*e)*exp_integral_e(-m, 3*(d*x + c)*f/d)/d - 3*(d*x + c)^(m + 1)*e^(c*f
/d - e)*exp_integral_e(-m, (d*x + c)*f/d)/d + 3*(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^(-3*c*f/d + 3*e)*exp_integral_e(-m, -3*(d*x + c)*f/d)/d)*a^3 + 3/4*((d*x + c)^(m
+ 1)*e^(2*c*f/d - 2*e)*exp_integral_e(-m, 2*(d*x + c)*f/d)/d + (d*x + c)^(m + 1)*e^(-2*c*f/d + 2*e)*exp_integ
ral_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^(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
^3 + (d*x + c)^(m + 1)*a^3/(d*(m + 1))

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Fricas [A]
time = 0.10, size = 380, normalized size = 0.93 \begin {gather*} \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 \, c f + 3 \, d e}{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 \, c f + 2 \, d e}{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 ) - c f + d e}{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 ) + c f - d e}{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 \, c f - 2 \, d e}{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 \, c f - 3 \, d e}{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 )}} \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))^3,x, algorithm="fricas")

[Out]

1/24*((-I*a^3*d*m - I*a^3*d)*e^(-(d*m*log(3*f/d) - 3*c*f + 3*d*e)/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*c*f + 2*d*e)/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) - c*f + d*e)/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) + c*f - d*e)/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*c*f - 2*d*e)
/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*c*f - 3*d*e)/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)

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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))**3,x)

[Out]

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

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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))^3,x, algorithm="giac")

[Out]

integrate((I*a*sinh(f*x + e) + a)^3*(d*x + c)^m, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (a+a\,\mathrm {sinh}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,{\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)^3*(c + d*x)^m,x)

[Out]

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

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