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

Optimal. Leaf size=162 \[ \frac {\left (2 a^2 c-3 a b d+b^2 c\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac {\left (a^2 (-d)+3 a b c-2 b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac {(b c-a d) \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \]

[Out]

(2*a^2*c-3*a*b*d+b^2*c)*arctan((b+a*tan(1/2*f*x+1/2*e))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)/f+1/2*(-a*d+b*c)*cos(
f*x+e)/(a^2-b^2)/f/(a+b*sin(f*x+e))^2+1/2*(-a^2*d+3*a*b*c-2*b^2*d)*cos(f*x+e)/(a^2-b^2)^2/f/(a+b*sin(f*x+e))

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Rubi [A]  time = 0.17, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2754, 12, 2660, 618, 204} \[ \frac {\left (2 a^2 c-3 a b d+b^2 c\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{5/2}}+\frac {\left (a^2 (-d)+3 a b c-2 b^2 d\right ) \cos (e+f x)}{2 f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac {(b c-a d) \cos (e+f x)}{2 f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*c + b^2*c - 3*a*b*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/((a^2 - b^2)^(5/2)*f) + ((b*c -
a*d)*Cos[e + f*x])/(2*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) + ((3*a*b*c - a^2*d - 2*b^2*d)*Cos[e + f*x])/(2*(
a^2 - b^2)^2*f*(a + b*Sin[e + f*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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])

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 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 2754

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

Rubi steps

\begin {align*} \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^3} \, dx &=\frac {(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}-\frac {\int \frac {-2 (a c-b d)+(b c-a d) \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=\frac {(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac {\int \frac {2 a^2 c+b^2 c-3 a b d}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac {\left (2 a^2 c+b^2 c-3 a b d\right ) \int \frac {1}{a+b \sin (e+f x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=\frac {(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac {\left (2 a^2 c+b^2 c-3 a b d\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 )}{\left (a^2-b^2\right )^2 f}\\ &=\frac {(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac {\left (2 \left (2 a^2 c+b^2 c-3 a b 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 )}{\left (a^2-b^2\right )^2 f}\\ &=\frac {\left (2 a^2 c+b^2 c-3 a b d\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} f}+\frac {(b c-a d) \cos (e+f x)}{2 \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {\left (3 a b c-a^2 d-2 b^2 d\right ) \cos (e+f x)}{2 \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 157, normalized size = 0.97 \[ \frac {\frac {2 \left (2 a^2 c-3 a b d+b^2 c\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (e+f x)\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}-\frac {\left (a^2 d-3 a b c+2 b^2 d\right ) \cos (e+f x)}{(a-b)^2 (a+b)^2 (a+b \sin (e+f x))}+\frac {(b c-a d) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*(2*a^2*c + b^2*c - 3*a*b*d)*ArcTan[(b + a*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + ((b*c -
a*d)*Cos[e + f*x])/((a - b)*(a + b)*(a + b*Sin[e + f*x])^2) - ((-3*a*b*c + a^2*d + 2*b^2*d)*Cos[e + f*x])/((a
- b)^2*(a + b)^2*(a + b*Sin[e + f*x])))/(2*f)

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fricas [B]  time = 0.91, size = 799, normalized size = 4.93 \[ \left [-\frac {2 \, {\left (3 \, {\left (a^{3} b^{2} - a b^{4}\right )} c - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left ({\left (3 \, a b^{3} d - {\left (2 \, a^{2} b^{2} + b^{4}\right )} c\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4}\right )} c - 3 \, {\left (a^{3} b + a b^{3}\right )} d - 2 \, {\left (3 \, a^{2} b^{2} d - {\left (2 \, a^{3} b + a b^{3}\right )} c\right )} \sin \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) + 2 \, {\left ({\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} c - {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d\right )} \cos \left (f x + e\right )}{4 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}, -\frac {{\left (3 \, {\left (a^{3} b^{2} - a b^{4}\right )} c - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} d\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left ({\left (3 \, a b^{3} d - {\left (2 \, a^{2} b^{2} + b^{4}\right )} c\right )} \cos \left (f x + e\right )^{2} + {\left (2 \, a^{4} + 3 \, a^{2} b^{2} + b^{4}\right )} c - 3 \, {\left (a^{3} b + a b^{3}\right )} d - 2 \, {\left (3 \, a^{2} b^{2} d - {\left (2 \, a^{3} b + a b^{3}\right )} c\right )} \sin \left (f x + e\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) + {\left ({\left (4 \, a^{4} b - 5 \, a^{2} b^{3} + b^{5}\right )} c - {\left (2 \, a^{5} - a^{3} b^{2} - a b^{4}\right )} d\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} f \sin \left (f x + e\right ) - {\left (a^{8} - 2 \, a^{6} b^{2} + 2 \, a^{2} b^{6} - b^{8}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.24, size = 429, normalized size = 2.65 \[ \frac {\frac {{\left (2 \, a^{2} c + b^{2} c - 3 \, a b d\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {5 \, a^{3} b^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, a b^{4} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{4} b d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 4 \, a^{4} b c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 7 \, a^{2} b^{3} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, b^{5} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a^{5} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 5 \, a^{3} b^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 2 \, a b^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 11 \, a^{3} b^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a b^{4} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 \, a^{4} b d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 4 \, a^{2} b^{3} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 4 \, a^{4} b c - a^{2} b^{3} c - 2 \, a^{5} d - a^{3} b^{2} d}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}^{2}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*a^2*c + b^2*c - 3*a*b*d)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sq
rt(a^2 - b^2)))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2*c*tan(1/2*f*x + 1/2*e)^3 - 2*a*b^4*c*ta
n(1/2*f*x + 1/2*e)^3 - 3*a^4*b*d*tan(1/2*f*x + 1/2*e)^3 + 4*a^4*b*c*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^3*c*tan(1
/2*f*x + 1/2*e)^2 - 2*b^5*c*tan(1/2*f*x + 1/2*e)^2 - 2*a^5*d*tan(1/2*f*x + 1/2*e)^2 - 5*a^3*b^2*d*tan(1/2*f*x
+ 1/2*e)^2 - 2*a*b^4*d*tan(1/2*f*x + 1/2*e)^2 + 11*a^3*b^2*c*tan(1/2*f*x + 1/2*e) - 2*a*b^4*c*tan(1/2*f*x + 1/
2*e) - 5*a^4*b*d*tan(1/2*f*x + 1/2*e) - 4*a^2*b^3*d*tan(1/2*f*x + 1/2*e) + 4*a^4*b*c - a^2*b^3*c - 2*a^5*d - a
^3*b^2*d)/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)^2))/f

