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

Optimal. Leaf size=161 \[ \frac {f^2 \sin ^2(c+d x)}{4 a d^3}-\frac {2 f^2 \sin (c+d x)}{a d^3}+\frac {2 f (e+f x) \cos (c+d x)}{a d^2}-\frac {f (e+f x) \sin (c+d x) \cos (c+d x)}{2 a d^2}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \sin (c+d x)}{a d}+\frac {e f x}{2 a d}+\frac {f^2 x^2}{4 a d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 161, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4523, 3296, 2637, 4404, 3310} \[ \frac {2 f (e+f x) \cos (c+d x)}{a d^2}-\frac {f (e+f x) \sin (c+d x) \cos (c+d x)}{2 a d^2}+\frac {f^2 \sin ^2(c+d x)}{4 a d^3}-\frac {2 f^2 \sin (c+d x)}{a d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 a d}+\frac {(e+f x)^2 \sin (c+d x)}{a d}+\frac {e f x}{2 a d}+\frac {f^2 x^2}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 4404

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

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

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Mathematica [A]  time = 1.09, size = 95, normalized size = 0.59 \[ \frac {\cos (2 (c+d x)) \left (2 d^2 (e+f x)^2-f^2\right )-4 \sin (c+d x) \left (d f (e+f x) \cos (c+d x)-2 \left (d^2 (e+f x)^2-2 f^2\right )\right )+16 d f (e+f x) \cos (c+d x)}{8 a d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(16*d*f*(e + f*x)*Cos[c + d*x] + (-f^2 + 2*d^2*(e + f*x)^2)*Cos[2*(c + d*x)] - 4*(-2*(-2*f^2 + d^2*(e + f*x)^2
) + d*f*(e + f*x)*Cos[c + d*x])*Sin[c + d*x])/(8*a*d^3)

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fricas [A]  time = 0.44, size = 149, normalized size = 0.93 \[ -\frac {d^{2} f^{2} x^{2} + 2 \, d^{2} e f x - {\left (2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} e f x + 2 \, d^{2} e^{2} - f^{2}\right )} \cos \left (d x + c\right )^{2} - 8 \, {\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right ) - 2 \, {\left (2 \, d^{2} f^{2} x^{2} + 4 \, d^{2} e f x + 2 \, d^{2} e^{2} - 4 \, f^{2} - {\left (d f^{2} x + d e f\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, a d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(d^2*f^2*x^2 + 2*d^2*e*f*x - (2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 - f^2)*cos(d*x + c)^2 - 8*(d*f^2*x
+ d*e*f)*cos(d*x + c) - 2*(2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 - 4*f^2 - (d*f^2*x + d*e*f)*cos(d*x + c))*s
in(d*x + c))/(a*d^3)

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

[Out]

