3.2.22 \(\int \frac {A+B \sin (e+f x)}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \, dx\) [122]

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

[Out]

5/64*(7*A-B)*cos(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)+1/24*(7*A-B)*sec(f*x+e)/a^2/c/f/(c-c*sin(f*x+e))^(3/2)+
5/128*(7*A-B)*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/a^2/c^(5/2)/f*2^(1/2)-5/48*(7*A-B
)*sec(f*x+e)/a^2/c^2/f/(c-c*sin(f*x+e))^(1/2)-1/3*(A-B)*sec(f*x+e)^3/a^2/c^2/f/(c-c*sin(f*x+e))^(1/2)

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Rubi [A]
time = 0.33, antiderivative size = 225, normalized size of antiderivative = 1.00, 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 = {3046, 2934, 2760, 2766, 2729, 2728, 212} \begin {gather*} \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 {(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) \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}} \end {gather*}

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 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 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 2760

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2766

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))), In
t[(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 2934

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 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+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)) \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 )}{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*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.96, size = 430, normalized size = 1.91 \begin {gather*} \frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (3 (11 A+3 B) \cos ^3(e+f x)+16 (-A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4+24 (-3 A+B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \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 (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \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} (7 A-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 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+24 (A+B) \sin \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 )^3+6 (11 A+3 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 ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3\right )}{192 a^2 f (1+\sin (e+f x))^2 (c-c \sin (e+f x))^{5/2}} \end {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. \(425\) vs. \(2(198)=396\).
time = 8.99, size = 426, normalized size = 1.89

method	result	size
default	\(-\frac {30 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{2}+105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}-15 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{2}-210 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{2}-210 A \,c^{\frac {7}{2}} \left (\sin ^{3}\left (f x +e \right )\right )+30 B \,c^{\frac {7}{2}} \left (\sin ^{3}\left (f x +e \right )\right )+70 A \,c^{\frac {7}{2}} \left (\sin ^{2}\left (f x +e \right )\right )-10 B \,c^{\frac {7}{2}} \left (\sin ^{2}\left (f x +e \right )\right )+322 A \,c^{\frac {7}{2}} \sin \left (f x +e \right )-46 B \,c^{\frac {7}{2}} \sin \left (f x +e \right )+105 A \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-15 B \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{2}-86 A \,c^{\frac {7}{2}}+122 B \,c^{\frac {7}{2}}}{384 c^{\frac {11}{2}} a^{2} \left (1+\sin \left (f x +e \right )\right ) \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}\)	\(426\)

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,method=_RETURNVERBOSE)

[Out]

-1/384/c^(11/2)/a^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+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))
*sin(f*x+e)^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))*
sin(f*x+e)^2*c^2-210*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))*
sin(f*x+e)*c^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)*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^2-86*A*c^(7/2)+122*B*c^(7/2))/(1+sin(f*x+e))/(sin(f*x+e)-1)/cos(f*x+e)/(c-c*sin(
f*x+e))^(1/2)/f

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Maxima [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+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 = 0.37, size = 300, normalized size = 1.33 \begin {gather*} -\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 {gather*}

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)] 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+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] Leaf count of result is larger than twice the leaf count of optimal. 787 vs. \(2 (208) = 416\).
time = 0.67, size = 787, normalized size = 3.50 \begin {gather*} \frac {\frac {60 \, \sqrt {2} {\left (7 \, A \sqrt {c} - B \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 )}{a^{2} 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 \sqrt {c} + B \sqrt {c} - \frac {24 \, A \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 {8 \, B \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 {210 \, A \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 {30 \, B \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}}{a^{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 {256 \, \sqrt {2} {\left (5 \, A \sqrt {c} - 2 \, B \sqrt {c} + \frac {9 \, A \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 \, B \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 {6 \, A \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 {3 \, B \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 )}}{a^{2} 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 )} - \frac {3 \, {\left (\frac {24 \, \sqrt {2} A a^{2} 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 {8 \, \sqrt {2} B a^{2} 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^{2} 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^{2} 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 )}}{a^{4} c^{6}}}{3072 \, f} \end {gather*}

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]

1/3072*(60*sqrt(2)*(7*A*sqrt(c) - B*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))/(a^2*c^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))) - 3*sqrt(2)*(A*sqrt(c) + B*sqrt(c) - 24*A*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) - 8*B*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) + 210*A*sqrt(c)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2/(c
os(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2 - 30*B*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/(a^2*c^3*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sg
n(sin(-1/4*pi + 1/2*f*x + 1/2*e))) + 256*sqrt(2)*(5*A*sqrt(c) - 2*B*sqrt(c) + 9*A*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*B*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) + 6*A*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 - 3*B*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)/(a^
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))) - 3*(24*sqrt(2)*A*a^2*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) + 8*sqrt(2)*B*a^2*c^(7/2)*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sg
n(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^2*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 - sqrt(2)*B
*a^2*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)/(a^4*c^6))/f

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

[Out]

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

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