3.2.4 \(\int (c+d x) (a+i a \sinh (e+f x))^2 \, dx\) [104]

Optimal. Leaf size=122 \[ \frac {1}{2} a^2 c x+\frac {1}{4} a^2 d x^2+\frac {a^2 (c+d x)^2}{2 d}+\frac {2 i a^2 (c+d x) \cosh (e+f x)}{f}-\frac {2 i a^2 d \sinh (e+f x)}{f^2}-\frac {a^2 (c+d x) \cosh (e+f x) \sinh (e+f x)}{2 f}+\frac {a^2 d \sinh ^2(e+f x)}{4 f^2} \]

[Out]

1/2*a^2*c*x+1/4*a^2*d*x^2+1/2*a^2*(d*x+c)^2/d+2*I*a^2*(d*x+c)*cosh(f*x+e)/f-2*I*a^2*d*sinh(f*x+e)/f^2-1/2*a^2*
(d*x+c)*cosh(f*x+e)*sinh(f*x+e)/f+1/4*a^2*d*sinh(f*x+e)^2/f^2

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Rubi [A]
time = 0.07, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3398, 3377, 2717, 3391} \begin {gather*} \frac {2 i a^2 (c+d x) \cosh (e+f x)}{f}-\frac {a^2 (c+d x) \sinh (e+f x) \cosh (e+f x)}{2 f}+\frac {a^2 (c+d x)^2}{2 d}+\frac {1}{2} a^2 c x+\frac {a^2 d \sinh ^2(e+f x)}{4 f^2}-\frac {2 i a^2 d \sinh (e+f x)}{f^2}+\frac {1}{4} a^2 d x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + I*a*Sinh[e + f*x])^2,x]

[Out]

(a^2*c*x)/2 + (a^2*d*x^2)/4 + (a^2*(c + d*x)^2)/(2*d) + ((2*I)*a^2*(c + d*x)*Cosh[e + f*x])/f - ((2*I)*a^2*d*S
inh[e + f*x])/f^2 - (a^2*(c + d*x)*Cosh[e + f*x]*Sinh[e + f*x])/(2*f) + (a^2*d*Sinh[e + f*x]^2)/(4*f^2)

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3377

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[(-(c + d*x)^m)*(Cos[e + f*x]/f), x]
+ Dist[d*(m/f), Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3391

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*((b*Sin[e + f*x])^n/(f^2*n^
2)), x] + (Dist[b^2*((n - 1)/n), Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[b*(c + d*x)*Cos[e + f*x
]*((b*Sin[e + f*x])^(n - 1)/(f*n)), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

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) (a+i a \sinh (e+f x))^2 \, dx &=\int \left (a^2 (c+d x)+2 i a^2 (c+d x) \sinh (e+f x)-a^2 (c+d x) \sinh ^2(e+f x)\right ) \, dx\\ &=\frac {a^2 (c+d x)^2}{2 d}+\left (2 i a^2\right ) \int (c+d x) \sinh (e+f x) \, dx-a^2 \int (c+d x) \sinh ^2(e+f x) \, dx\\ &=\frac {a^2 (c+d x)^2}{2 d}+\frac {2 i a^2 (c+d x) \cosh (e+f x)}{f}-\frac {a^2 (c+d x) \cosh (e+f x) \sinh (e+f x)}{2 f}+\frac {a^2 d \sinh ^2(e+f x)}{4 f^2}+\frac {1}{2} a^2 \int (c+d x) \, dx-\frac {\left (2 i a^2 d\right ) \int \cosh (e+f x) \, dx}{f}\\ &=\frac {1}{2} a^2 c x+\frac {1}{4} a^2 d x^2+\frac {a^2 (c+d x)^2}{2 d}+\frac {2 i a^2 (c+d x) \cosh (e+f x)}{f}-\frac {2 i a^2 d \sinh (e+f x)}{f^2}-\frac {a^2 (c+d x) \cosh (e+f x) \sinh (e+f x)}{2 f}+\frac {a^2 d \sinh ^2(e+f x)}{4 f^2}\\ \end {align*}

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Mathematica [A]
time = 0.56, size = 86, normalized size = 0.70 \begin {gather*} \frac {a^2 (16 i f (c+d x) \cosh (e+f x)+d \cosh (2 (e+f x))-2 (3 (e+f x) (d e-2 c f-d f x)+8 i d \sinh (e+f x)+f (c+d x) \sinh (2 (e+f x))))}{8 f^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*(a + I*a*Sinh[e + f*x])^2,x]

