3.165 \(\int (c+d x) (a+b \sinh (e+f x))^2 \, dx\)

Optimal. Leaf size=116 \[ \frac {a^2 (c+d x)^2}{2 d}+\frac {2 a b (c+d x) \cosh (e+f x)}{f}-\frac {2 a b d \sinh (e+f x)}{f^2}+\frac {b^2 (c+d x) \sinh (e+f x) \cosh (e+f x)}{2 f}-\frac {1}{2} b^2 c x-\frac {b^2 d \sinh ^2(e+f x)}{4 f^2}-\frac {1}{4} b^2 d x^2 \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3317, 3296, 2637, 3310} \[ \frac {a^2 (c+d x)^2}{2 d}+\frac {2 a b (c+d x) \cosh (e+f x)}{f}-\frac {2 a b d \sinh (e+f x)}{f^2}+\frac {b^2 (c+d x) \sinh (e+f x) \cosh (e+f x)}{2 f}-\frac {1}{2} b^2 c x-\frac {b^2 d \sinh ^2(e+f x)}{4 f^2}-\frac {1}{4} b^2 d x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2637

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

Rule 3296

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 3310

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 3317

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

________________________________________________________________________________________

Mathematica [A]  time = 0.72, size = 98, normalized size = 0.84 \[ -\frac {2 \left (2 a^2-b^2\right ) (e+f x) (d (e-f x)-2 c f)-16 a b f (c+d x) \cosh (e+f x)+16 a b d \sinh (e+f x)-2 b^2 f (c+d x) \sinh (2 (e+f x))+b^2 d \cosh (2 (e+f x))}{8 f^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(2*(2*a^2 - b^2)*(e + f*x)*(-2*c*f + d*(e - f*x)) - 16*a*b*f*(c + d*x)*Cosh[e + f*x] + b^2*d*Cosh[2*(e +
f*x)] + 16*a*b*d*Sinh[e + f*x] - 2*b^2*f*(c + d*x)*Sinh[2*(e + f*x)])/f^2

________________________________________________________________________________________

fricas [A]  time = 0.61, size = 128, normalized size = 1.10 \[ \frac {2 \, {\left (2 \, a^{2} - b^{2}\right )} d f^{2} x^{2} + 4 \, {\left (2 \, a^{2} - b^{2}\right )} c f^{2} x - b^{2} d \cosh \left (f x + e\right )^{2} - b^{2} d \sinh \left (f x + e\right )^{2} + 16 \, {\left (a b d f x + a b c f\right )} \cosh \left (f x + e\right ) - 4 \, {\left (4 \, a b d - {\left (b^{2} d f x + b^{2} c f\right )} \cosh \left (f x + e\right )\right )} \sinh \left (f x + e\right )}{8 \, f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.59, size = 163, normalized size = 1.41 \[ \frac {1}{2} \, a^{2} d x^{2} - \frac {1}{4} \, b^{2} d x^{2} + a^{2} c x - \frac {1}{2} \, b^{2} c x + \frac {{\left (2 \, b^{2} d f x + 2 \, b^{2} c f - b^{2} d\right )} e^{\left (2 \, f x + 2 \, e\right )}}{16 \, f^{2}} + \frac {{\left (a b d f x + a b c f - a b d\right )} e^{\left (f x + e\right )}}{f^{2}} + \frac {{\left (a b d f x + a b c f + a b d\right )} e^{\left (-f x - e\right )}}{f^{2}} - \frac {{\left (2 \, b^{2} d f x + 2 \, b^{2} c f + b^{2} d\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{16 \, f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.03, size = 208, normalized size = 1.79 \[ \frac {\frac {d \,a^{2} \left (f x +e \right )^{2}}{2 f}+\frac {2 d a b \left (\left (f x +e \right ) \cosh \left (f x +e \right )-\sinh \left (f x +e \right )\right )}{f}+\frac {d \,b^{2} \left (\frac {\left (f x +e \right ) \cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {\left (f x +e \right )^{2}}{4}-\frac {\left (\cosh ^{2}\left (f x +e \right )\right )}{4}\right )}{f}-\frac {d e \,a^{2} \left (f x +e \right )}{f}-\frac {2 d e a b \cosh \left (f x +e \right )}{f}-\frac {d e \,b^{2} \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {f x}{2}-\frac {e}{2}\right )}{f}+c \,a^{2} \left (f x +e \right )+2 c a b \cosh \left (f x +e \right )+c \,b^{2} \left (\frac {\cosh \left (f x +e \right ) \sinh \left (f x +e \right )}{2}-\frac {f x}{2}-\frac {e}{2}\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 164, normalized size = 1.41 \[ \frac {1}{2} \, a^{2} d x^{2} - \frac {1}{16} \, {\left (4 \, x^{2} - \frac {{\left (2 \, f x e^{\left (2 \, e\right )} - e^{\left (2 \, e\right )}\right )} e^{\left (2 \, f x\right )}}{f^{2}} + \frac {{\left (2 \, f x + 1\right )} e^{\left (-2 \, f x - 2 \, e\right )}}{f^{2}}\right )} b^{2} d - \frac {1}{8} \, b^{2} c {\left (4 \, x - \frac {e^{\left (2 \, f x + 2 \, e\right )}}{f} + \frac {e^{\left (-2 \, f x - 2 \, e\right )}}{f}\right )} + a^{2} c x + a b d {\left (\frac {{\left (f x e^{e} - e^{e}\right )} e^{\left (f x\right )}}{f^{2}} + \frac {{\left (f x + 1\right )} e^{\left (-f x - e\right )}}{f^{2}}\right )} + \frac {2 \, a b c \cosh \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

