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

Optimal. Leaf size=205 \[ \frac{d \left (-2 a^2 d^2+2 a b c d+b^2 \left (-\left (c^2-d^2\right )\right )\right ) \cos (e+f x)}{b^2 f \left (a^2-b^2\right )}+\frac{2 (b c-a d)^2 \left (2 a^2 d+a b c-3 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{d^2 x (3 b c-2 a d)}{b^3} \]

[Out]

(d^2*(3*b*c - 2*a*d)*x)/b^3 + (2*(b*c - a*d)^2*(a*b*c + 2*a^2*d - 3*b^2*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqr
t[a^2 - b^2]])/(b^3*(a^2 - b^2)^(3/2)*f) + (d*(2*a*b*c*d - 2*a^2*d^2 - b^2*(c^2 - d^2))*Cos[e + f*x])/(b^2*(a^
2 - b^2)*f) + ((b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(b*(a^2 - b^2)*f*(a + b*Sin[e + f*x]))

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Rubi [A]  time = 0.459016, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2792, 3023, 2735, 2660, 618, 204} \[ \frac{d \left (-2 a^2 d^2+2 a b c d+b^2 \left (-\left (c^2-d^2\right )\right )\right ) \cos (e+f x)}{b^2 f \left (a^2-b^2\right )}+\frac{2 (b c-a d)^2 \left (2 a^2 d+a b c-3 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^3 f \left (a^2-b^2\right )^{3/2}}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b f \left (a^2-b^2\right ) (a+b \sin (e+f x))}+\frac{d^2 x (3 b c-2 a d)}{b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(3*b*c - 2*a*d)*x)/b^3 + (2*(b*c - a*d)^2*(a*b*c + 2*a^2*d - 3*b^2*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqr
t[a^2 - b^2]])/(b^3*(a^2 - b^2)^(3/2)*f) + (d*(2*a*b*c*d - 2*a^2*d^2 - b^2*(c^2 - d^2))*Cos[e + f*x])/(b^2*(a^
2 - b^2)*f) + ((b*c - a*d)^2*Cos[e + f*x]*(c + d*Sin[e + f*x]))/(b*(a^2 - b^2)*f*(a + b*Sin[e + f*x]))

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^3}{(a+b \sin (e+f x))^2} \, dx &=\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\int \frac{3 b^2 c^2 d+a^2 d^3-a b c \left (c^2+3 d^2\right )-d \left (a^2 c d-3 b^2 c d+a b \left (c^2+d^2\right )\right ) \sin (e+f x)+d \left (2 a b c d-2 a^2 d^2-b^2 \left (c^2-d^2\right )\right ) \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx}{b \left (a^2-b^2\right )}\\ &=\frac{d \left (2 a b c d-2 a^2 d^2-b^2 \left (c^2-d^2\right )\right ) \cos (e+f x)}{b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\int \frac{b \left (3 b^2 c^2 d+a^2 d^3-a b c \left (c^2+3 d^2\right )\right )-\left (a^2-b^2\right ) d^2 (3 b c-2 a d) \sin (e+f x)}{a+b \sin (e+f x)} \, dx}{b^2 \left (a^2-b^2\right )}\\ &=\frac{d^2 (3 b c-2 a d) x}{b^3}+\frac{d \left (2 a b c d-2 a^2 d^2-b^2 \left (c^2-d^2\right )\right ) \cos (e+f x)}{b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\left ((b c-a d)^2 \left (a b c+2 a^2 d-3 b^2 d\right )\right ) \int \frac{1}{a+b \sin (e+f x)} \, dx}{b^3 \left (a^2-b^2\right )}\\ &=\frac{d^2 (3 b c-2 a d) x}{b^3}+\frac{d \left (2 a b c d-2 a^2 d^2-b^2 \left (c^2-d^2\right )\right ) \cos (e+f x)}{b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}+\frac{\left (2 (b c-a d)^2 \left (a b c+2 a^2 d-3 b^2 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{1}{2} (e+f x)\right )\right )}{b^3 \left (a^2-b^2\right ) f}\\ &=\frac{d^2 (3 b c-2 a d) x}{b^3}+\frac{d \left (2 a b c d-2 a^2 d^2-b^2 \left (c^2-d^2\right )\right ) \cos (e+f x)}{b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}-\frac{\left (4 (b c-a d)^2 \left (a b c+2 a^2 d-3 b^2 d\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{1}{2} (e+f x)\right )\right )}{b^3 \left (a^2-b^2\right ) f}\\ &=\frac{d^2 (3 b c-2 a d) x}{b^3}+\frac{2 (b c-a d)^2 \left (a b c+2 a^2 d-3 b^2 d\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{b^3 \left (a^2-b^2\right )^{3/2} f}+\frac{d \left (2 a b c d-2 a^2 d^2-b^2 \left (c^2-d^2\right )\right ) \cos (e+f x)}{b^2 \left (a^2-b^2\right ) f}+\frac{(b c-a d)^2 \cos (e+f x) (c+d \sin (e+f x))}{b \left (a^2-b^2\right ) f (a+b \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 1.09986, size = 151, normalized size = 0.74 \[ \frac{\frac{2 (b c-a d)^2 \left (2 a^2 d+a b c-3 b^2 d\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (e+f x)\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+d^2 (e+f x) (3 b c-2 a d)+\frac{b (b c-a d)^3 \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}-b d^3 \cos (e+f x)}{b^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(d^2*(3*b*c - 2*a*d)*(e + f*x) + (2*(b*c - a*d)^2*(a*b*c + 2*a^2*d - 3*b^2*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/
Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - b*d^3*Cos[e + f*x] + (b*(b*c - a*d)^3*Cos[e + f*x])/((a - b)*(a + b)*(a
+ b*Sin[e + f*x])))/(b^3*f)