[Out]

(a^2*((16*I)*f*(c + d*x)*Cosh[e + f*x] + d*Cosh[2*(e + f*x)] - 2*(3*(e + f*x)*(d*e - 2*c*f - d*f*x) + (8*I)*d*
Sinh[e + f*x] + f*(c + d*x)*Sinh[2*(e + f*x)])))/(8*f^2)

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Maple [A]
time = 0.62, size = 215, normalized size = 1.76

method	result	size
risch	\(\frac {3 d \,a^{2} x^{2}}{4}+\frac {3 a^{2} c x}{2}-\frac {a^{2} \left (2 d x f +2 c f -d \right ) {\mathrm e}^{2 f x +2 e}}{16 f^{2}}+\frac {i a^{2} \left (d x f +c f -d \right ) {\mathrm e}^{f x +e}}{f^{2}}+\frac {i a^{2} \left (d x f +c f +d \right ) {\mathrm e}^{-f x -e}}{f^{2}}+\frac {a^{2} \left (2 d x f +2 c f +d \right ) {\mathrm e}^{-2 f x -2 e}}{16 f^{2}}\)	\(129\)
derivativedivides	\(\frac {\frac {d \,a^{2} \left (f x +e \right )^{2}}{2 f}+\frac {2 i d \,a^{2} \left (\left (f x +e \right ) \cosh \left (f x +e \right )-\sinh \left (f x +e \right )\right )}{f}-\frac {d \,a^{2} \left (\frac {\left (f x +e \right ) \cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\cosh ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-\frac {d e \,a^{2} \left (f x +e \right )}{f}-\frac {2 i d e \,a^{2} \cosh \left (f x +e \right )}{f}+\frac {d e \,a^{2} \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {f x}{2}-\frac {e}{2}\right )}{f}+a^{2} c \left (f x +e \right )+2 i c \,a^{2} \cosh \left (f x +e \right )-a^{2} c \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {f x}{2}-\frac {e}{2}\right )}{f}\)	\(215\)
default	\(\frac {\frac {d \,a^{2} \left (f x +e \right )^{2}}{2 f}+\frac {2 i d \,a^{2} \left (\left (f x +e \right ) \cosh \left (f x +e \right )-\sinh \left (f x +e \right )\right )}{f}-\frac {d \,a^{2} \left (\frac {\left (f x +e \right ) \cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\cosh ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-\frac {d e \,a^{2} \left (f x +e \right )}{f}-\frac {2 i d e \,a^{2} \cosh \left (f x +e \right )}{f}+\frac {d e \,a^{2} \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {f x}{2}-\frac {e}{2}\right )}{f}+a^{2} c \left (f x +e \right )+2 i c \,a^{2} \cosh \left (f x +e \right )-a^{2} c \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {f x}{2}-\frac {e}{2}\right )}{f}\)	\(215\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(a+I*a*sinh(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(1/2*d/f*a^2*(f*x+e)^2+2*I*d/f*a^2*((f*x+e)*cosh(f*x+e)-sinh(f*x+e))-d/f*a^2*(1/2*(f*x+e)*cosh(f*x+e)*sinh
(f*x+e)-1/4*(f*x+e)^2-1/4*cosh(f*x+e)^2)-d/f*e*a^2*(f*x+e)-2*I*d/f*e*a^2*cosh(f*x+e)+d/f*e*a^2*(1/2*cosh(f*x+e
)*sinh(f*x+e)-1/2*f*x-1/2*e)+a^2*c*(f*x+e)+2*I*c*a^2*cosh(f*x+e)-a^2*c*(1/2*cosh(f*x+e)*sinh(f*x+e)-1/2*f*x-1/
2*e))

