3.157 \(\int (c+d x)^3 (a+b \sinh (e+f x)) \, dx\)

Optimal. Leaf size=89 \[ \frac {a (c+d x)^4}{4 d}+\frac {6 b d^2 (c+d x) \cosh (e+f x)}{f^3}-\frac {3 b d (c+d x)^2 \sinh (e+f x)}{f^2}+\frac {b (c+d x)^3 \cosh (e+f x)}{f}-\frac {6 b d^3 \sinh (e+f x)}{f^4} \]

[Out]

1/4*a*(d*x+c)^4/d+6*b*d^2*(d*x+c)*cosh(f*x+e)/f^3+b*(d*x+c)^3*cosh(f*x+e)/f-6*b*d^3*sinh(f*x+e)/f^4-3*b*d*(d*x
+c)^2*sinh(f*x+e)/f^2

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Rubi [A]  time = 0.14, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3317, 3296, 2637} \[ \frac {a (c+d x)^4}{4 d}+\frac {6 b d^2 (c+d x) \cosh (e+f x)}{f^3}-\frac {3 b d (c+d x)^2 \sinh (e+f x)}{f^2}+\frac {b (c+d x)^3 \cosh (e+f x)}{f}-\frac {6 b d^3 \sinh (e+f x)}{f^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(c + d*x)^4)/(4*d) + (6*b*d^2*(c + d*x)*Cosh[e + f*x])/f^3 + (b*(c + d*x)^3*Cosh[e + f*x])/f - (6*b*d^3*Sin
h[e + f*x])/f^4 - (3*b*d*(c + d*x)^2*Sinh[e + f*x])/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 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)^3 (a+b \sinh (e+f x)) \, dx &=\int \left (a (c+d x)^3+b (c+d x)^3 \sinh (e+f x)\right ) \, dx\\ &=\frac {a (c+d x)^4}{4 d}+b \int (c+d x)^3 \sinh (e+f x) \, dx\\ &=\frac {a (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \cosh (e+f x)}{f}-\frac {(3 b d) \int (c+d x)^2 \cosh (e+f x) \, dx}{f}\\ &=\frac {a (c+d x)^4}{4 d}+\frac {b (c+d x)^3 \cosh (e+f x)}{f}-\frac {3 b d (c+d x)^2 \sinh (e+f x)}{f^2}+\frac {\left (6 b d^2\right ) \int (c+d x) \sinh (e+f x) \, dx}{f^2}\\ &=\frac {a (c+d x)^4}{4 d}+\frac {6 b d^2 (c+d x) \cosh (e+f x)}{f^3}+\frac {b (c+d x)^3 \cosh (e+f x)}{f}-\frac {3 b d (c+d x)^2 \sinh (e+f x)}{f^2}-\frac {\left (6 b d^3\right ) \int \cosh (e+f x) \, dx}{f^3}\\ &=\frac {a (c+d x)^4}{4 d}+\frac {6 b d^2 (c+d x) \cosh (e+f x)}{f^3}+\frac {b (c+d x)^3 \cosh (e+f x)}{f}-\frac {6 b d^3 \sinh (e+f x)}{f^4}-\frac {3 b d (c+d x)^2 \sinh (e+f x)}{f^2}\\ \end {align*}

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Mathematica [A]  time = 0.46, size = 123, normalized size = 1.38 \[ \frac {1}{4} a x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right )-\frac {3 b d \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+2\right )\right ) \sinh (e+f x)}{f^4}+\frac {b (c+d x) \left (c^2 f^2+2 c d f^2 x+d^2 \left (f^2 x^2+6\right )\right ) \cosh (e+f x)}{f^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x^3))/4 + (b*(c + d*x)*(c^2*f^2 + 2*c*d*f^2*x + d^2*(6 + f^2*x^2))
*Cosh[e + f*x])/f^3 - (3*b*d*(c^2*f^2 + 2*c*d*f^2*x + d^2*(2 + f^2*x^2))*Sinh[e + f*x])/f^4

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fricas [A]  time = 0.52, size = 168, normalized size = 1.89 \[ \frac {a d^{3} f^{4} x^{4} + 4 \, a c d^{2} f^{4} x^{3} + 6 \, a c^{2} d f^{4} x^{2} + 4 \, a c^{3} f^{4} x + 4 \, {\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + b c^{3} f^{3} + 6 \, b c d^{2} f + 3 \, {\left (b c^{2} d f^{3} + 2 \, b d^{3} f\right )} x\right )} \cosh \left (f x + e\right ) - 12 \, {\left (b d^{3} f^{2} x^{2} + 2 \, b c d^{2} f^{2} x + b c^{2} d f^{2} + 2 \, b d^{3}\right )} \sinh \left (f x + e\right )}{4 \, f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(a+b*sinh(f*x+e)),x, algorithm="fricas")

