3.122 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=225 \[ -\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{5 (7 A-B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}+\frac{5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}} \]

[Out]

(5*(7*A - B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) +
(5*(7*A - B)*Cos[e + f*x])/(64*a^2*c*f*(c - c*Sin[e + f*x])^(3/2)) + ((7*A - B)*Sec[e + f*x])/(24*a^2*c*f*(c -
c*Sin[e + f*x])^(3/2)) - (5*(7*A - B)*Sec[e + f*x])/(48*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((A - B)*Sec[e
+ f*x]^3)/(3*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.483725, antiderivative size = 225, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used = {2967, 2855, 2681, 2687, 2650, 2649, 206} \[ -\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{5 (7 A-B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}+\frac{5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*(7*A - B)*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) +
(5*(7*A - B)*Cos[e + f*x])/(64*a^2*c*f*(c - c*Sin[e + f*x])^(3/2)) + ((7*A - B)*Sec[e + f*x])/(24*a^2*c*f*(c -
c*Sin[e + f*x])^(3/2)) - (5*(7*A - B)*Sec[e + f*x])/(48*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]]) - ((A - B)*Sec[e
+ f*x]^3)/(3*a^2*c^2*f*Sqrt[c - c*Sin[e + f*x]])

Rule 2967

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]))

Rule 2855

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*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2681

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

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2650

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 2649

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 206

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

Rubi steps

\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx &=\frac{\int \frac{\sec ^4(e+f x) (A+B \sin (e+f x))}{\sqrt{c-c \sin (e+f x)}} \, dx}{a^2 c^2}\\ &=-\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{(7 A-B) \int \frac{\sec ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx}{6 a^2 c}\\ &=\frac{(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{(5 (7 A-B)) \int \frac{\sec ^2(e+f x)}{\sqrt{c-c \sin (e+f x)}} \, dx}{48 a^2 c^2}\\ &=\frac{(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{(5 (7 A-B)) \int \frac{1}{(c-c \sin (e+f x))^{3/2}} \, dx}{32 a^2 c}\\ &=\frac{5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}+\frac{(5 (7 A-B)) \int \frac{1}{\sqrt{c-c \sin (e+f x)}} \, dx}{128 a^2 c^2}\\ &=\frac{5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(5 (7 A-B)) \operatorname{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 )}{64 a^2 c^2 f}\\ &=\frac{5 (7 A-B) \tanh ^{-1}\left (\frac{\sqrt{c} \cos (e+f x)}{\sqrt{2} \sqrt{c-c \sin (e+f x)}}\right )}{64 \sqrt{2} a^2 c^{5/2} f}+\frac{5 (7 A-B) \cos (e+f x)}{64 a^2 c f (c-c \sin (e+f x))^{3/2}}+\frac{(7 A-B) \sec (e+f x)}{24 a^2 c f (c-c \sin (e+f x))^{3/2}}-\frac{5 (7 A-B) \sec (e+f x)}{48 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}-\frac{(A-B) \sec ^3(e+f x)}{3 a^2 c^2 f \sqrt{c-c \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 1.46215, size = 430, normalized size = 1.91 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (3 (11 A+3 B) \cos ^3(e+f x)+24 (B-3 A) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4+16 (B-A) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4+6 (11 A+3 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+12 (A+B) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )+24 (A+B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+(-15-15 i) \sqrt [4]{-1} (7 A-B) \tan ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac{1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4\right )}{192 a^2 f (\sin (e+f x)+1)^2 (c-c \sin (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(3*(11*A + 3*B)*Cos[e + f*x]^3 +
16*(-A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4 + 24*(-3*A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 12*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2])^3 - (15 + 15*I)*(-1)^(1/4)*(7*A - B)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(C
os[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 24*(A + B)*Sin[(e + f*x)/2]*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 6*(11*A + 3*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2
]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3))/(192*a^2*f*(1 + Sin[e + f*x])^2*(c - c*Sin[e + f*x])^(5/2))

