3.257 \(\int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=99 \[ \frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^4}{4 a f} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4523, 32, 3296, 2637} \[ -\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {6 f^3 \sin (c+d x)}{a d^4}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {(e+f x)^4}{4 a f} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Cos[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

(e + f*x)^4/(4*a*f) - (6*f^2*(e + f*x)*Cos[c + d*x])/(a*d^3) + ((e + f*x)^3*Cos[c + d*x])/(a*d) + (6*f^3*Sin[c
+ d*x])/(a*d^4) - (3*f*(e + f*x)^2*Sin[c + d*x])/(a*d^2)

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 4523

Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol
] :> Dist[1/a, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] - Dist[1/b, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2)*S
in[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 1] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x)^3 \, dx}{a}-\frac {\int (e+f x)^3 \sin (c+d x) \, dx}{a}\\ &=\frac {(e+f x)^4}{4 a f}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {(3 f) \int (e+f x)^2 \cos (c+d x) \, dx}{a d}\\ &=\frac {(e+f x)^4}{4 a f}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {\left (6 f^2\right ) \int (e+f x) \sin (c+d x) \, dx}{a d^2}\\ &=\frac {(e+f x)^4}{4 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {\left (6 f^3\right ) \int \cos (c+d x) \, dx}{a d^3}\\ &=\frac {(e+f x)^4}{4 a f}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 102, normalized size = 1.03 \[ \frac {-12 f \sin (c+d x) \left (d^2 (e+f x)^2-2 f^2\right )+4 d (e+f x) \cos (c+d x) \left (d^2 (e+f x)^2-6 f^2\right )+d^4 x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )}{4 a d^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Cos[c + d*x]^2)/(a + a*Sin[c + d*x]),x]

[Out]

(d^4*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3) + 4*d*(e + f*x)*(-6*f^2 + d^2*(e + f*x)^2)*Cos[c + d*x] - 1
2*f*(-2*f^2 + d^2*(e + f*x)^2)*Sin[c + d*x])/(4*a*d^4)

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fricas [A]  time = 0.46, size = 157, normalized size = 1.59 \[ \frac {d^{4} f^{3} x^{4} + 4 \, d^{4} e f^{2} x^{3} + 6 \, d^{4} e^{2} f x^{2} + 4 \, d^{4} e^{3} x + 4 \, {\left (d^{3} f^{3} x^{3} + 3 \, d^{3} e f^{2} x^{2} + d^{3} e^{3} - 6 \, d e f^{2} + 3 \, {\left (d^{3} e^{2} f - 2 \, d f^{3}\right )} x\right )} \cos \left (d x + c\right ) - 12 \, {\left (d^{2} f^{3} x^{2} + 2 \, d^{2} e f^{2} x + d^{2} e^{2} f - 2 \, f^{3}\right )} \sin \left (d x + c\right )}{4 \, a d^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cos(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cos(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.12, size = 436, normalized size = 4.40 \[ -\frac {f^{3} \left (-\left (d x +c \right )^{3} \cos \left (d x +c \right )+3 \left (d x +c \right )^{2} \sin \left (d x +c \right )-6 \sin \left (d x +c \right )+6 \left (d x +c \right ) \cos \left (d x +c \right )\right )-3 c \,f^{3} \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )+3 f^{2} e d \left (-\left (d x +c \right )^{2} \cos \left (d x +c \right )+2 \cos \left (d x +c \right )+2 \left (d x +c \right ) \sin \left (d x +c \right )\right )+3 c^{2} f^{3} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )-6 c d e \,f^{2} \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )+3 d^{2} e^{2} f \left (\sin \left (d x +c \right )-\left (d x +c \right ) \cos \left (d x +c \right )\right )+c^{3} f^{3} \cos \left (d x +c \right )-3 c^{2} d e \,f^{2} \cos \left (d x +c \right )+3 c \,d^{2} e^{2} f \cos \left (d x +c \right )-d^{3} e^{3} \cos \left (d x +c \right )-\frac {f^{3} \left (d x +c \right )^{4}}{4}+c \,f^{3} \left (d x +c \right )^{3}-f^{2} e d \left (d x +c \right )^{3}-\frac {3 c^{2} f^{3} \left (d x +c \right )^{2}}{2}+3 c d e \,f^{2} \left (d x +c \right )^{2}-\frac {3 d^{2} e^{2} f \left (d x +c \right )^{2}}{2}+c^{3} f^{3} \left (d x +c \right )-3 c^{2} d e \,f^{2} \left (d x +c \right )+3 c \,d^{2} e^{2} f \left (d x +c \right )-d^{3} e^{3} \left (d x +c \right )}{d^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*cos(d*x+c)^2/(a+a*sin(d*x+c)),x)