Timed out

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maple [B]  time = 0.11, size = 339, normalized size = 2.11 \[ -\frac {f^{2} \left (-\frac {\left (d x +c \right )^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\left (d x +c \right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )-\frac {\left (d x +c \right )^{2}}{4}-\frac {\left (\sin ^{2}\left (d x +c \right )\right )}{4}\right )-2 c \,f^{2} \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )+2 d e f \left (-\frac {\left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{2}+\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{4}+\frac {d x}{4}+\frac {c}{4}\right )-\frac {c^{2} f^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}+c d e f \left (\cos ^{2}\left (d x +c \right )\right )-\frac {d^{2} e^{2} \left (\cos ^{2}\left (d x +c \right )\right )}{2}-f^{2} \left (\left (d x +c \right )^{2} \sin \left (d x +c \right )-2 \sin \left (d x +c \right )+2 \left (d x +c \right ) \cos \left (d x +c \right )\right )+2 c \,f^{2} \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )-2 d e f \left (\cos \left (d x +c \right )+\left (d x +c \right ) \sin \left (d x +c \right )\right )-\sin \left (d x +c \right ) c^{2} f^{2}+2 \sin \left (d x +c \right ) c d e f -\sin \left (d x +c \right ) d^{2} e^{2}}{d^{3} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.92, size = 289, normalized size = 1.80 \[ -\frac {\frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} e^{2}}{a} - \frac {8 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c e f}{a d} + \frac {4 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right )\right )} c^{2} f^{2}}{a d^{2}} - \frac {2 \, {\left (2 \, {\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, {\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} e f}{a d} + \frac {2 \, {\left (2 \, {\left (d x + c\right )} \cos \left (2 \, d x + 2 \, c\right ) + 8 \, {\left (d x + c\right )} \sin \left (d x + c\right ) + 8 \, \cos \left (d x + c\right ) - \sin \left (2 \, d x + 2 \, c\right )\right )} c f^{2}}{a d^{2}} - \frac {{\left ({\left (2 \, {\left (d x + c\right )}^{2} - 1\right )} \cos \left (2 \, d x + 2 \, c\right ) + 16 \, {\left (d x + c\right )} \cos \left (d x + c\right ) - 2 \, {\left (d x + c\right )} \sin \left (2 \, d x + 2 \, c\right ) + 8 \, {\left ({\left (d x + c\right )}^{2} - 2\right )} \sin \left (d x + c\right )\right )} f^{2}}{a d^{2}}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.20, size = 187, normalized size = 1.16 \[ \frac {8\,d^2\,e^2\,\sin \left (c+d\,x\right )-f^2\,\cos \left (2\,c+2\,d\,x\right )-16\,f^2\,\sin \left (c+d\,x\right )+2\,d^2\,e^2\,\cos \left (2\,c+2\,d\,x\right )+8\,d^2\,f^2\,x^2\,\sin \left (c+d\,x\right )-2\,d\,e\,f\,\sin \left (2\,c+2\,d\,x\right )+16\,d\,f^2\,x\,\cos \left (c+d\,x\right )+2\,d^2\,f^2\,x^2\,\cos \left (2\,c+2\,d\,x\right )-2\,d\,f^2\,x\,\sin \left (2\,c+2\,d\,x\right )+16\,d\,e\,f\,\cos \left (c+d\,x\right )+4\,d^2\,e\,f\,x\,\cos \left (2\,c+2\,d\,x\right )+16\,d^2\,e\,f\,x\,\sin \left (c+d\,x\right )}{8\,a\,d^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*d^2*e^2*sin(c + d*x) - f^2*cos(2*c + 2*d*x) - 16*f^2*sin(c + d*x) + 2*d^2*e^2*cos(2*c + 2*d*x) + 8*d^2*f^2*
x^2*sin(c + d*x) - 2*d*e*f*sin(2*c + 2*d*x) + 16*d*f^2*x*cos(c + d*x) + 2*d^2*f^2*x^2*cos(2*c + 2*d*x) - 2*d*f
^2*x*sin(2*c + 2*d*x) + 16*d*e*f*cos(c + d*x) + 4*d^2*e*f*x*cos(2*c + 2*d*x) + 16*d^2*e*f*x*sin(c + d*x))/(8*a
*d^3)

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sympy [A]  time = 12.35, size = 1528, normalized size = 9.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((8*d**2*e**2*tan(c/2 + d*x/2)**3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*
d**3) - 8*d**2*e**2*tan(c/2 + d*x/2)**2/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**
3) + 8*d**2*e**2*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 2
*d**2*e*f*x*tan(c/2 + d*x/2)**4/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 16*
d**2*e*f*x*tan(c/2 + d*x/2)**3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) - 12*d
**2*e*f*x*tan(c/2 + d*x/2)**2/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 16*d*
*2*e*f*x*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 2*d**2*e*
f*x/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + d**2*f**2*x**2*tan(c/2 + d*x/2)
**4/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 8*d**2*f**2*x**2*tan(c/2 + d*x/
2)**3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) - 6*d**2*f**2*x**2*tan(c/2 + d*
x/2)**2/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 8*d**2*f**2*x**2*tan(c/2 +
d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + d**2*f**2*x**2/(4*a*d**3*tan
(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 4*d*e*f*tan(c/2 + d*x/2)**3/(4*a*d**3*tan(c/2 +
d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 16*d*e*f*tan(c/2 + d*x/2)**2/(4*a*d**3*tan(c/2 + d*x/2)
**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) - 4*d*e*f*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*
d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 16*d*e*f/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 +
4*a*d**3) - 8*d*f**2*x*tan(c/2 + d*x/2)**4/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a
*d**3) + 4*d*f**2*x*tan(c/2 + d*x/2)**3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**
3) - 4*d*f**2*x*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 8*
d*f**2*x/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) - 16*f**2*tan(c/2 + d*x/2)**
3/(4*a*d**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) + 4*f**2*tan(c/2 + d*x/2)**2/(4*a*d
**3*tan(c/2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3) - 16*f**2*tan(c/2 + d*x/2)/(4*a*d**3*tan(c/
2 + d*x/2)**4 + 8*a*d**3*tan(c/2 + d*x/2)**2 + 4*a*d**3), Ne(d, 0)), ((e**2*x + e*f*x**2 + f**2*x**3/3)*cos(c)
**3/(a*sin(c) + a), True))

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