3.1 \(\int (c+d x)^4 \sin (a+b x) \, dx\)

Optimal. Leaf size=92 \[ -\frac {24 d^4 \cos (a+b x)}{b^5}-\frac {24 d^3 (c+d x) \sin (a+b x)}{b^4}+\frac {12 d^2 (c+d x)^2 \cos (a+b x)}{b^3}+\frac {4 d (c+d x)^3 \sin (a+b x)}{b^2}-\frac {(c+d x)^4 \cos (a+b x)}{b} \]

[Out]

-24*d^4*cos(b*x+a)/b^5+12*d^2*(d*x+c)^2*cos(b*x+a)/b^3-(d*x+c)^4*cos(b*x+a)/b-24*d^3*(d*x+c)*sin(b*x+a)/b^4+4*
d*(d*x+c)^3*sin(b*x+a)/b^2

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Rubi [A]  time = 0.09, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3296, 2638} \[ -\frac {24 d^3 (c+d x) \sin (a+b x)}{b^4}+\frac {12 d^2 (c+d x)^2 \cos (a+b x)}{b^3}+\frac {4 d (c+d x)^3 \sin (a+b x)}{b^2}-\frac {24 d^4 \cos (a+b x)}{b^5}-\frac {(c+d x)^4 \cos (a+b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^4*Sin[a + b*x],x]

[Out]

(-24*d^4*Cos[a + b*x])/b^5 + (12*d^2*(c + d*x)^2*Cos[a + b*x])/b^3 - ((c + d*x)^4*Cos[a + b*x])/b - (24*d^3*(c
+ d*x)*Sin[a + b*x])/b^4 + (4*d*(c + d*x)^3*Sin[a + b*x])/b^2

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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]

Rubi steps

\begin {align*} \int (c+d x)^4 \sin (a+b x) \, dx &=-\frac {(c+d x)^4 \cos (a+b x)}{b}+\frac {(4 d) \int (c+d x)^3 \cos (a+b x) \, dx}{b}\\ &=-\frac {(c+d x)^4 \cos (a+b x)}{b}+\frac {4 d (c+d x)^3 \sin (a+b x)}{b^2}-\frac {\left (12 d^2\right ) \int (c+d x)^2 \sin (a+b x) \, dx}{b^2}\\ &=\frac {12 d^2 (c+d x)^2 \cos (a+b x)}{b^3}-\frac {(c+d x)^4 \cos (a+b x)}{b}+\frac {4 d (c+d x)^3 \sin (a+b x)}{b^2}-\frac {\left (24 d^3\right ) \int (c+d x) \cos (a+b x) \, dx}{b^3}\\ &=\frac {12 d^2 (c+d x)^2 \cos (a+b x)}{b^3}-\frac {(c+d x)^4 \cos (a+b x)}{b}-\frac {24 d^3 (c+d x) \sin (a+b x)}{b^4}+\frac {4 d (c+d x)^3 \sin (a+b x)}{b^2}+\frac {\left (24 d^4\right ) \int \sin (a+b x) \, dx}{b^4}\\ &=-\frac {24 d^4 \cos (a+b x)}{b^5}+\frac {12 d^2 (c+d x)^2 \cos (a+b x)}{b^3}-\frac {(c+d x)^4 \cos (a+b x)}{b}-\frac {24 d^3 (c+d x) \sin (a+b x)}{b^4}+\frac {4 d (c+d x)^3 \sin (a+b x)}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 77, normalized size = 0.84 \[ \frac {4 b d (c+d x) \sin (a+b x) \left (b^2 (c+d x)^2-6 d^2\right )-\cos (a+b x) \left (b^4 (c+d x)^4-12 b^2 d^2 (c+d x)^2+24 d^4\right )}{b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^4*Sin[a + b*x],x]

[Out]

(-((24*d^4 - 12*b^2*d^2*(c + d*x)^2 + b^4*(c + d*x)^4)*Cos[a + b*x]) + 4*b*d*(c + d*x)*(-6*d^2 + b^2*(c + d*x)
^2)*Sin[a + b*x])/b^5