[Out]

1/4*(a*d^3*f^4*x^4 + 4*a*c*d^2*f^4*x^3 + 6*a*c^2*d*f^4*x^2 + 4*a*c^3*f^4*x + 4*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*
x^2 + b*c^3*f^3 + 6*b*c*d^2*f + 3*(b*c^2*d*f^3 + 2*b*d^3*f)*x)*cosh(f*x + e) - 12*(b*d^3*f^2*x^2 + 2*b*c*d^2*f
^2*x + b*c^2*d*f^2 + 2*b*d^3)*sinh(f*x + e))/f^4

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giac [B]  time = 0.16, size = 260, normalized size = 2.92 \[ \frac {1}{4} \, a d^{3} x^{4} + a c d^{2} x^{3} + \frac {3}{2} \, a c^{2} d x^{2} + a c^{3} x + \frac {{\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x - 3 \, b d^{3} f^{2} x^{2} + b c^{3} f^{3} - 6 \, b c d^{2} f^{2} x - 3 \, b c^{2} d f^{2} + 6 \, b d^{3} f x + 6 \, b c d^{2} f - 6 \, b d^{3}\right )} e^{\left (f x + e\right )}}{2 \, f^{4}} + \frac {{\left (b d^{3} f^{3} x^{3} + 3 \, b c d^{2} f^{3} x^{2} + 3 \, b c^{2} d f^{3} x + 3 \, b d^{3} f^{2} x^{2} + b c^{3} f^{3} + 6 \, b c d^{2} f^{2} x + 3 \, b c^{2} d f^{2} + 6 \, b d^{3} f x + 6 \, b c d^{2} f + 6 \, b d^{3}\right )} e^{\left (-f x - e\right )}}{2 \, f^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(a+b*sinh(f*x+e)),x, algorithm="giac")

[Out]

1/4*a*d^3*x^4 + a*c*d^2*x^3 + 3/2*a*c^2*d*x^2 + a*c^3*x + 1/2*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f
^3*x - 3*b*d^3*f^2*x^2 + b*c^3*f^3 - 6*b*c*d^2*f^2*x - 3*b*c^2*d*f^2 + 6*b*d^3*f*x + 6*b*c*d^2*f - 6*b*d^3)*e^
(f*x + e)/f^4 + 1/2*(b*d^3*f^3*x^3 + 3*b*c*d^2*f^3*x^2 + 3*b*c^2*d*f^3*x + 3*b*d^3*f^2*x^2 + b*c^3*f^3 + 6*b*c
*d^2*f^2*x + 3*b*c^2*d*f^2 + 6*b*d^3*f*x + 6*b*c*d^2*f + 6*b*d^3)*e^(-f*x - e)/f^4

