3.306 \(\int \frac {(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^{9/2}} \, dx\)

Optimal. Leaf size=190 \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {3 a^2 \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}} \]

[Out]

1/4*a^2*c*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(11/2)-1/8*a^2*cos(f*x+e)/c/f/(c-c*sin(f*x+e))^(7/2)+1/64*a^2*cos(f*
x+e)/c^2/f/(c-c*sin(f*x+e))^(5/2)+3/256*a^2*cos(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(3/2)+3/512*a^2*arctanh(1/2*cos(
f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(9/2)/f*2^(1/2)

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Rubi [A]  time = 0.30, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2736, 2680, 2650, 2649, 206} \[ \frac {3 a^2 \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {a^2 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(3*a^2*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(256*Sqrt[2]*c^(9/2)*f) + (a^2*c*Co
s[e + f*x]^3)/(4*f*(c - c*Sin[e + f*x])^(11/2)) - (a^2*Cos[e + f*x])/(8*c*f*(c - c*Sin[e + f*x])^(7/2)) + (a^2
*Cos[e + f*x])/(64*c^2*f*(c - c*Sin[e + f*x])^(5/2)) + (3*a^2*Cos[e + f*x])/(256*c^3*f*(c - c*Sin[e + f*x])^(3
/2))

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])

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 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 2680

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 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[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}{(c-c \sin (e+f x))^{9/2}} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {1}{8} \left (3 a^2\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx\\ &=\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}}+\frac {a^2 \int \frac {1}{(c-c \sin (e+f x))^{5/2}} \, dx}{16 c^2}\\ &=\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}}+\frac {a^2 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {\left (3 a^2\right ) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{128 c^3}\\ &=\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}}+\frac {a^2 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{512 c^4}\\ &=\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}}+\frac {a^2 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (3 a^2\right ) \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 )}{256 c^4 f}\\ &=\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{256 \sqrt {2} c^{9/2} f}+\frac {a^2 c \cos ^3(e+f x)}{4 f (c-c \sin (e+f x))^{11/2}}-\frac {a^2 \cos (e+f x)}{8 c f (c-c \sin (e+f x))^{7/2}}+\frac {a^2 \cos (e+f x)}{64 c^2 f (c-c \sin (e+f x))^{5/2}}+\frac {3 a^2 \cos (e+f x)}{256 c^3 f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 1.46, size = 371, normalized size = 1.95 \[ \frac {a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (256 \sin \left (\frac {1}{2} (e+f x)\right )+3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7+6 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6+4 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+8 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-96 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-192 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+128 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(-3-3 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8\right )}{256 f (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^(9/2),x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(128*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 96*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^3 + 4*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7 -
(3 + 3*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2])^8 + 256*Sin[(e + f*x)/2] - 192*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 8*(Cos[(e + f*x)
/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] + 6*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[(e + f*x)/2])*(1 +
Sin[e + f*x])^2)/(256*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(c - c*Sin[e + f*x])^(9/2))

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fricas [B]  time = 0.47, size = 523, normalized size = 2.75 \[ \frac {3 \, \sqrt {2} {\left (a^{2} \cos \left (f x + e\right )^{5} + 5 \, a^{2} \cos \left (f x + e\right )^{4} - 8 \, a^{2} \cos \left (f x + e\right )^{3} - 20 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + 16 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right )^{4} - 4 \, a^{2} \cos \left (f x + e\right )^{3} - 12 \, a^{2} \cos \left (f x + e\right )^{2} + 8 \, a^{2} \cos \left (f x + e\right ) + 16 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {c} \log \left (-\frac {c \cos \left (f x + e\right )^{2} + 2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} \sqrt {c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )} + 3 \, c \cos \left (f x + e\right ) + {\left (c \cos \left (f x + e\right ) - 2 \, c\right )} \sin \left (f x + e\right ) + 2 \, c}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{4} + 13 \, a^{2} \cos \left (f x + e\right )^{3} + 86 \, a^{2} \cos \left (f x + e\right )^{2} - 52 \, a^{2} \cos \left (f x + e\right ) - 128 \, a^{2} - {\left (3 \, a^{2} \cos \left (f x + e\right )^{3} - 10 \, a^{2} \cos \left (f x + e\right )^{2} + 76 \, a^{2} \cos \left (f x + e\right ) + 128 \, a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{1024 \, {\left (c^{5} f \cos \left (f x + e\right )^{5} + 5 \, c^{5} f \cos \left (f x + e\right )^{4} - 8 \, c^{5} f \cos \left (f x + e\right )^{3} - 20 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f - {\left (c^{5} f \cos \left (f x + e\right )^{4} - 4 \, c^{5} f \cos \left (f x + e\right )^{3} - 12 \, c^{5} f \cos \left (f x + e\right )^{2} + 8 \, c^{5} f \cos \left (f x + e\right ) + 16 \, c^{5} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="fricas")