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Maxima [A]
time = 0.28, size = 176, normalized size = 1.44 \begin {gather*} \frac {1}{2} \, a^{2} d x^{2} + \frac {1}{16} \, {\left (4 \, x^{2} - \frac {{\left (2 \, f x e^{\left (2 \, e\right )} - e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f^{2}} + \frac {{\left (2 \, f x + 1\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{f^{2}}\right )} a^{2} d + \frac {1}{8} \, a^{2} c {\left (4 \, x - \frac {e^{\left (2 \, f x + 2 \, e\right )}}{f} + \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{f}\right )} + a^{2} c x + i \, a^{2} d {\left (\frac {{\left (f x e^{e} - e^{e}\right )} e^{\left (f x\right )}}{f^{2}} + \frac {{\left (f x + 1\right )} e^{\left (-f x - e\right )}}{f^{2}}\right )} + \frac {2 i \, a^{2} c \cosh \left (f x + e\right )}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+I*a*sinh(f*x+e))^2,x, algorithm="maxima")

[Out]

1/2*a^2*d*x^2 + 1/16*(4*x^2 - (2*f*x*e^(2*e) - e^(2*e))*e^(2*f*x)/f^2 + (2*f*x + 1)*e^(-2*f*x - 2*e)/f^2)*a^2*
d + 1/8*a^2*c*(4*x - e^(2*f*x + 2*e)/f + e^(-2*f*x - 2*e)/f) + a^2*c*x + I*a^2*d*((f*x*e^e - e^e)*e^(f*x)/f^2
+ (f*x + 1)*e^(-f*x - e)/f^2) + 2*I*a^2*c*cosh(f*x + e)/f

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Fricas [A]
time = 0.36, size = 169, normalized size = 1.39 \begin {gather*} \frac {{\left (2 \, a^{2} d f x + 2 \, a^{2} c f + a^{2} d - {\left (2 \, a^{2} d f x + 2 \, a^{2} c f - a^{2} d\right )} e^{\left (4 \, f x + 4 \, e\right )} - 16 \, {\left (-i \, a^{2} d f x - i \, a^{2} c f + i \, a^{2} d\right )} e^{\left (3 \, f x + 3 \, e\right )} + 12 \, {\left (a^{2} d f^{2} x^{2} + 2 \, a^{2} c f^{2} x\right )} e^{\left (2 \, f x + 2 \, e\right )} - 16 \, {\left (-i \, a^{2} d f x - i \, a^{2} c f - i \, a^{2} d\right )} e^{\left (f x + e\right )}\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+I*a*sinh(f*x+e))^2,x, algorithm="fricas")

[Out]

1/16*(2*a^2*d*f*x + 2*a^2*c*f + a^2*d - (2*a^2*d*f*x + 2*a^2*c*f - a^2*d)*e^(4*f*x + 4*e) - 16*(-I*a^2*d*f*x -
I*a^2*c*f + I*a^2*d)*e^(3*f*x + 3*e) + 12*(a^2*d*f^2*x^2 + 2*a^2*c*f^2*x)*e^(2*f*x + 2*e) - 16*(-I*a^2*d*f*x
- I*a^2*c*f - I*a^2*d)*e^(f*x + e))*e^(-2*f*x - 2*e)/f^2

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Sympy [A]
time = 0.38, size = 357, normalized size = 2.93 \begin {gather*} \frac {3 a^{2} c x}{2} + \frac {3 a^{2} d x^{2}}{4} + \begin {cases} \frac {\left (\left (32 a^{2} c f^{7} e^{e} + 32 a^{2} d f^{7} x e^{e} + 16 a^{2} d f^{6} e^{e}\right ) e^{- 2 f x} + \left (- 32 a^{2} c f^{7} e^{5 e} - 32 a^{2} d f^{7} x e^{5 e} + 16 a^{2} d f^{6} e^{5 e}\right ) e^{2 f x} + \left (256 i a^{2} c f^{7} e^{2 e} + 256 i a^{2} d f^{7} x e^{2 e} + 256 i a^{2} d f^{6} e^{2 e}\right ) e^{- f x} + \left (256 i a^{2} c f^{7} e^{4 e} + 256 i a^{2} d f^{7} x e^{4 e} - 256 i a^{2} d f^{6} e^{4 e}\right ) e^{f x}\right ) e^{- 3 e}}{256 f^{8}} & \text {for}\: f^{8} e^{3 e} \neq 0 \\\frac {x^{2} \left (- a^{2} d e^{4 e} + 4 i a^{2} d e^{3 e} - 4 i a^{2} d e^{e} - a^{2} d\right ) e^{- 2 e}}{8} + \frac {x \left (- a^{2} c e^{4 e} + 4 i a^{2} c e^{3 e} - 4 i a^{2} c e^{e} - a^{2} c\right ) e^{- 2 e}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+I*a*sinh(f*x+e))**2,x)