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Maple [B]  time = 0.101, size = 842, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*d^3/b^2/(1+tan(1/2*f*x+1/2*e)^2)-4/f*d^3/b^3*arctan(tan(1/2*f*x+1/2*e))*a+6/f*d^2/b^2*arctan(tan(1/2*f*x+
1/2*e))*c-2/f/b/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a^2*tan(1/2*f*x+1/2*e)*d^3+6/f/(ta
n(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a*tan(1/2*f*x+1/2*e)*c*d^2-6/f*b/(tan(1/2*f*x+1/2*e)^
2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*tan(1/2*f*x+1/2*e)*c^2*d+2/f*b^2/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x
+1/2*e)*b+a)/(a^2-b^2)/a*tan(1/2*f*x+1/2*e)*c^3-2/f/b^2/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2
-b^2)*a^3*d^3+6/f/b/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a^2*c*d^2-6/f/(tan(1/2*f*x+1/2
*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a^2-b^2)*a*c^2*d+2/f*b/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)/(a
^2-b^2)*c^3+4/f/b^3/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^4*d^3-6/f/b^2/(
a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^3*c*d^2-6/f/b/(a^2-b^2)^(3/2)*arctan
(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*a^2*d^3+2/f/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2
*e)+2*b)/(a^2-b^2)^(1/2))*a*c^3+12/f/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*
a*c*d^2-6/f*b/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2))*c^2*d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.54654, size = 2078, normalized size = 10.14 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2*(a^6 - 2*a^4*b^2 + a^2*b^4)*d^3)*f*x - (a^2*b^3*c^3 - 3*a*b^4
*c^2*d - 3*(a^4*b - 2*a^2*b^3)*c*d^2 + (2*a^5 - 3*a^3*b^2)*d^3 + (a*b^4*c^3 - 3*b^5*c^2*d - 3*(a^3*b^2 - 2*a*b
^4)*c*d^2 + (2*a^4*b - 3*a^2*b^3)*d^3)*sin(f*x + e))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*
b*sin(f*x + e) - a^2 - b^2 + 2*(a*cos(f*x + e)*sin(f*x + e) + b*cos(f*x + e))*sqrt(-a^2 + b^2))/(b^2*cos(f*x +
e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)) + 2*((a^2*b^4 - b^6)*c^3 - 3*(a^3*b^3 - a*b^5)*c^2*d + 3*(a^4*b^2 - a