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fricas [A]  time = 0.87, size = 170, normalized size = 1.85 \[ -\frac {{\left (b^{4} d^{4} x^{4} + 4 \, b^{4} c d^{3} x^{3} + b^{4} c^{4} - 12 \, b^{2} c^{2} d^{2} + 24 \, d^{4} + 6 \, {\left (b^{4} c^{2} d^{2} - 2 \, b^{2} d^{4}\right )} x^{2} + 4 \, {\left (b^{4} c^{3} d - 6 \, b^{2} c d^{3}\right )} x\right )} \cos \left (b x + a\right ) - 4 \, {\left (b^{3} d^{4} x^{3} + 3 \, b^{3} c d^{3} x^{2} + b^{3} c^{3} d - 6 \, b c d^{3} + 3 \, {\left (b^{3} c^{2} d^{2} - 2 \, b d^{4}\right )} x\right )} \sin \left (b x + a\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*sin(b*x+a),x, algorithm="fricas")

[Out]

-((b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + b^4*c^4 - 12*b^2*c^2*d^2 + 24*d^4 + 6*(b^4*c^2*d^2 - 2*b^2*d^4)*x^2 + 4*(b^
4*c^3*d - 6*b^2*c*d^3)*x)*cos(b*x + a) - 4*(b^3*d^4*x^3 + 3*b^3*c*d^3*x^2 + b^3*c^3*d - 6*b*c*d^3 + 3*(b^3*c^2
*d^2 - 2*b*d^4)*x)*sin(b*x + a))/b^5

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giac [A]  time = 0.32, size = 171, normalized size = 1.86 \[ -\frac {{\left (b^{4} d^{4} x^{4} + 4 \, b^{4} c d^{3} x^{3} + 6 \, b^{4} c^{2} d^{2} x^{2} + 4 \, b^{4} c^{3} d x + b^{4} c^{4} - 12 \, b^{2} d^{4} x^{2} - 24 \, b^{2} c d^{3} x - 12 \, b^{2} c^{2} d^{2} + 24 \, d^{4}\right )} \cos \left (b x + a\right )}{b^{5}} + \frac {4 \, {\left (b^{3} d^{4} x^{3} + 3 \, b^{3} c d^{3} x^{2} + 3 \, b^{3} c^{2} d^{2} x + b^{3} c^{3} d - 6 \, b d^{4} x - 6 \, b c d^{3}\right )} \sin \left (b x + a\right )}{b^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*sin(b*x+a),x, algorithm="giac")

[Out]

-(b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + b^4*c^4 - 12*b^2*d^4*x^2 - 24*b^2*c*d^3*
x - 12*b^2*c^2*d^2 + 24*d^4)*cos(b*x + a)/b^5 + 4*(b^3*d^4*x^3 + 3*b^3*c*d^3*x^2 + 3*b^3*c^2*d^2*x + b^3*c^3*d
- 6*b*d^4*x - 6*b*c*d^3)*sin(b*x + a)/b^5

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maple [B]  time = 0.02, size = 551, normalized size = 5.99 \[ \frac {\frac {d^{4} \left (-\left (b x +a \right )^{4} \cos \left (b x +a \right )+4 \left (b x +a \right )^{3} \sin \left (b x +a \right )+12 \left (b x +a \right )^{2} \cos \left (b x +a \right )-24 \cos \left (b x +a \right )-24 \left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{4}}-\frac {4 a \,d^{4} \left (-\left (b x +a \right )^{3} \cos \left (b x +a \right )+3 \left (b x +a \right )^{2} \sin \left (b x +a \right )-6 \sin \left (b x +a \right )+6 \left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{4}}+\frac {4 c \,d^{3} \left (-\left (b x +a \right )^{3} \cos \left (b x +a \right )+3 \left (b x +a \right )^{2} \sin \left (b x +a \right )-6 \sin \left (b x +a \right )+6 \left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{3}}+\frac {6 a^{2} d^{4} \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{4}}-\frac {12 a c \,d^{3} \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{3}}+\frac {6 c^{2} d^{2} \left (-\left (b x +a \right )^{2} \cos \left (b x +a \right )+2 \cos \left (b x +a \right )+2 \left (b x +a \right ) \sin \left (b x +a \right )\right )}{b^{2}}-\frac {4 a^{3} d^{4} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{4}}+\frac {12 a^{2} c \,d^{3} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{3}}-\frac {12 a \,c^{2} d^{2} \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b^{2}}+\frac {4 c^{3} d \left (\sin \left (b x +a \right )-\left (b x +a \right ) \cos \left (b x +a \right )\right )}{b}-\frac {a^{4} d^{4} \cos \left (b x +a \right )}{b^{4}}+\frac {4 a^{3} c \,d^{3} \cos \left (b x +a \right )}{b^{3}}-\frac {6 a^{2} c^{2} d^{2} \cos \left (b x +a \right )}{b^{2}}+\frac {4 a \,c^{3} d \cos \left (b x +a \right )}{b}-c^{4} \cos \left (b x +a \right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^4*sin(b*x+a),x)

