3.3.37 \(\int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx\) [237]

Optimal. Leaf size=89 \[ -\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))} \]

[Out]

-A*x/a^3-2/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^3+7/5*A*cos(d*x+c)/a^3/d/(1+sin(d*x+c))^2-13/5*A*cos(d*x+c)/a^3
/d/(1+sin(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.094, Rules used = {3045, 2729, 2727} \begin {gather*} -\frac {13 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)}+\frac {7 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^2}-\frac {2 A \cos (c+d x)}{5 a^3 d (\sin (c+d x)+1)^3}-\frac {A x}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sin[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

-((A*x)/a^3) - (2*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x])^3) + (7*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*
x])^2) - (13*A*Cos[c + d*x])/(5*a^3*d*(1 + Sin[c + d*x]))

Rule 2727

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> Simp[-Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2729

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

Rule 3045

Int[sin[(e_.) + (f_.)*(x_)]^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Int[ExpandTrig[sin[e + f*x]^n*(a + b*sin[e + f*x])^m*(A + B*sin[e + f*x]), x], x] /; Fr
eeQ[{a, b, e, f, A, B}, x] && EqQ[A*b + a*B, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin {align*} \int \frac {\sin ^2(c+d x) (A-A \sin (c+d x))}{(a+a \sin (c+d x))^3} \, dx &=\int \left (-\frac {A}{a^3}+\frac {2 A}{a^3 (1+\sin (c+d x))^3}-\frac {5 A}{a^3 (1+\sin (c+d x))^2}+\frac {4 A}{a^3 (1+\sin (c+d x))}\right ) \, dx\\ &=-\frac {A x}{a^3}+\frac {(2 A) \int \frac {1}{(1+\sin (c+d x))^3} \, dx}{a^3}+\frac {(4 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{a^3}-\frac {(5 A) \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{a^3}\\ &=-\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {5 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))^2}-\frac {4 A \cos (c+d x)}{a^3 d (1+\sin (c+d x))}+\frac {(4 A) \int \frac {1}{(1+\sin (c+d x))^2} \, dx}{5 a^3}-\frac {(5 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{3 a^3}\\ &=-\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {7 A \cos (c+d x)}{3 a^3 d (1+\sin (c+d x))}+\frac {(4 A) \int \frac {1}{1+\sin (c+d x)} \, dx}{15 a^3}\\ &=-\frac {A x}{a^3}-\frac {2 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^3}+\frac {7 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))^2}-\frac {13 A \cos (c+d x)}{5 a^3 d (1+\sin (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(189\) vs. \(2(89)=178\).
time = 0.60, size = 189, normalized size = 2.12 \begin {gather*} \frac {A \left (-50 d x \cos \left (\frac {d x}{2}\right )+110 \cos \left (c+\frac {d x}{2}\right )-90 \cos \left (c+\frac {3 d x}{2}\right )+25 d x \cos \left (2 c+\frac {3 d x}{2}\right )+5 d x \cos \left (2 c+\frac {5 d x}{2}\right )+150 \sin \left (\frac {d x}{2}\right )-50 d x \sin \left (c+\frac {d x}{2}\right )-25 d x \sin \left (c+\frac {3 d x}{2}\right )+40 \sin \left (2 c+\frac {3 d x}{2}\right )-26 \sin \left (2 c+\frac {5 d x}{2}\right )+5 d x \sin \left (3 c+\frac {5 d x}{2}\right )\right )}{20 a^3 d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sin[c + d*x]^2*(A - A*Sin[c + d*x]))/(a + a*Sin[c + d*x])^3,x]

[Out]

(A*(-50*d*x*Cos[(d*x)/2] + 110*Cos[c + (d*x)/2] - 90*Cos[c + (3*d*x)/2] + 25*d*x*Cos[2*c + (3*d*x)/2] + 5*d*x*
Cos[2*c + (5*d*x)/2] + 150*Sin[(d*x)/2] - 50*d*x*Sin[c + (d*x)/2] - 25*d*x*Sin[c + (3*d*x)/2] + 40*Sin[2*c + (
3*d*x)/2] - 26*Sin[2*c + (5*d*x)/2] + 5*d*x*Sin[3*c + (5*d*x)/2]))/(20*a^3*d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2])^5)