[Out]

-1/d^4/a*(f^3*(-(d*x+c)^3*cos(d*x+c)+3*(d*x+c)^2*sin(d*x+c)-6*sin(d*x+c)+6*(d*x+c)*cos(d*x+c))-3*c*f^3*(-(d*x+
c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin(d*x+c))+3*f^2*e*d*(-(d*x+c)^2*cos(d*x+c)+2*cos(d*x+c)+2*(d*x+c)*sin
(d*x+c))+3*c^2*f^3*(sin(d*x+c)-(d*x+c)*cos(d*x+c))-6*c*d*e*f^2*(sin(d*x+c)-(d*x+c)*cos(d*x+c))+3*d^2*e^2*f*(si
n(d*x+c)-(d*x+c)*cos(d*x+c))+c^3*f^3*cos(d*x+c)-3*c^2*d*e*f^2*cos(d*x+c)+3*c*d^2*e^2*f*cos(d*x+c)-d^3*e^3*cos(
d*x+c)-1/4*f^3*(d*x+c)^4+c*f^3*(d*x+c)^3-f^2*e*d*(d*x+c)^3-3/2*c^2*f^3*(d*x+c)^2+3*c*d*e*f^2*(d*x+c)^2-3/2*d^2
*e^2*f*(d*x+c)^2+c^3*f^3*(d*x+c)-3*c^2*d*e*f^2*(d*x+c)+3*c*d^2*e^2*f*(d*x+c)-d^3*e^3*(d*x+c))

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maxima [B]  time = 0.73, size = 534, normalized size = 5.39 \[ -\frac {8 \, c^{3} f^{3} {\left (\frac {1}{a d^{3} + \frac {a d^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a d^{3}}\right )} - 24 \, c^{2} e f^{2} {\left (\frac {1}{a d^{2} + \frac {a d^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a d^{2}}\right )} + 24 \, c e^{2} f {\left (\frac {1}{a d + \frac {a d \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}} + \frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a d}\right )} - 8 \, e^{3} {\left (\frac {\arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {1}{a + \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}\right )} - \frac {6 \, {\left ({\left (d x + c\right )}^{2} + 2 \, {\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right )\right )} e^{2} f}{a d} + \frac {12 \, {\left ({\left (d x + c\right )}^{2} + 2 \, {\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right )\right )} c e f^{2}}{a d^{2}} - \frac {6 \, {\left ({\left (d x + c\right )}^{2} + 2 \, {\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right )\right )} c^{2} f^{3}}{a d^{3}} - \frac {4 \, {\left ({\left (d x + c\right )}^{3} + 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 6 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} e f^{2}}{a d^{2}} + \frac {4 \, {\left ({\left (d x + c\right )}^{3} + 3 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \cos \left (d x + c\right ) - 6 \, {\left (d x + c\right )} \sin \left (d x + c\right )\right )} c f^{3}}{a d^{3}} - \frac {{\left ({\left (d x + c\right )}^{4} + 4 \, {\left ({\left (d x + c\right )}^{3} - 6 \, d x - 6 \, c\right )} \cos \left (d x + c\right ) - 12 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} f^{3}}{a d^{3}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cos(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/4*(8*c^3*f^3*(1/(a*d^3 + a*d^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) + arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/(a*d^3)) - 24*c^2*e*f^2*(1/(a*d^2 + a*d^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) + arctan(sin(d*x + c)/(cos(d
*x + c) + 1))/(a*d^2)) + 24*c*e^2*f*(1/(a*d + a*d*sin(d*x + c)^2/(cos(d*x + c) + 1)^2) + arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/(a*d)) - 8*e^3*(arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 1/(a + a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2)) - 6*((d*x + c)^2 + 2*(d*x + c)*cos(d*x + c) - 2*sin(d*x + c))*e^2*f/(a*d) + 12*((d*x + c)^2 +
2*(d*x + c)*cos(d*x + c) - 2*sin(d*x + c))*c*e*f^2/(a*d^2) - 6*((d*x + c)^2 + 2*(d*x + c)*cos(d*x + c) - 2*sin
(d*x + c))*c^2*f^3/(a*d^3) - 4*((d*x + c)^3 + 3*((d*x + c)^2 - 2)*cos(d*x + c) - 6*(d*x + c)*sin(d*x + c))*e*f
^2/(a*d^2) + 4*((d*x + c)^3 + 3*((d*x + c)^2 - 2)*cos(d*x + c) - 6*(d*x + c)*sin(d*x + c))*c*f^3/(a*d^3) - ((d
*x + c)^4 + 4*((d*x + c)^3 - 6*d*x - 6*c)*cos(d*x + c) - 12*((d*x + c)^2 - 2)*sin(d*x + c))*f^3/(a*d^3))/d