[Out]

1/1024*(3*sqrt(2)*(a^2*cos(f*x + e)^5 + 5*a^2*cos(f*x + e)^4 - 8*a^2*cos(f*x + e)^3 - 20*a^2*cos(f*x + e)^2 +
8*a^2*cos(f*x + e) + 16*a^2 - (a^2*cos(f*x + e)^4 - 4*a^2*cos(f*x + e)^3 - 12*a^2*cos(f*x + e)^2 + 8*a^2*cos(f
*x + e) + 16*a^2)*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*(3*a^2*cos(f*x + e)^4 + 13*a^2*cos(f*x + e)^3 +
86*a^2*cos(f*x + e)^2 - 52*a^2*cos(f*x + e) - 128*a^2 - (3*a^2*cos(f*x + e)^3 - 10*a^2*cos(f*x + e)^2 + 76*a^2
*cos(f*x + e) + 128*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^5*f*cos(f*x + e)^5 + 5*c^5*f*cos(f*x + e)
^4 - 8*c^5*f*cos(f*x + e)^3 - 20*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*f - (c^5*f*cos(f*x + e)^
4 - 4*c^5*f*cos(f*x + e)^3 - 12*c^5*f*cos(f*x + e)^2 + 8*c^5*f*cos(f*x + e) + 16*c^5*f)*sin(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*2*(1/512*(-509*a^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*ta
n((f*x+exp(1))/2)^2+c))^15-1491*a^2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^14-
7411*a^2*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^13-7937*a^2*sqrt(c)*c*(-sqrt(c)*tan(
(f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^12+711*a^2*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+
exp(1))/2)^2+c))^11+22673*a^2*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^10+84
57*a^2*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^9-23829*a^2*sqrt(c)*c^3*(-sqrt(c)*ta
n((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-8391*a^2*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*
x+exp(1))/2)^2+c))^7+18823*a^2*sqrt(c)*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^6-36
57*a^2*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5-10627*a^2*sqrt(c)*c^5*(-sqrt(c)*ta
n((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+5693*a^2*c^6*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*
x+exp(1))/2)^2+c))^3+115*a^2*c^7*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))-2565*a^2*sqrt(
c)*c^6*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-39*a^2*sqrt(c)*c^7)/c^4/(-(-sqrt(c)*ta
n((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+e
xp(1))/2)^2+c))+c)^8/sign(tan((f*x+exp(1))/2)-1)+3/512*a^2*atan((-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c)+sqrt(c*t
an((f*x+exp(1))/2)^2+c))/sqrt(2)/sqrt(-c))/sqrt(2)/c^4/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1))

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maple [A]  time = 1.30, size = 299, normalized size = 1.57 \[ \frac {a^{2} \left (6 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}} c^{\frac {5}{2}}-3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{4}\left (f x +e \right )\right ) c^{6}-44 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{\frac {7}{2}}+12 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{6}-88 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {9}{2}}-18 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{6}+48 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {11}{2}}+12 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{6}-3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{6}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{512 c^{\frac {21}{2}} \left (\sin \left (f x +e \right )-1\right )^{3} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x)

[Out]

1/512/c^(21/2)*a^2*(6*(c*(1+sin(f*x+e)))^(7/2)*c^(5/2)-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/
c^(1/2))*sin(f*x+e)^4*c^6-44*(c*(1+sin(f*x+e)))^(5/2)*c^(7/2)+12*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*
2^(1/2)/c^(1/2))*sin(f*x+e)^3*c^6-88*(c*(1+sin(f*x+e)))^(3/2)*c^(9/2)-18*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^6+48*(c*(1+sin(f*x+e)))^(1/2)*c^(11/2)+12*2^(1/2)*arctanh(1/2*(c*(1+si
n(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)*c^6-3*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2
))*c^6)*(c*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)^3/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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^(9/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^2/(-c*sin(f*x + e) + c)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(9/2),x)

[Out]

int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**(9/2),x)

[Out]

Timed out

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