mupad [B]  time = 0.15, size = 135, normalized size = 1.16 \[ \frac {a^2\,d\,x^2}{2}-\frac {b^2\,d\,x^2}{4}+a^2\,c\,x-\frac {b^2\,c\,x}{2}-\frac {b^2\,d\,{\mathrm {cosh}\left (e+f\,x\right )}^2}{4\,f^2}+\frac {b^2\,c\,\mathrm {cosh}\left (e+f\,x\right )\,\mathrm {sinh}\left (e+f\,x\right )}{2\,f}+\frac {2\,a\,b\,c\,\mathrm {cosh}\left (e+f\,x\right )}{f}-\frac {2\,a\,b\,d\,\mathrm {sinh}\left (e+f\,x\right )}{f^2}+\frac {2\,a\,b\,d\,x\,\mathrm {cosh}\left (e+f\,x\right )}{f}+\frac {b^2\,d\,x\,\mathrm {cosh}\left (e+f\,x\right )\,\mathrm {sinh}\left (e+f\,x\right )}{2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.65, size = 219, normalized size = 1.89 \[ \begin {cases} a^{2} c x + \frac {a^{2} d x^{2}}{2} + \frac {2 a b c \cosh {\left (e + f x \right )}}{f} + \frac {2 a b d x \cosh {\left (e + f x \right )}}{f} - \frac {2 a b d \sinh {\left (e + f x \right )}}{f^{2}} + \frac {b^{2} c x \sinh ^{2}{\left (e + f x \right )}}{2} - \frac {b^{2} c x \cosh ^{2}{\left (e + f x \right )}}{2} + \frac {b^{2} c \sinh {\left (e + f x \right )} \cosh {\left (e + f x \right )}}{2 f} + \frac {b^{2} d x^{2} \sinh ^{2}{\left (e + f x \right )}}{4} - \frac {b^{2} d x^{2} \cosh ^{2}{\left (e + f x \right )}}{4} + \frac {b^{2} d x \sinh {\left (e + f x \right )} \cosh {\left (e + f x \right )}}{2 f} - \frac {b^{2} d \sinh ^{2}{\left (e + f x \right )}}{4 f^{2}} & \text {for}\: f \neq 0 \\\left (a + b \sinh {\relax (e )}\right )^{2} \left (c x + \frac {d x^{2}}{2}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a**2*c*x + a**2*d*x**2/2 + 2*a*b*c*cosh(e + f*x)/f + 2*a*b*d*x*cosh(e + f*x)/f - 2*a*b*d*sinh(e + f
*x)/f**2 + b**2*c*x*sinh(e + f*x)**2/2 - b**2*c*x*cosh(e + f*x)**2/2 + b**2*c*sinh(e + f*x)*cosh(e + f*x)/(2*f
) + b**2*d*x**2*sinh(e + f*x)**2/4 - b**2*d*x**2*cosh(e + f*x)**2/4 + b**2*d*x*sinh(e + f*x)*cosh(e + f*x)/(2*
f) - b**2*d*sinh(e + f*x)**2/(4*f**2), Ne(f, 0)), ((a + b*sinh(e))**2*(c*x + d*x**2/2), True))

________________________________________________________________________________________