^2*b^4)*c*d^2 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*d^3)*cos(f*x + e) - 2*((a^4*b^2 - 2*a^2*b^4 + b^6)*d^3*cos(f*x +
e) - (3*(a^4*b^2 - 2*a^2*b^4 + b^6)*c*d^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5)*d^3)*f*x)*sin(f*x + e))/((a^4*b^4 -
2*a^2*b^6 + b^8)*f*sin(f*x + e) + (a^5*b^3 - 2*a^3*b^5 + a*b^7)*f), ((3*(a^5*b - 2*a^3*b^3 + a*b^5)*c*d^2 - 2
*(a^6 - 2*a^4*b^2 + a^2*b^4)*d^3)*f*x - (a^2*b^3*c^3 - 3*a*b^4*c^2*d - 3*(a^4*b - 2*a^2*b^3)*c*d^2 + (2*a^5 -
3*a^3*b^2)*d^3 + (a*b^4*c^3 - 3*b^5*c^2*d - 3*(a^3*b^2 - 2*a*b^4)*c*d^2 + (2*a^4*b - 3*a^2*b^3)*d^3)*sin(f*x +
e))*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + ((a^2*b^4 - b^6)*c^3 - 3*(
a^3*b^3 - a*b^5)*c^2*d + 3*(a^4*b^2 - a^2*b^4)*c*d^2 - (2*a^5*b - 3*a^3*b^3 + a*b^5)*d^3)*cos(f*x + e) - ((a^4
*b^2 - 2*a^2*b^4 + b^6)*d^3*cos(f*x + e) - (3*(a^4*b^2 - 2*a^2*b^4 + b^6)*c*d^2 - 2*(a^5*b - 2*a^3*b^3 + a*b^5
)*d^3)*f*x)*sin(f*x + e))/((a^4*b^4 - 2*a^2*b^6 + b^8)*f*sin(f*x + e) + (a^5*b^3 - 2*a^3*b^5 + a*b^7)*f)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.50792, size = 782, normalized size = 3.81 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(a*b^3*c^3 - 3*b^4*c^2*d - 3*a^3*b*c*d^2 + 6*a*b^3*c*d^2 + 2*a^4*d^3 - 3*a^2*b^2*d^3)*(pi*floor(1/2*(f*x +
e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/((a^2*b^3 - b^5)*sqrt(a^2 - b^2))
+ 2*(b^4*c^3*tan(1/2*f*x + 1/2*e)^3 - 3*a*b^3*c^2*d*tan(1/2*f*x + 1/2*e)^3 + 3*a^2*b^2*c*d^2*tan(1/2*f*x + 1/2
*e)^3 - a^3*b*d^3*tan(1/2*f*x + 1/2*e)^3 + a*b^3*c^3*tan(1/2*f*x + 1/2*e)^2 - 3*a^2*b^2*c^2*d*tan(1/2*f*x + 1/
2*e)^2 + 3*a^3*b*c*d^2*tan(1/2*f*x + 1/2*e)^2 - 2*a^4*d^3*tan(1/2*f*x + 1/2*e)^2 + a^2*b^2*d^3*tan(1/2*f*x + 1
/2*e)^2 + b^4*c^3*tan(1/2*f*x + 1/2*e) - 3*a*b^3*c^2*d*tan(1/2*f*x + 1/2*e) + 3*a^2*b^2*c*d^2*tan(1/2*f*x + 1/
2*e) - 3*a^3*b*d^3*tan(1/2*f*x + 1/2*e) + 2*a*b^3*d^3*tan(1/2*f*x + 1/2*e) + a*b^3*c^3 - 3*a^2*b^2*c^2*d + 3*a
^3*b*c*d^2 - 2*a^4*d^3 + a^2*b^2*d^3)/((a^3*b^2 - a*b^4)*(a*tan(1/2*f*x + 1/2*e)^4 + 2*b*tan(1/2*f*x + 1/2*e)^
3 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)) + (3*b*c*d^2 - 2*a*d^3)*(f*x + e)/b^3)/f