3.1.33 \(\int \frac {(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx\) [33]

Optimal. Leaf size=112 \[ -\frac {a^2 B x}{c^3}+\frac {a^2 (A+B) c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3}+\frac {2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )} \]

[Out]

-a^2*B*x/c^3+1/5*a^2*(A+B)*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^5-2/3*a^2*B*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^3+2
*a^2*B*cos(f*x+e)/f/(c^3-c^3*sin(f*x+e))

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Rubi [A]
time = 0.19, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3046, 2938, 2759, 8} \begin {gather*} \frac {a^2 c^2 (A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac {2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac {a^2 B x}{c^3}-\frac {2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*B*x)/c^3) + (a^2*(A + B)*c^2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^5) - (2*a^2*B*Cos[e + f*x]^3)/(3
*f*(c - c*Sin[e + f*x])^3) + (2*a^2*B*Cos[e + f*x])/(f*(c^3 - c^3*Sin[e + f*x]))

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2759

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[2*g*(g
*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(2*m + p + 1))), x] + Dist[g^2*((p - 1)/(b^2*(2*m +
p + 1))), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2938

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

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rubi steps

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

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(278\) vs. \(2(112)=224\).
time = 0.48, size = 278, normalized size = 2.48 \begin {gather*} \frac {a^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (12 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-4 (3 A+8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-15 B (e+f x) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-8 (3 A+8 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2 \sin \left (\frac {1}{2} (e+f x)\right )+2 (3 A+43 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^2}{15 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 (c-c \sin (e+f x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(12*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 4*(3*A + 8*B)*(
Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - 15*B*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 24*(A + B)*S
in[(e + f*x)/2] - 8*(3*A + 8*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 2*(3*A + 43*B)*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(15*f*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2])^4*(c - c*Sin[e + f*x])^3)

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Maple [A]
time = 0.40, size = 126, normalized size = 1.12

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {24 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}\right )}{f \,c^{3}}\)	\(126\)
default	\(\frac {2 a^{2} \left (-B \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {A +B}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {16 A +16 B}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {24 A +16 B}{3 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32 A +32 B}{4 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}-\frac {4 A}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}\right )}{f \,c^{3}}\)	\(126\)
risch	\(-\frac {a^{2} B x}{c^{3}}+\frac {-4 A \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}+2 A \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-\frac {100 B \,a^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{3}-24 i B \,a^{2} {\mathrm e}^{3 i \left (f x +e \right )}+\frac {56 i B \,a^{2} {\mathrm e}^{i \left (f x +e \right )}}{3}+10 B \,a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 a^{2} A}{5}+\frac {86 B \,a^{2}}{15}}{\left ({\mathrm e}^{i \left (f x +e \right )}-i\right )^{5} f \,c^{3}}\)	\(139\)
norman	\(\frac {\frac {a^{2} x B}{c}+\frac {8 B \,a^{2} \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {48 B \,a^{2} \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {6 a^{2} A +46 B \,a^{2}}{15 c f}+\frac {40 B \,a^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{3 c f}+\frac {64 B \,a^{2} \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}+\frac {112 B \,a^{2} \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {\left (2 a^{2} A +2 B \,a^{2}\right ) \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c f}-\frac {\left (30 a^{2} A +86 B \,a^{2}\right ) \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3 c f}-\frac {2 \left (38 a^{2} A +198 B \,a^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{5 c f}-\frac {\left (78 a^{2} A +478 B \,a^{2}\right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}-\frac {2 \left (138 a^{2} A +578 B \,a^{2}\right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{15 c f}-\frac {5 a^{2} x B \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{c}+\frac {13 a^{2} x B \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {25 a^{2} x B \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {38 a^{2} x B \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {46 a^{2} x B \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {46 a^{2} x B \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {38 a^{2} x B \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {25 a^{2} x B \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {13 a^{2} x B \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}+\frac {5 a^{2} x B \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}-\frac {a^{2} x B \left (\tan ^{11}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c}}{\left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{3} c^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}\)	\(566\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f*a^2/c^3*(-B*arctan(tan(1/2*f*x+1/2*e))-(A+B)/(tan(1/2*f*x+1/2*e)-1)-1/5*(16*A+16*B)/(tan(1/2*f*x+1/2*e)-1)
^5-1/3*(24*A+16*B)/(tan(1/2*f*x+1/2*e)-1)^3-1/4*(32*A+32*B)/(tan(1/2*f*x+1/2*e)-1)^4-4*A/(tan(1/2*f*x+1/2*e)-1
)^2)