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Maple [B]  time = 1.369, size = 426, normalized size = 1.9 \begin{align*} -{\frac{1}{384\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( -210\,A{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}+30\,B{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}+70\,A{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}-10\,B{c}^{7/2} \left ( \sin \left ( fx+e \right ) \right ) ^{2}+322\,A{c}^{7/2}\sin \left ( fx+e \right ) -46\,B{c}^{7/2}\sin \left ( fx+e \right ) +105\,A \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{2}-15\,B \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \left ( \sin \left ( fx+e \right ) \right ) ^{2}{c}^{2}-86\,A{c}^{7/2}+122\,B{c}^{7/2}+105\,A \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}-15\,B \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ){c}^{2}-210\,A \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{2}+30\,B \left ( c \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) ^{3/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c \left ( 1+\sin \left ( fx+e \right ) \right ) }\sqrt{2}}{\sqrt{c}}} \right ) \sin \left ( fx+e \right ){c}^{2} \right ){c}^{-{\frac{11}{2}}}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/384/c^(11/2)/a^2*(-210*A*c^(7/2)*sin(f*x+e)^3+30*B*c^(7/2)*sin(f*x+e)^3+70*A*c^(7/2)*sin(f*x+e)^2-10*B*c^(7
/2)*sin(f*x+e)^2+322*A*c^(7/2)*sin(f*x+e)-46*B*c^(7/2)*sin(f*x+e)+105*A*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arcta
nh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2-15*B*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arctan
h(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^2*c^2-86*A*c^(7/2)+122*B*c^(7/2)+105*A*(c*(1+sin(f*
x+e)))^(3/2)*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2-15*B*(c*(1+sin(f*x+e)))^(3/2)*2
^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c^2-210*A*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arctan
h(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^2+30*B*(c*(1+sin(f*x+e)))^(3/2)*2^(1/2)*arctanh(1
/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^2)/(1+sin(f*x+e))/(-1+sin(f*x+e))/cos(f*x+e)/(c-c*si
n(f*x+e))^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]  time = 1.85341, size = 717, normalized size = 3.19 \begin{align*} -\frac{15 \, \sqrt{2}{\left ({\left (7 \, A - B\right )} \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) -{\left (7 \, A - B\right )} \cos \left (f x + e\right )^{3}\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 (5 \,{\left (7 \, A - B\right )} \cos \left (f x + e\right )^{2} -{\left (15 \,{\left (7 \, A - B\right )} \cos \left (f x + e\right )^{2} + 56 \, A - 8 \, B\right )} \sin \left (f x + e\right ) + 8 \, A - 56 \, B\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{768 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/768*(15*sqrt(2)*((7*A - B)*cos(f*x + e)^3*sin(f*x + e) - (7*A - B)*cos(f*x + e)^3)*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*(5*(7*A - B)*cos(f*x + e)^2 - (15*(7*A - B)*cos(f*x + e)^2 + 56*A - 8*B)*sin(f*x + e) + 8*A - 56*B)*sqrt
(-c*sin(f*x + e) + c))/(a^2*c^3*f*cos(f*x + e)^3*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 3.83316, size = 1805, normalized size = 8.02 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(15*sqrt(2)*(7*A - B)*arctan(-1/2*sqrt(2)*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2
+ c) - sqrt(c))/sqrt(-c))/(a^2*sqrt(-c)*c^2*sgn(tan(1/2*f*x + 1/2*e) - 1)) + 16*(15*(sqrt(c)*tan(1/2*f*x + 1/2
*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*A - 9*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^
2 + c))^5*B + 33*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*A*sqrt(c) - 15*(sqrt(c)
*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*B*sqrt(c) - 22*(sqrt(c)*tan(1/2*f*x + 1/2*e) - s
qrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*A*c + 10*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 +
c))^3*B*c - 66*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*A*c^(3/2) + 30*(sqrt(c)*t
an(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*B*c^(3/2) + 51*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqr
t(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^2 - 21*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c)
)*B*c^2 - 11*A*c^(5/2) + 5*B*c^(5/2))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2
+ 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^3*a^2*c^2*sgn(tan(1/2*f*x
+ 1/2*e) - 1)) + 6*(53*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*A + 29*(sqrt(c)*
tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*B - 179*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*ta
n(1/2*f*x + 1/2*e)^2 + c))^6*A*sqrt(c) - 75*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c)
)^6*B*sqrt(c) + 127*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*A*c + 55*(sqrt(c)*ta
n(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*B*c + 195*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*ta
n(1/2*f*x + 1/2*e)^2 + c))^4*A*c^(3/2) + 91*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c)
)^4*B*c^(3/2) + 7*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*A*c^2 - (sqrt(c)*tan(1
/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B*c^2 - 121*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan
(1/2*f*x + 1/2*e)^2 + c))^2*A*c^(5/2) - 65*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))
^2*B*c^(5/2) - 67*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^3 - 27*(sqrt(c)*tan(
1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B*c^3 - 15*A*c^(7/2) - 7*B*c^(7/2))/(((sqrt(c)*tan(1/2*
f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2 - 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x +
1/2*e)^2 + c))*sqrt(c) - c)^4*a^2*c^2*sgn(tan(1/2*f*x + 1/2*e) - 1)))/f