[Out]

3*a**2*c*x/2 + 3*a**2*d*x**2/4 + Piecewise((((32*a**2*c*f**7*exp(e) + 32*a**2*d*f**7*x*exp(e) + 16*a**2*d*f**6
*exp(e))*exp(-2*f*x) + (-32*a**2*c*f**7*exp(5*e) - 32*a**2*d*f**7*x*exp(5*e) + 16*a**2*d*f**6*exp(5*e))*exp(2*
f*x) + (256*I*a**2*c*f**7*exp(2*e) + 256*I*a**2*d*f**7*x*exp(2*e) + 256*I*a**2*d*f**6*exp(2*e))*exp(-f*x) + (2
56*I*a**2*c*f**7*exp(4*e) + 256*I*a**2*d*f**7*x*exp(4*e) - 256*I*a**2*d*f**6*exp(4*e))*exp(f*x))*exp(-3*e)/(25
6*f**8), Ne(f**8*exp(3*e), 0)), (x**2*(-a**2*d*exp(4*e) + 4*I*a**2*d*exp(3*e) - 4*I*a**2*d*exp(e) - a**2*d)*ex
p(-2*e)/8 + x*(-a**2*c*exp(4*e) + 4*I*a**2*c*exp(3*e) - 4*I*a**2*c*exp(e) - a**2*c)*exp(-2*e)/4, True))

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Giac [A]
time = 0.45, size = 155, normalized size = 1.27 \begin {gather*} \frac {3}{4} \, a^{2} d x^{2} + \frac {3}{2} \, a^{2} c x - \frac {{\left (2 \, a^{2} d f x + 2 \, a^{2} c f - a^{2} d\right )} e^{\left (2 \, f x + 2 \, e\right )}}{16 \, f^{2}} + \frac {{\left (i \, a^{2} d f x + i \, a^{2} c f - i \, a^{2} d\right )} e^{\left (f x + e\right )}}{f^{2}} + \frac {{\left (i \, a^{2} d f x + i \, a^{2} c f + i \, a^{2} d\right )} e^{\left (-f x - e\right )}}{f^{2}} + \frac {{\left (2 \, a^{2} d f x + 2 \, a^{2} c f + a^{2} d\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(a+I*a*sinh(f*x+e))^2,x, algorithm="giac")

[Out]

3/4*a^2*d*x^2 + 3/2*a^2*c*x - 1/16*(2*a^2*d*f*x + 2*a^2*c*f - a^2*d)*e^(2*f*x + 2*e)/f^2 + (I*a^2*d*f*x + I*a^
2*c*f - I*a^2*d)*e^(f*x + e)/f^2 + (I*a^2*d*f*x + I*a^2*c*f + I*a^2*d)*e^(-f*x - e)/f^2 + 1/16*(2*a^2*d*f*x +
2*a^2*c*f + a^2*d)*e^(-2*f*x - 2*e)/f^2

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Mupad [B]
time = 0.35, size = 104, normalized size = 0.85 \begin {gather*} \frac {a^2\,\left (6\,d\,x^2+12\,c\,x\right )}{8}-\frac {\frac {a^2\,\left (-d\,\mathrm {cosh}\left (2\,e+2\,f\,x\right )+d\,\mathrm {sinh}\left (e+f\,x\right )\,16{}\mathrm {i}\right )}{8}-\frac {a^2\,f\,\left (c\,\mathrm {cosh}\left (e+f\,x\right )\,16{}\mathrm {i}-2\,c\,\mathrm {sinh}\left (2\,e+2\,f\,x\right )-2\,d\,x\,\mathrm {sinh}\left (2\,e+2\,f\,x\right )+d\,x\,\mathrm {cosh}\left (e+f\,x\right )\,16{}\mathrm {i}\right )}{8}}{f^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sinh(e + f*x)*1i)^2*(c + d*x),x)

[Out]

(a^2*(12*c*x + 6*d*x^2))/8 - ((a^2*(d*sinh(e + f*x)*16i - d*cosh(2*e + 2*f*x)))/8 - (a^2*f*(c*cosh(e + f*x)*16
i - 2*c*sinh(2*e + 2*f*x) - 2*d*x*sinh(2*e + 2*f*x) + d*x*cosh(e + f*x)*16i))/8)/f^2

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