[Out]

1/b*(1/b^4*d^4*(-(b*x+a)^4*cos(b*x+a)+4*(b*x+a)^3*sin(b*x+a)+12*(b*x+a)^2*cos(b*x+a)-24*cos(b*x+a)-24*(b*x+a)*
sin(b*x+a))-4/b^4*a*d^4*(-(b*x+a)^3*cos(b*x+a)+3*(b*x+a)^2*sin(b*x+a)-6*sin(b*x+a)+6*(b*x+a)*cos(b*x+a))+4/b^3
*c*d^3*(-(b*x+a)^3*cos(b*x+a)+3*(b*x+a)^2*sin(b*x+a)-6*sin(b*x+a)+6*(b*x+a)*cos(b*x+a))+6/b^4*a^2*d^4*(-(b*x+a
)^2*cos(b*x+a)+2*cos(b*x+a)+2*(b*x+a)*sin(b*x+a))-12/b^3*a*c*d^3*(-(b*x+a)^2*cos(b*x+a)+2*cos(b*x+a)+2*(b*x+a)
*sin(b*x+a))+6/b^2*c^2*d^2*(-(b*x+a)^2*cos(b*x+a)+2*cos(b*x+a)+2*(b*x+a)*sin(b*x+a))-4/b^4*a^3*d^4*(sin(b*x+a)
-(b*x+a)*cos(b*x+a))+12/b^3*a^2*c*d^3*(sin(b*x+a)-(b*x+a)*cos(b*x+a))-12/b^2*a*c^2*d^2*(sin(b*x+a)-(b*x+a)*cos
(b*x+a))+4/b*c^3*d*(sin(b*x+a)-(b*x+a)*cos(b*x+a))-1/b^4*a^4*d^4*cos(b*x+a)+4/b^3*a^3*c*d^3*cos(b*x+a)-6/b^2*a
^2*c^2*d^2*cos(b*x+a)+4/b*a*c^3*d*cos(b*x+a)-c^4*cos(b*x+a))