________________________________________________________________________________________

Maple [A]
time = 0.30, size = 96, normalized size = 1.08

method	result	size
risch	\(-\frac {A x}{a^{3}}-\frac {2 \left (-75 A \,{\mathrm e}^{2 i \left (d x +c \right )}+55 i A \,{\mathrm e}^{3 i \left (d x +c \right )}+20 A \,{\mathrm e}^{4 i \left (d x +c \right )}-45 i A \,{\mathrm e}^{i \left (d x +c \right )}+13 A \right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{5}}\)	\(85\)
derivativedivides	\(\frac {8 A \left (-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\right )}{d \,a^{3}}\)	\(96\)
default	\(\frac {8 A \left (-\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4}-\frac {2}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}-\frac {1}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}\right )}{d \,a^{3}}\)	\(96\)
norman	\(\frac {-\frac {A x}{a}-\frac {16 A}{5 a d}-\frac {5 A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a}-\frac {13 A x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {25 A x \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {38 A x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {46 A x \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {46 A x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {38 A x \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {25 A x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {13 A x \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {5 A x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {A x \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {376 A \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {388 A \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {158 A \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 a d}-\frac {28 A \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {52 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {72 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {44 A \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {10 A \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {2 A \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {14 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} a^{2} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}\)	\(444\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

8/d*A/a^3*(-1/4*arctan(tan(1/2*d*x+1/2*c))-2/5/(tan(1/2*d*x+1/2*c)+1)^5+1/(tan(1/2*d*x+1/2*c)+1)^4-1/2/(tan(1/
2*d*x+1/2*c)+1)^3-1/4/(tan(1/2*d*x+1/2*c)+1)^2-1/4/(tan(1/2*d*x+1/2*c)+1))

________________________________________________________________________________________

Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 392 vs. \(2 (83) = 166\).
time = 0.49, size = 392, normalized size = 4.40 \begin {gather*} -\frac {2 \, {\left (A {\left (\frac {\frac {95 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {145 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {75 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + 22}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}} + \frac {15 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + \frac {2 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}\right )}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

-2/15*(A*((95*sin(d*x + c)/(cos(d*x + c) + 1) + 145*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 75*sin(d*x + c)^3/(c
os(d*x + c) + 1)^3 + 15*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 22)/(a^3 + 5*a^3*sin(d*x + c)/(cos(d*x + c) + 1)
+ 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 5*a^3*sin(d*x + c
)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5) + 15*arctan(sin(d*x + c)/(cos(d*x + c) + 1
))/a^3) + 2*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)/(a^3 + 5*a^3*si
n(d*x + c)/(cos(d*x + c) + 1) + 10*a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 10*a^3*sin(d*x + c)^3/(cos(d*x +
c) + 1)^3 + 5*a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a^3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5))/d

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 204 vs. \(2 (83) = 166\).
time = 0.36, size = 204, normalized size = 2.29 \begin {gather*} -\frac {{\left (5 \, A d x + 13 \, A\right )} \cos \left (d x + c\right )^{3} - 20 \, A d x + 3 \, {\left (5 \, A d x - 2 \, A\right )} \cos \left (d x + c\right )^{2} - {\left (10 \, A d x + 21 \, A\right )} \cos \left (d x + c\right ) - {\left (20 \, A d x - {\left (5 \, A d x - 13 \, A\right )} \cos \left (d x + c\right )^{2} + {\left (10 \, A d x + 19 \, A\right )} \cos \left (d x + c\right ) - 2 \, A\right )} \sin \left (d x + c\right ) - 2 \, A}{5 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/5*((5*A*d*x + 13*A)*cos(d*x + c)^3 - 20*A*d*x + 3*(5*A*d*x - 2*A)*cos(d*x + c)^2 - (10*A*d*x + 21*A)*cos(d*
x + c) - (20*A*d*x - (5*A*d*x - 13*A)*cos(d*x + c)^2 + (10*A*d*x + 19*A)*cos(d*x + c) - 2*A)*sin(d*x + c) - 2*
A)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*cos(d*x + c)^2 - 2
*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1268 vs. \(2 (85) = 170\).
time = 9.09, size = 1268, normalized size = 14.25 \begin {gather*} \begin {cases} - \frac {5 A d x \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {25 A d x \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {50 A d x \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {50 A d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {25 A d x \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {5 A d x}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {10 A \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {50 A \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {110 A \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {70 A \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} - \frac {16 A}{5 a^{3} d \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 50 a^{3} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a^{3} d \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 5 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (- A \sin {\left (c \right )} + A\right ) \sin ^{2}{\left (c \right )}}{\left (a \sin {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**2*(A-A*sin(d*x+c))/(a+a*sin(d*x+c))**3,x)