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1237 vs. \(2 (117) = 234\).
time = 0.53, size = 1237, normalized size = 11.04 \begin {gather*} -\frac {2 \, {\left (B a^{2} {\left (\frac {\frac {95 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {145 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {75 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 22}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {15 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{3}}\right )} + \frac {A a^{2} {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 7\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {6 \, A a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {3 \, B a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {2 \, A a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {4 \, B a^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 1\right )}}{c^{3} - \frac {5 \, c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {10 \, c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(B*a^2*((95*sin(f*x + e)/(cos(f*x + e) + 1) - 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^
3/(cos(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 22)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e)
+ 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e)
+ 1))/c^3) + A*a^2*(20*sin(f*x + e)/(cos(f*x + e) + 1) - 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 - 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 7)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x +
e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin
(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 6*A*a^2*(5*sin(f*x + e)/(cos(f*x
+ e) + 1) - 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 1)/(c^3 - 5*c^3*s
in(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*a^2*(5
*sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin
(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e
) + 1)^5) + 2*A*a^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c^3 - 5*
c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(
f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 4*B*a
^2*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c^3 - 5*c^3*sin(f*x + e)/
(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 +
5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (117) = 234\).
time = 0.37, size = 289, normalized size = 2.58 \begin {gather*} \frac {60 \, B a^{2} f x - {\left (15 \, B a^{2} f x - {\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \, {\left (A + B\right )} a^{2} - {\left (45 \, B a^{2} f x - {\left (9 \, A - 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B a^{2} f x - {\left (A + 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) - {\left (60 \, B a^{2} f x + 12 \, {\left (A + B\right )} a^{2} - {\left (15 \, B a^{2} f x + {\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, B a^{2} f x + {\left (A - 9 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f - {\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(60*B*a^2*f*x - (15*B*a^2*f*x - (3*A + 43*B)*a^2)*cos(f*x + e)^3 - 12*(A + B)*a^2 - (45*B*a^2*f*x - (9*A
- 11*B)*a^2)*cos(f*x + e)^2 + 6*(5*B*a^2*f*x - (A + 11*B)*a^2)*cos(f*x + e) - (60*B*a^2*f*x + 12*(A + B)*a^2 -
(15*B*a^2*f*x + (3*A + 43*B)*a^2)*cos(f*x + e)^2 + 6*(5*B*a^2*f*x + (A - 9*B)*a^2)*cos(f*x + e))*sin(f*x + e)
)/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*
c^3*f*cos(f*x + e) - 4*c^3*f)*sin(f*x + e))

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1647 vs. \(2 (102) = 204\).
time = 9.58, size = 1647, normalized size = 14.71 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**3,x)

[Out]

Piecewise((-30*A*a**2*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150
*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 60*A*
a**2*tan(e/2 + f*x/2)**2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 +
f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 6*A*a**2/(15*c**3*f*ta
n(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/
2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 15*B*a**2*f*x*tan(e/2 + f*x/2)**5/(15*c**3*f*tan(e/2 + f*x/2
)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c*
*3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 75*B*a**2*f*x*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**
3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2
+ f*x/2) - 15*c**3*f) - 150*B*a**2*f*x*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2
+ f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 1
5*c**3*f) + 150*B*a**2*f*x*tan(e/2 + f*x/2)**2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4
+ 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) -
75*B*a**2*f*x*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan
(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 15*B*a**2*f*x/(1
5*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan
(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 30*B*a**2*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2
+ f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2
+ 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 120*B*a**2*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75
*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(
e/2 + f*x/2) - 15*c**3*f) - 340*B*a**2*tan(e/2 + f*x/2)**2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2
+ f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 1
5*c**3*f) + 200*B*a**2*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c
**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 46*B*a*
*2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*
f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f), Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)**
2/(-c*sin(e) + c)**3, True))

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Giac [A]
time = 0.48, size = 159, normalized size = 1.42 \begin {gather*} -\frac {\frac {15 \, {\left (f x + e\right )} B a^{2}}{c^{3}} + \frac {2 \, {\left (15 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 60 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, A a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 170 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 100 \, B a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, A a^{2} + 23 \, B a^{2}\right )}}{c^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{5}}}{15 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(15*(f*x + e)*B*a^2/c^3 + 2*(15*A*a^2*tan(1/2*f*x + 1/2*e)^4 + 15*B*a^2*tan(1/2*f*x + 1/2*e)^4 - 60*B*a^
2*tan(1/2*f*x + 1/2*e)^3 + 30*A*a^2*tan(1/2*f*x + 1/2*e)^2 + 170*B*a^2*tan(1/2*f*x + 1/2*e)^2 - 100*B*a^2*tan(
1/2*f*x + 1/2*e) + 3*A*a^2 + 23*B*a^2)/(c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f

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Mupad [B]
time = 14.68, size = 233, normalized size = 2.08 \begin {gather*} -\frac {B\,a^2\,x}{c^3}-\frac {\frac {a^2\,\left (6\,A+46\,B-15\,B\,\left (e+f\,x\right )\right )}{15}-{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {a^2\,\left (120\,B-150\,B\,\left (e+f\,x\right )\right )}{15}+10\,B\,a^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (\frac {a^2\,\left (30\,A+30\,B-75\,B\,\left (e+f\,x\right )\right )}{15}+5\,B\,a^2\,\left (e+f\,x\right )\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (60\,A+340\,B-150\,B\,\left (e+f\,x\right )\right )}{15}+10\,B\,a^2\,\left (e+f\,x\right )\right )-\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {a^2\,\left (200\,B-75\,B\,\left (e+f\,x\right )\right )}{15}+5\,B\,a^2\,\left (e+f\,x\right )\right )+B\,a^2\,\left (e+f\,x\right )}{c^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )-1\right )}^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (B*a^2*x)/c^3 - ((a^2*(6*A + 46*B - 15*B*(e + f*x)))/15 - tan(e/2 + (f*x)/2)^3*((a^2*(120*B - 150*B*(e + f*x
)))/15 + 10*B*a^2*(e + f*x)) + tan(e/2 + (f*x)/2)^4*((a^2*(30*A + 30*B - 75*B*(e + f*x)))/15 + 5*B*a^2*(e + f*
x)) + tan(e/2 + (f*x)/2)^2*((a^2*(60*A + 340*B - 150*B*(e + f*x)))/15 + 10*B*a^2*(e + f*x)) - tan(e/2 + (f*x)/
2)*((a^2*(200*B - 75*B*(e + f*x)))/15 + 5*B*a^2*(e + f*x)) + B*a^2*(e + f*x))/(c^3*f*(tan(e/2 + (f*x)/2) - 1)^
5)

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