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maxima [B]  time = 0.35, size = 490, normalized size = 5.33 \[ -\frac {c^{4} \cos \left (b x + a\right ) - \frac {4 \, a c^{3} d \cos \left (b x + a\right )}{b} + \frac {6 \, a^{2} c^{2} d^{2} \cos \left (b x + a\right )}{b^{2}} - \frac {4 \, a^{3} c d^{3} \cos \left (b x + a\right )}{b^{3}} + \frac {a^{4} d^{4} \cos \left (b x + a\right )}{b^{4}} + \frac {4 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} c^{3} d}{b} - \frac {12 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a c^{2} d^{2}}{b^{2}} + \frac {12 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a^{2} c d^{3}}{b^{3}} - \frac {4 \, {\left ({\left (b x + a\right )} \cos \left (b x + a\right ) - \sin \left (b x + a\right )\right )} a^{3} d^{4}}{b^{4}} + \frac {6 \, {\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} c^{2} d^{2}}{b^{2}} - \frac {12 \, {\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} a c d^{3}}{b^{3}} + \frac {6 \, {\left ({\left ({\left (b x + a\right )}^{2} - 2\right )} \cos \left (b x + a\right ) - 2 \, {\left (b x + a\right )} \sin \left (b x + a\right )\right )} a^{2} d^{4}}{b^{4}} + \frac {4 \, {\left ({\left ({\left (b x + a\right )}^{3} - 6 \, b x - 6 \, a\right )} \cos \left (b x + a\right ) - 3 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} c d^{3}}{b^{3}} - \frac {4 \, {\left ({\left ({\left (b x + a\right )}^{3} - 6 \, b x - 6 \, a\right )} \cos \left (b x + a\right ) - 3 \, {\left ({\left (b x + a\right )}^{2} - 2\right )} \sin \left (b x + a\right )\right )} a d^{4}}{b^{4}} + \frac {{\left ({\left ({\left (b x + a\right )}^{4} - 12 \, {\left (b x + a\right )}^{2} + 24\right )} \cos \left (b x + a\right ) - 4 \, {\left ({\left (b x + a\right )}^{3} - 6 \, b x - 6 \, a\right )} \sin \left (b x + a\right )\right )} d^{4}}{b^{4}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^4*sin(b*x+a),x, algorithm="maxima")

[Out]

-(c^4*cos(b*x + a) - 4*a*c^3*d*cos(b*x + a)/b + 6*a^2*c^2*d^2*cos(b*x + a)/b^2 - 4*a^3*c*d^3*cos(b*x + a)/b^3
+ a^4*d^4*cos(b*x + a)/b^4 + 4*((b*x + a)*cos(b*x + a) - sin(b*x + a))*c^3*d/b - 12*((b*x + a)*cos(b*x + a) -
sin(b*x + a))*a*c^2*d^2/b^2 + 12*((b*x + a)*cos(b*x + a) - sin(b*x + a))*a^2*c*d^3/b^3 - 4*((b*x + a)*cos(b*x
+ a) - sin(b*x + a))*a^3*d^4/b^4 + 6*(((b*x + a)^2 - 2)*cos(b*x + a) - 2*(b*x + a)*sin(b*x + a))*c^2*d^2/b^2 -
12*(((b*x + a)^2 - 2)*cos(b*x + a) - 2*(b*x + a)*sin(b*x + a))*a*c*d^3/b^3 + 6*(((b*x + a)^2 - 2)*cos(b*x + a
) - 2*(b*x + a)*sin(b*x + a))*a^2*d^4/b^4 + 4*(((b*x + a)^3 - 6*b*x - 6*a)*cos(b*x + a) - 3*((b*x + a)^2 - 2)*
sin(b*x + a))*c*d^3/b^3 - 4*(((b*x + a)^3 - 6*b*x - 6*a)*cos(b*x + a) - 3*((b*x + a)^2 - 2)*sin(b*x + a))*a*d^
4/b^4 + (((b*x + a)^4 - 12*(b*x + a)^2 + 24)*cos(b*x + a) - 4*((b*x + a)^3 - 6*b*x - 6*a)*sin(b*x + a))*d^4/b^
4)/b

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mupad [B]  time = 0.79, size = 221, normalized size = 2.40 \[ \frac {4\,x\,\cos \left (a+b\,x\right )\,\left (6\,c\,d^3-b^2\,c^3\,d\right )}{b^3}-\frac {4\,\sin \left (a+b\,x\right )\,\left (6\,c\,d^3-b^2\,c^3\,d\right )}{b^4}-\frac {d^4\,x^4\,\cos \left (a+b\,x\right )}{b}-\frac {\cos \left (a+b\,x\right )\,\left (b^4\,c^4-12\,b^2\,c^2\,d^2+24\,d^4\right )}{b^5}+\frac {4\,d^4\,x^3\,\sin \left (a+b\,x\right )}{b^2}-\frac {12\,x\,\sin \left (a+b\,x\right )\,\left (2\,d^4-b^2\,c^2\,d^2\right )}{b^4}+\frac {6\,x^2\,\cos \left (a+b\,x\right )\,\left (2\,d^4-b^2\,c^2\,d^2\right )}{b^3}-\frac {4\,c\,d^3\,x^3\,\cos \left (a+b\,x\right )}{b}+\frac {12\,c\,d^3\,x^2\,\sin \left (a+b\,x\right )}{b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)*(c + d*x)^4,x)