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mupad [B]  time = 3.02, size = 184, normalized size = 1.86 \[ \frac {e^3\,x+\frac {3\,e^2\,f\,x^2}{2}+e\,f^2\,x^3+\frac {f^3\,x^4}{4}}{a}-\frac {d\,\left (6\,x\,\cos \left (c+d\,x\right )\,f^3+6\,e\,\cos \left (c+d\,x\right )\,f^2\right )+d^2\,\left (3\,f^3\,x^2\,\sin \left (c+d\,x\right )+3\,e^2\,f\,\sin \left (c+d\,x\right )+6\,e\,f^2\,x\,\sin \left (c+d\,x\right )\right )-d^3\,\left (e^3\,\cos \left (c+d\,x\right )+f^3\,x^3\,\cos \left (c+d\,x\right )+3\,e^2\,f\,x\,\cos \left (c+d\,x\right )+3\,e\,f^2\,x^2\,\cos \left (c+d\,x\right )\right )-6\,f^3\,\sin \left (c+d\,x\right )}{a\,d^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(e + f*x)^3)/(a + a*sin(c + d*x)),x)

[Out]

(e^3*x + (f^3*x^4)/4 + (3*e^2*f*x^2)/2 + e*f^2*x^3)/a - (d*(6*e*f^2*cos(c + d*x) + 6*f^3*x*cos(c + d*x)) + d^2
*(3*f^3*x^2*sin(c + d*x) + 3*e^2*f*sin(c + d*x) + 6*e*f^2*x*sin(c + d*x)) - d^3*(e^3*cos(c + d*x) + f^3*x^3*co
s(c + d*x) + 3*e^2*f*x*cos(c + d*x) + 3*e*f^2*x^2*cos(c + d*x)) - 6*f^3*sin(c + d*x))/(a*d^4)

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sympy [A]  time = 10.04, size = 984, normalized size = 9.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*cos(d*x+c)**2/(a+a*sin(d*x+c)),x)

[Out]

Piecewise((4*d**4*e**3*x*tan(c/2 + d*x/2)**2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + 4*d**4*e**3*x/(4*a*d*
*4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + 6*d**4*e**2*f*x**2*tan(c/2 + d*x/2)**2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*
a*d**4) + 6*d**4*e**2*f*x**2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + 4*d**4*e*f**2*x**3*tan(c/2 + d*x/2)**
2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + 4*d**4*e*f**2*x**3/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + d
**4*f**3*x**4*tan(c/2 + d*x/2)**2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + d**4*f**3*x**4/(4*a*d**4*tan(c/2
+ d*x/2)**2 + 4*a*d**4) + 8*d**3*e**3/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) - 12*d**3*e**2*f*x*tan(c/2 +
d*x/2)**2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + 12*d**3*e**2*f*x/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**
4) - 12*d**3*e*f**2*x**2*tan(c/2 + d*x/2)**2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + 12*d**3*e*f**2*x**2/(
4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) - 4*d**3*f**3*x**3*tan(c/2 + d*x/2)**2/(4*a*d**4*tan(c/2 + d*x/2)**2
+ 4*a*d**4) + 4*d**3*f**3*x**3/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) - 24*d**2*e**2*f*tan(c/2 + d*x/2)/(4*
a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) - 48*d**2*e*f**2*x*tan(c/2 + d*x/2)/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a
*d**4) - 24*d**2*f**3*x**2*tan(c/2 + d*x/2)/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) - 48*d*e*f**2/(4*a*d**4*
tan(c/2 + d*x/2)**2 + 4*a*d**4) + 24*d*f**3*x*tan(c/2 + d*x/2)**2/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) -
24*d*f**3*x/(4*a*d**4*tan(c/2 + d*x/2)**2 + 4*a*d**4) + 48*f**3*tan(c/2 + d*x/2)/(4*a*d**4*tan(c/2 + d*x/2)**2
+ 4*a*d**4), Ne(d, 0)), ((e**3*x + 3*e**2*f*x**2/2 + e*f**2*x**3 + f**3*x**4/4)*cos(c)**2/(a*sin(c) + a), Tru
e))

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