[Out]

Piecewise((-5*A*d*x*tan(c/2 + d*x/2)**5/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**
3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 25*A*d*x*ta
n(c/2 + d*x/2)**4/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**
3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 50*A*d*x*tan(c/2 + d*x/2)**3/(5*a
**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2
+ d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 50*A*d*x*tan(c/2 + d*x/2)**2/(5*a**3*d*tan(c/2 + d*x/2)
**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*
d*tan(c/2 + d*x/2) + 5*a**3*d) - 25*A*d*x*tan(c/2 + d*x/2)/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 +
d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a*
*3*d) - 5*A*d*x/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3
+ 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 10*A*tan(c/2 + d*x/2)**4/(5*a**3*d*
tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/
2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 50*A*tan(c/2 + d*x/2)**3/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a
**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2
+ d*x/2) + 5*a**3*d) - 110*A*tan(c/2 + d*x/2)**2/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4
+ 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 70
*A*tan(c/2 + d*x/2)/(5*a**3*d*tan(c/2 + d*x/2)**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)
**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*d*tan(c/2 + d*x/2) + 5*a**3*d) - 16*A/(5*a**3*d*tan(c/2 + d*x/2)
**5 + 25*a**3*d*tan(c/2 + d*x/2)**4 + 50*a**3*d*tan(c/2 + d*x/2)**3 + 50*a**3*d*tan(c/2 + d*x/2)**2 + 25*a**3*
d*tan(c/2 + d*x/2) + 5*a**3*d), Ne(d, 0)), (x*(-A*sin(c) + A)*sin(c)**2/(a*sin(c) + a)**3, True))

________________________________________________________________________________________

Giac [A]
time = 0.51, size = 93, normalized size = 1.04 \begin {gather*} -\frac {\frac {5 \, {\left (d x + c\right )} A}{a^{3}} + \frac {2 \, {\left (5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 25 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 55 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 8 \, A\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{5}}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5*(5*(d*x + c)*A/a^3 + 2*(5*A*tan(1/2*d*x + 1/2*c)^4 + 25*A*tan(1/2*d*x + 1/2*c)^3 + 55*A*tan(1/2*d*x + 1/2
*c)^2 + 35*A*tan(1/2*d*x + 1/2*c) + 8*A)/(a^3*(tan(1/2*d*x + 1/2*c) + 1)^5))/d

________________________________________________________________________________________

Mupad [B]
time = 14.92, size = 178, normalized size = 2.00 \begin {gather*} \frac {\left (5\,A\,\left (c+d\,x\right )-\frac {A\,\left (25\,c+25\,d\,x+10\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (10\,A\,\left (c+d\,x\right )-\frac {A\,\left (50\,c+50\,d\,x+50\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (10\,A\,\left (c+d\,x\right )-\frac {A\,\left (50\,c+50\,d\,x+110\right )}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (5\,A\,\left (c+d\,x\right )-\frac {A\,\left (25\,c+25\,d\,x+70\right )}{5}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+A\,\left (c+d\,x\right )-\frac {A\,\left (5\,c+5\,d\,x+16\right )}{5}}{a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^5}-\frac {A\,x}{a^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^2*(A - A*sin(c + d*x)))/(a + a*sin(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*(5*A*(c + d*x) - (A*(25*c + 25*d*x + 70))/5) + tan(c/2 + (d*x)/2)^4*(5*A*(c + d*x) - (A*(2
5*c + 25*d*x + 10))/5) + tan(c/2 + (d*x)/2)^3*(10*A*(c + d*x) - (A*(50*c + 50*d*x + 50))/5) + tan(c/2 + (d*x)/
2)^2*(10*A*(c + d*x) - (A*(50*c + 50*d*x + 110))/5) + A*(c + d*x) - (A*(5*c + 5*d*x + 16))/5)/(a^3*d*(tan(c/2
+ (d*x)/2) + 1)^5) - (A*x)/a^3

________________________________________________________________________________________