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maple [B]  time = 0.02, size = 482, normalized size = 5.42 \[ \frac {\frac {d^{3} a \left (f x +e \right )^{4}}{4 f^{3}}+\frac {d^{3} b \left (\left (f x +e \right )^{3} \cosh \left (f x +e \right )-3 \left (f x +e \right )^{2} \sinh \left (f x +e \right )+6 \left (f x +e \right ) \cosh \left (f x +e \right )-6 \sinh \left (f x +e \right )\right )}{f^{3}}-\frac {d^{3} e a \left (f x +e \right )^{3}}{f^{3}}-\frac {3 d^{3} e b \left (\left (f x +e \right )^{2} \cosh \left (f x +e \right )-2 \left (f x +e \right ) \sinh \left (f x +e \right )+2 \cosh \left (f x +e \right )\right )}{f^{3}}+\frac {3 d^{3} e^{2} a \left (f x +e \right )^{2}}{2 f^{3}}+\frac {3 d^{3} e^{2} b \left (\left (f x +e \right ) \cosh \left (f x +e \right )-\sinh \left (f x +e \right )\right )}{f^{3}}-\frac {d^{3} e^{3} a \left (f x +e \right )}{f^{3}}-\frac {d^{3} e^{3} b \cosh \left (f x +e \right )}{f^{3}}+\frac {d^{2} c a \left (f x +e \right )^{3}}{f^{2}}+\frac {3 c \,d^{2} b \left (\left (f x +e \right )^{2} \cosh \left (f x +e \right )-2 \left (f x +e \right ) \sinh \left (f x +e \right )+2 \cosh \left (f x +e \right )\right )}{f^{2}}-\frac {3 d^{2} e c a \left (f x +e \right )^{2}}{f^{2}}-\frac {6 c \,d^{2} e b \left (\left (f x +e \right ) \cosh \left (f x +e \right )-\sinh \left (f x +e \right )\right )}{f^{2}}+\frac {3 d^{2} e^{2} c a \left (f x +e \right )}{f^{2}}+\frac {3 c \,d^{2} e^{2} b \cosh \left (f x +e \right )}{f^{2}}+\frac {3 d \,c^{2} a \left (f x +e \right )^{2}}{2 f}+\frac {3 c^{2} d b \left (\left (f x +e \right ) \cosh \left (f x +e \right )-\sinh \left (f x +e \right )\right )}{f}-\frac {3 d e \,c^{2} a \left (f x +e \right )}{f}-\frac {3 c^{2} d e b \cosh \left (f x +e \right )}{f}+c^{3} a \left (f x +e \right )+b \,c^{3} \cosh \left (f x +e \right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(1/4/f^3*d^3*a*(f*x+e)^4+1/f^3*d^3*b*((f*x+e)^3*cosh(f*x+e)-3*(f*x+e)^2*sinh(f*x+e)+6*(f*x+e)*cosh(f*x+e)-
6*sinh(f*x+e))-1/f^3*d^3*e*a*(f*x+e)^3-3/f^3*d^3*e*b*((f*x+e)^2*cosh(f*x+e)-2*(f*x+e)*sinh(f*x+e)+2*cosh(f*x+e
))+3/2/f^3*d^3*e^2*a*(f*x+e)^2+3/f^3*d^3*e^2*b*((f*x+e)*cosh(f*x+e)-sinh(f*x+e))-d^3*e^3/f^3*a*(f*x+e)-1/f^3*d
^3*e^3*b*cosh(f*x+e)+1/f^2*d^2*c*a*(f*x+e)^3+3/f^2*c*d^2*b*((f*x+e)^2*cosh(f*x+e)-2*(f*x+e)*sinh(f*x+e)+2*cosh
(f*x+e))-3/f^2*d^2*e*c*a*(f*x+e)^2-6/f^2*c*d^2*e*b*((f*x+e)*cosh(f*x+e)-sinh(f*x+e))+3*d^2*e^2/f^2*c*a*(f*x+e)
+3/f^2*c*d^2*e^2*b*cosh(f*x+e)+3/2/f*d*c^2*a*(f*x+e)^2+3/f*c^2*d*b*((f*x+e)*cosh(f*x+e)-sinh(f*x+e))-3*d*e/f*c
^2*a*(f*x+e)-3/f*c^2*d*e*b*cosh(f*x+e)+c^3*a*(f*x+e)+b*c^3*cosh(f*x+e))

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maxima [B]  time = 0.33, size = 234, normalized size = 2.63 \[ \frac {1}{4} \, a d^{3} x^{4} + a c d^{2} x^{3} + \frac {3}{2} \, a c^{2} d x^{2} + a c^{3} x + \frac {3}{2} \, b c^{2} d {\left (\frac {{\left (f x e^{e} - e^{e}\right )} e^{\left (f x\right )}}{f^{2}} + \frac {{\left (f x + 1\right )} e^{\left (-f x - e\right )}}{f^{2}}\right )} + \frac {3}{2} \, b c d^{2} {\left (\frac {{\left (f^{2} x^{2} e^{e} - 2 \, f x e^{e} + 2 \, e^{e}\right )} e^{\left (f x\right )}}{f^{3}} + \frac {{\left (f^{2} x^{2} + 2 \, f x + 2\right )} e^{\left (-f x - e\right )}}{f^{3}}\right )} + \frac {1}{2} \, b d^{3} {\left (\frac {{\left (f^{3} x^{3} e^{e} - 3 \, f^{2} x^{2} e^{e} + 6 \, f x e^{e} - 6 \, e^{e}\right )} e^{\left (f x\right )}}{f^{4}} + \frac {{\left (f^{3} x^{3} + 3 \, f^{2} x^{2} + 6 \, f x + 6\right )} e^{\left (-f x - e\right )}}{f^{4}}\right )} + \frac {b c^{3} \cosh \left (f x + e\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^3*(a+b*sinh(f*x+e)),x, algorithm="maxima")

[Out]

1/4*a*d^3*x^4 + a*c*d^2*x^3 + 3/2*a*c^2*d*x^2 + a*c^3*x + 3/2*b*c^2*d*((f*x*e^e - e^e)*e^(f*x)/f^2 + (f*x + 1)
*e^(-f*x - e)/f^2) + 3/2*b*c*d^2*((f^2*x^2*e^e - 2*f*x*e^e + 2*e^e)*e^(f*x)/f^3 + (f^2*x^2 + 2*f*x + 2)*e^(-f*
x - e)/f^3) + 1/2*b*d^3*((f^3*x^3*e^e - 3*f^2*x^2*e^e + 6*f*x*e^e - 6*e^e)*e^(f*x)/f^4 + (f^3*x^3 + 3*f^2*x^2
+ 6*f*x + 6)*e^(-f*x - e)/f^4) + b*c^3*cosh(f*x + e)/f