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maple [B]  time = 0.24, size = 1291, normalized size = 7.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b*a^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^3*d+5/f/
(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^2*a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^3*c-2/f/(tan(
1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^4/a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^3*c-2/f/(tan(1/2*f
*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^3*tan(1/2*f*x+1/2*e)^2*d+4/f/(tan(1/2*f*x+1/2*
e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^2*tan(1/2*f*x+1/2*e)^2*b*c-5/f/(tan(1/2*f*x+1/2*e)^2*
a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*tan(1/2*f*x+1/2*e)^2*b^2*d+7/f/(tan(1/2*f*x+1/2*e)^2*a+2*t
an(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)^2*b^3*c-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*
f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)^2*b^4*d-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1
/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)/a^2*tan(1/2*f*x+1/2*e)^2*b^5*c-5/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*
e)*b+a)^2*b*a^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*d+11/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a
)^2*b^2*a/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*c-4/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^3
/(a^4-2*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*d-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2*b^4/a/(a^4-2
*a^2*b^2+b^4)*tan(1/2*f*x+1/2*e)*c-2/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)
*a^3*d+4/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a^2*b*c-1/f/(tan(1/2*f*x+1/
2*e)^2*a+2*tan(1/2*f*x+1/2*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*a*b^2*d-1/f/(tan(1/2*f*x+1/2*e)^2*a+2*tan(1/2*f*x+1/2
*e)*b+a)^2/(a^4-2*a^2*b^2+b^4)*b^3*c+2/f/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e
)+2*b)/(a^2-b^2)^(1/2))*a^2*c-3/f/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/
(a^2-b^2)^(1/2))*a*b*d+1/f/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^
2)^(1/2))*b^2*c

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 9.79, size = 477, normalized size = 2.94 \[ \frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )\,\left (2\,c\,a^2-3\,d\,a\,b+c\,b^2\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,c\,a^2-3\,d\,a\,b+c\,b^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,c\,a^2-3\,d\,a\,b+c\,b^2}\right )\,\left (2\,c\,a^2-3\,d\,a\,b+c\,b^2\right )}{f\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {2\,d\,a^3-4\,c\,a^2\,b+d\,a\,b^2+c\,b^3}{a^4-2\,a^2\,b^2+b^4}+\frac {b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (3\,d\,a^3-5\,c\,a^2\,b+2\,c\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,d\,a^3-11\,c\,a^2\,b+4\,d\,a\,b^2+2\,c\,b^3\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a^2+2\,b^2\right )\,\left (2\,d\,a^3-4\,c\,a^2\,b+d\,a\,b^2+c\,b^3\right )}{a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(((((2*a^4*b + 2*b^5 - 4*a^2*b^3)*(2*a^2*c + b^2*c - 3*a*b*d))/(2*(a + b)^(5/2)*(a - b)^(5/2)*(a^4 + b^4
- 2*a^2*b^2)) + (a*tan(e/2 + (f*x)/2)*(2*a^2*c + b^2*c - 3*a*b*d))/((a + b)^(5/2)*(a - b)^(5/2)))*(a^4 + b^4 -
2*a^2*b^2))/(2*a^2*c + b^2*c - 3*a*b*d))*(2*a^2*c + b^2*c - 3*a*b*d))/(f*(a + b)^(5/2)*(a - b)^(5/2)) - ((2*a
^3*d + b^3*c - 4*a^2*b*c + a*b^2*d)/(a^4 + b^4 - 2*a^2*b^2) + (b*tan(e/2 + (f*x)/2)^3*(3*a^3*d + 2*b^3*c - 5*a
^2*b*c))/(a*(a^4 + b^4 - 2*a^2*b^2)) + (b*tan(e/2 + (f*x)/2)*(5*a^3*d + 2*b^3*c - 11*a^2*b*c + 4*a*b^2*d))/(a*
(a^4 + b^4 - 2*a^2*b^2)) + (tan(e/2 + (f*x)/2)^2*(a^2 + 2*b^2)*(2*a^3*d + b^3*c - 4*a^2*b*c + a*b^2*d))/(a^2*(
a^4 + b^4 - 2*a^2*b^2)))/(f*(tan(e/2 + (f*x)/2)^2*(2*a^2 + 4*b^2) + a^2*tan(e/2 + (f*x)/2)^4 + a^2 + 4*a*b*tan
(e/2 + (f*x)/2)^3 + 4*a*b*tan(e/2 + (f*x)/2)))

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sympy [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((c+d*sin(f*x+e))/(a+b*sin(f*x+e))**3,x)

[Out]

Timed out

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