3.2.4 \(\int \frac {(a+a \sin (e+f x))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx\) [104]

Optimal. Leaf size=225 \[ \frac {5 a^3 (3 A+11 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}} \]

[Out]

1/4*a^3*(A+B)*c^3*cos(f*x+e)^7/f/(c-c*sin(f*x+e))^(11/2)-1/8*a^3*(3*A+11*B)*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^
(7/2)-5/24*a^3*(3*A+11*B)*cos(f*x+e)^3/c/f/(c-c*sin(f*x+e))^(3/2)+5/4*a^3*(3*A+11*B)*arctanh(1/2*cos(f*x+e)*c^
(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(5/2)/f*2^(1/2)-5/4*a^3*(3*A+11*B)*cos(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(
1/2)

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Rubi [A]
time = 0.38, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3046, 2938, 2759, 2758, 2728, 212} \begin {gather*} \frac {5 a^3 (3 A+11 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {a^3 c^3 (A+B) \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {a^3 c (3 A+11 B) \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a^3*(3*A + 11*B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(2*Sqrt[2]*c^(5/2)*f)
+ (a^3*(A + B)*c^3*Cos[e + f*x]^7)/(4*f*(c - c*Sin[e + f*x])^(11/2)) - (a^3*(3*A + 11*B)*c*Cos[e + f*x]^5)/(8*
f*(c - c*Sin[e + f*x])^(7/2)) - (5*a^3*(3*A + 11*B)*Cos[e + f*x]^3)/(24*c*f*(c - c*Sin[e + f*x])^(3/2)) - (5*a
^3*(3*A + 11*B)*Cos[e + f*x])/(4*c^2*f*Sqrt[c - c*Sin[e + f*x]])

Rule 212

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

Rule 2728

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

Rule 2758

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

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))^3 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{5/2}} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{11/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {1}{8} \left (a^3 (3 A+11 B) c^2\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}+\frac {1}{16} \left (5 a^3 (3 A+11 B)\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{5/2}} \, dx\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}+\frac {\left (5 a^3 (3 A+11 B)\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{8 c}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}+\frac {\left (5 a^3 (3 A+11 B)\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{4 c^2}\\ &=\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}-\frac {\left (5 a^3 (3 A+11 B)\right ) \text {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{2 c^2 f}\\ &=\frac {5 a^3 (3 A+11 B) \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{2 \sqrt {2} c^{5/2} f}+\frac {a^3 (A+B) c^3 \cos ^7(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^3 (3 A+11 B) c \cos ^5(e+f x)}{8 f (c-c \sin (e+f x))^{7/2}}-\frac {5 a^3 (3 A+11 B) \cos ^3(e+f x)}{24 c f (c-c \sin (e+f x))^{3/2}}-\frac {5 a^3 (3 A+11 B) \cos (e+f x)}{4 c^2 f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.49, size = 434, normalized size = 1.93 \begin {gather*} \frac {a^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3 \left (12 (A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-3 (9 A+17 B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-(15+15 i) \sqrt [4]{-1} (3 A+11 B) \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1+\tan \left (\frac {1}{4} (e+f x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-6 (2 A+11 B) \cos \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+2 B \cos \left (\frac {3}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+24 (A+B) \sin \left (\frac {1}{2} (e+f x)\right )-6 (9 A+17 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 )-6 (2 A+11 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 )-2 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 {3}{2} (e+f x)\right )\right )}{6 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 (c-c \sin (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(12*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2]) - 3*(9*A + 17*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - (15 + 15*I)*(-1)^(1/4)*(3*A + 11*B)*ArcTan[(1/2
+ I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 - 6*(2*A + 11*B)*Cos[(e + f
*x)/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 2*B*Cos[(3*(e + f*x))/2]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]
)^4 + 24*(A + B)*Sin[(e + f*x)/2] - 6*(9*A + 17*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] -
6*(2*A + 11*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] - 2*B*(Cos[(e + f*x)/2] - Sin[(e + f*x
)/2])^4*Sin[(3*(e + f*x))/2]))/(6*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(433\) vs. \(2(198)=396\).
time = 9.44, size = 434, normalized size = 1.93