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mupad [B]  time = 0.24, size = 187, normalized size = 2.10 \[ \frac {\mathrm {cosh}\left (e+f\,x\right )\,\left (b\,c^3\,f^2+6\,b\,c\,d^2\right )}{f^3}-\frac {3\,\mathrm {sinh}\left (e+f\,x\right )\,\left (b\,c^2\,d\,f^2+2\,b\,d^3\right )}{f^4}+\frac {a\,d^3\,x^4}{4}+a\,c^3\,x+\frac {3\,x\,\mathrm {cosh}\left (e+f\,x\right )\,\left (b\,c^2\,d\,f^2+2\,b\,d^3\right )}{f^3}+\frac {3\,a\,c^2\,d\,x^2}{2}+a\,c\,d^2\,x^3+\frac {b\,d^3\,x^3\,\mathrm {cosh}\left (e+f\,x\right )}{f}-\frac {3\,b\,d^3\,x^2\,\mathrm {sinh}\left (e+f\,x\right )}{f^2}-\frac {6\,b\,c\,d^2\,x\,\mathrm {sinh}\left (e+f\,x\right )}{f^2}+\frac {3\,b\,c\,d^2\,x^2\,\mathrm {cosh}\left (e+f\,x\right )}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cosh(e + f*x)*(b*c^3*f^2 + 6*b*c*d^2))/f^3 - (3*sinh(e + f*x)*(2*b*d^3 + b*c^2*d*f^2))/f^4 + (a*d^3*x^4)/4 +
a*c^3*x + (3*x*cosh(e + f*x)*(2*b*d^3 + b*c^2*d*f^2))/f^3 + (3*a*c^2*d*x^2)/2 + a*c*d^2*x^3 + (b*d^3*x^3*cosh(
e + f*x))/f - (3*b*d^3*x^2*sinh(e + f*x))/f^2 - (6*b*c*d^2*x*sinh(e + f*x))/f^2 + (3*b*c*d^2*x^2*cosh(e + f*x)
)/f

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sympy [A]  time = 1.41, size = 264, normalized size = 2.97 \[ \begin {cases} a c^{3} x + \frac {3 a c^{2} d x^{2}}{2} + a c d^{2} x^{3} + \frac {a d^{3} x^{4}}{4} + \frac {b c^{3} \cosh {\left (e + f x \right )}}{f} + \frac {3 b c^{2} d x \cosh {\left (e + f x \right )}}{f} - \frac {3 b c^{2} d \sinh {\left (e + f x \right )}}{f^{2}} + \frac {3 b c d^{2} x^{2} \cosh {\left (e + f x \right )}}{f} - \frac {6 b c d^{2} x \sinh {\left (e + f x \right )}}{f^{2}} + \frac {6 b c d^{2} \cosh {\left (e + f x \right )}}{f^{3}} + \frac {b d^{3} x^{3} \cosh {\left (e + f x \right )}}{f} - \frac {3 b d^{3} x^{2} \sinh {\left (e + f x \right )}}{f^{2}} + \frac {6 b d^{3} x \cosh {\left (e + f x \right )}}{f^{3}} - \frac {6 b d^{3} \sinh {\left (e + f x \right )}}{f^{4}} & \text {for}\: f \neq 0 \\\left (a + b \sinh {\relax (e )}\right ) \left (c^{3} x + \frac {3 c^{2} d x^{2}}{2} + c d^{2} x^{3} + \frac {d^{3} x^{4}}{4}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**3*(a+b*sinh(f*x+e)),x)

[Out]

Piecewise((a*c**3*x + 3*a*c**2*d*x**2/2 + a*c*d**2*x**3 + a*d**3*x**4/4 + b*c**3*cosh(e + f*x)/f + 3*b*c**2*d*
x*cosh(e + f*x)/f - 3*b*c**2*d*sinh(e + f*x)/f**2 + 3*b*c*d**2*x**2*cosh(e + f*x)/f - 6*b*c*d**2*x*sinh(e + f*
x)/f**2 + 6*b*c*d**2*cosh(e + f*x)/f**3 + b*d**3*x**3*cosh(e + f*x)/f - 3*b*d**3*x**2*sinh(e + f*x)/f**2 + 6*b
*d**3*x*cosh(e + f*x)/f**3 - 6*b*d**3*sinh(e + f*x)/f**4, Ne(f, 0)), ((a + b*sinh(e))*(c**3*x + 3*c**2*d*x**2/
2 + c*d**2*x**3 + d**3*x**4/4), True))

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