[Out]

(4*x*cos(a + b*x)*(6*c*d^3 - b^2*c^3*d))/b^3 - (4*sin(a + b*x)*(6*c*d^3 - b^2*c^3*d))/b^4 - (d^4*x^4*cos(a + b
*x))/b - (cos(a + b*x)*(24*d^4 + b^4*c^4 - 12*b^2*c^2*d^2))/b^5 + (4*d^4*x^3*sin(a + b*x))/b^2 - (12*x*sin(a +
b*x)*(2*d^4 - b^2*c^2*d^2))/b^4 + (6*x^2*cos(a + b*x)*(2*d^4 - b^2*c^2*d^2))/b^3 - (4*c*d^3*x^3*cos(a + b*x))
/b + (12*c*d^3*x^2*sin(a + b*x))/b^2

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sympy [A]  time = 3.43, size = 311, normalized size = 3.38 \[ \begin {cases} - \frac {c^{4} \cos {\left (a + b x \right )}}{b} - \frac {4 c^{3} d x \cos {\left (a + b x \right )}}{b} - \frac {6 c^{2} d^{2} x^{2} \cos {\left (a + b x \right )}}{b} - \frac {4 c d^{3} x^{3} \cos {\left (a + b x \right )}}{b} - \frac {d^{4} x^{4} \cos {\left (a + b x \right )}}{b} + \frac {4 c^{3} d \sin {\left (a + b x \right )}}{b^{2}} + \frac {12 c^{2} d^{2} x \sin {\left (a + b x \right )}}{b^{2}} + \frac {12 c d^{3} x^{2} \sin {\left (a + b x \right )}}{b^{2}} + \frac {4 d^{4} x^{3} \sin {\left (a + b x \right )}}{b^{2}} + \frac {12 c^{2} d^{2} \cos {\left (a + b x \right )}}{b^{3}} + \frac {24 c d^{3} x \cos {\left (a + b x \right )}}{b^{3}} + \frac {12 d^{4} x^{2} \cos {\left (a + b x \right )}}{b^{3}} - \frac {24 c d^{3} \sin {\left (a + b x \right )}}{b^{4}} - \frac {24 d^{4} x \sin {\left (a + b x \right )}}{b^{4}} - \frac {24 d^{4} \cos {\left (a + b x \right )}}{b^{5}} & \text {for}\: b \neq 0 \\\left (c^{4} x + 2 c^{3} d x^{2} + 2 c^{2} d^{2} x^{3} + c d^{3} x^{4} + \frac {d^{4} x^{5}}{5}\right ) \sin {\relax (a )} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**4*sin(b*x+a),x)

[Out]

Piecewise((-c**4*cos(a + b*x)/b - 4*c**3*d*x*cos(a + b*x)/b - 6*c**2*d**2*x**2*cos(a + b*x)/b - 4*c*d**3*x**3*
cos(a + b*x)/b - d**4*x**4*cos(a + b*x)/b + 4*c**3*d*sin(a + b*x)/b**2 + 12*c**2*d**2*x*sin(a + b*x)/b**2 + 12
*c*d**3*x**2*sin(a + b*x)/b**2 + 4*d**4*x**3*sin(a + b*x)/b**2 + 12*c**2*d**2*cos(a + b*x)/b**3 + 24*c*d**3*x*
cos(a + b*x)/b**3 + 12*d**4*x**2*cos(a + b*x)/b**3 - 24*c*d**3*sin(a + b*x)/b**4 - 24*d**4*x*sin(a + b*x)/b**4
- 24*d**4*cos(a + b*x)/b**5, Ne(b, 0)), ((c**4*x + 2*c**3*d*x**2 + 2*c**2*d**2*x**3 + c*d**3*x**4 + d**4*x**5
/5)*sin(a), True))

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