method	result	size
default	\(-\frac {a^{3} \left (\sin \left (f x +e \right ) \left (-90 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+48 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-330 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+16 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}+240 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}\right )+\left (-45 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+24 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}-165 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+8 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}+120 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+90 A \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+54 A \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}-132 A \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}+330 B \sqrt {2}\, \arctanh \left (\frac {\sqrt {c +c \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}+86 B \left (c +c \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {c}-420 B \sqrt {c +c \sin \left (f x +e \right )}\, c^{\frac {3}{2}}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{12 c^{\frac {9}{2}} \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(434\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/12/c^(9/2)*a^3*(sin(f*x+e)*(-90*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+48*A*(c+c
*sin(f*x+e))^(1/2)*c^(3/2)-330*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+16*B*(c+c*sin
(f*x+e))^(3/2)*c^(1/2)+240*B*(c+c*sin(f*x+e))^(1/2)*c^(3/2))+(-45*A*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)
*2^(1/2)/c^(1/2))*c^2+24*A*(c+c*sin(f*x+e))^(1/2)*c^(3/2)-165*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(
1/2)/c^(1/2))*c^2+8*B*(c+c*sin(f*x+e))^(3/2)*c^(1/2)+120*B*(c+c*sin(f*x+e))^(1/2)*c^(3/2))*cos(f*x+e)^2+90*A*2
^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+54*A*(c+c*sin(f*x+e))^(3/2)*c^(1/2)-132*A*(c+c*
sin(f*x+e))^(1/2)*c^(3/2)+330*B*2^(1/2)*arctanh(1/2*(c+c*sin(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^2+86*B*(c+c*sin(
f*x+e))^(3/2)*c^(1/2)-420*B*(c+c*sin(f*x+e))^(1/2)*c^(3/2))*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)/cos(f*x+e)
/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 537 vs. \(2 (208) = 416\).
time = 0.38, size = 537, normalized size = 2.39 \begin {gather*} \frac {15 \, \sqrt {2} {\left ({\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right ) - 4 \, {\left (3 \, A + 11 \, B\right )} a^{3} - {\left ({\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} - 2 \, {\left (3 \, A + 11 \, B\right )} a^{3} \cos \left (f x + e\right ) - 4 \, {\left (3 \, A + 11 \, B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left (4 \, B a^{3} \cos \left (f x + e\right )^{4} - 4 \, {\left (3 \, A + 14 \, B\right )} a^{3} \cos \left (f x + e\right )^{3} + 3 \, {\left (13 \, A + 37 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (13 \, A + 53 \, B\right )} a^{3} \cos \left (f x + e\right ) - 12 \, {\left (A + B\right )} a^{3} - {\left (4 \, B a^{3} \cos \left (f x + e\right )^{3} + 12 \, {\left (A + 5 \, B\right )} a^{3} \cos \left (f x + e\right )^{2} + 3 \, {\left (17 \, A + 57 \, B\right )} a^{3} \cos \left (f x + e\right ) + 12 \, {\left (A + B\right )} a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{24 \, {\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))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/24*(15*sqrt(2)*((3*A + 11*B)*a^3*cos(f*x + e)^3 + 3*(3*A + 11*B)*a^3*cos(f*x + e)^2 - 2*(3*A + 11*B)*a^3*cos
(f*x + e) - 4*(3*A + 11*B)*a^3 - ((3*A + 11*B)*a^3*cos(f*x + e)^2 - 2*(3*A + 11*B)*a^3*cos(f*x + e) - 4*(3*A +
11*B)*a^3)*sin(f*x + e))*sqrt(c)*log(-(c*cos(f*x + e)^2 + 2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*sqrt(c)*(cos(f*
x + e) + sin(f*x + e) + 1) + 3*c*cos(f*x + e) + (c*cos(f*x + e) - 2*c)*sin(f*x + e) + 2*c)/(cos(f*x + e)^2 + (
cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(4*B*a^3*cos(f*x + e)^4 - 4*(3*A + 14*B)*a^3*cos(f*x +
e)^3 + 3*(13*A + 37*B)*a^3*cos(f*x + e)^2 + 3*(13*A + 53*B)*a^3*cos(f*x + e) - 12*(A + B)*a^3 - (4*B*a^3*cos(
f*x + e)^3 + 12*(A + 5*B)*a^3*cos(f*x + e)^2 + 3*(17*A + 57*B)*a^3*cos(f*x + e) + 12*(A + B)*a^3)*sin(f*x + e)
)*sqrt(-c*sin(f*x + e) + c))/(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 [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 815 vs. \(2 (208) = 416\).
time = 0.63, size = 815, normalized size = 3.62 \begin {gather*} \frac {\frac {60 \, \sqrt {2} {\left (3 \, A a^{3} \sqrt {c} + 11 \, B a^{3} \sqrt {c}\right )} \log \left (-\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1}\right )}{c^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} - \frac {3 \, \sqrt {2} {\left (A a^{3} \sqrt {c} + B a^{3} \sqrt {c} + \frac {16 \, A a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {32 \, B a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {90 \, A a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {330 \, B a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}{c^{3} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} + \frac {3 \, {\left (\frac {16 \, \sqrt {2} A a^{3} c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {32 \, \sqrt {2} B a^{3} c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {\sqrt {2} A a^{3} c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {\sqrt {2} B a^{3} c^{\frac {7}{2}} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{c^{6}} - \frac {128 \, \sqrt {2} {\left (3 \, A a^{3} \sqrt {c} + 17 \, B a^{3} \sqrt {c} - \frac {6 \, A a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - \frac {30 \, B a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {3 \, A a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {21 \, B a^{3} \sqrt {c} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2}}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{c^{3} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{96 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/96*(60*sqrt(2)*(3*A*a^3*sqrt(c) + 11*B*a^3*sqrt(c))*log(-(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi +
1/2*f*x + 1/2*e) + 1))/(c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(A*a^3*sqrt(c) + B*a^3*sqrt(c) +
16*A*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 32*B*a^3*sqrt(c)
*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 90*A*a^3*sqrt(c)*(cos(-1/4*pi + 1
/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 330*B*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2
*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2/(c^3*(cos(-1/4*pi +
1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 3*(16*sqrt(2)*A*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*
f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 32*sqrt(2)*B*a^3*
c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*
e) + 1) + sqrt(2)*A*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(co
s(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + sqrt(2)*B*a^3*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/
4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/c^6 - 128*sqrt(2)*(3*A*a^3*sqrt(c) + 17*B*a^3
*sqrt(c) - 6*A*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 30*B*a^
3*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 3*A*a^3*sqrt(c)*(cos(-1/
4*pi + 1/2*f*x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 + 21*B*a^3*sqrt(c)*(cos(-1/4*pi + 1/2*f*
x + 1/2*e) - 1)^2/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(c^3*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4
*pi + 1/2*f*x + 1/2*e) + 1) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \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))^3)/(c - c*sin(e + f*x))^(5/2),x)

[Out]

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

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