3.316 \(\int \frac {(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^{11/2}} \, dx\)

Optimal. Leaf size=225 \[ -\frac {3 a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{512 \sqrt {2} c^{11/2} f}-\frac {3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}+\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}} \]

[Out]

1/5*a^3*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^(15/2)-1/8*a^3*cos(f*x+e)^3/f/(c-c*sin(f*x+e))^(11/2)+1/16*a^3*cos
(f*x+e)/c^2/f/(c-c*sin(f*x+e))^(7/2)-1/128*a^3*cos(f*x+e)/c^3/f/(c-c*sin(f*x+e))^(5/2)-3/512*a^3*cos(f*x+e)/c^
4/f/(c-c*sin(f*x+e))^(3/2)-3/1024*a^3*arctanh(1/2*cos(f*x+e)*c^(1/2)*2^(1/2)/(c-c*sin(f*x+e))^(1/2))/c^(11/2)/
f*2^(1/2)

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Rubi [A]  time = 0.39, antiderivative size = 225, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}-\frac {a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {3 a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{512 \sqrt {2} c^{11/2} f}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(-3*a^3*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(512*Sqrt[2]*c^(11/2)*f) + (a^3*c^
2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^(15/2)) - (a^3*Cos[e + f*x]^3)/(8*f*(c - c*Sin[e + f*x])^(11/2)) +
(a^3*Cos[e + f*x])/(16*c^2*f*(c - c*Sin[e + f*x])^(7/2)) - (a^3*Cos[e + f*x])/(128*c^3*f*(c - c*Sin[e + f*x])
^(5/2)) - (3*a^3*Cos[e + f*x])/(512*c^4*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))^3}{(c-c \sin (e+f x))^{11/2}} \, dx &=\left (a^3 c^3\right ) \int \frac {\cos ^6(e+f x)}{(c-c \sin (e+f x))^{17/2}} \, dx\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {1}{2} \left (a^3 c\right ) \int \frac {\cos ^4(e+f x)}{(c-c \sin (e+f x))^{13/2}} \, dx\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac {\left (3 a^3\right ) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{9/2}} \, dx}{16 c}\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {a^3 \int \frac {1}{(c-c \sin (e+f x))^{5/2}} \, dx}{32 c^3}\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac {\left (3 a^3\right ) \int \frac {1}{(c-c \sin (e+f x))^{3/2}} \, dx}{256 c^4}\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}-\frac {\left (3 a^3\right ) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{1024 c^5}\\ &=\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}+\frac {\left (3 a^3\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 )}{512 c^5 f}\\ &=-\frac {3 a^3 \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{512 \sqrt {2} c^{11/2} f}+\frac {a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^{15/2}}-\frac {a^3 \cos ^3(e+f x)}{8 f (c-c \sin (e+f x))^{11/2}}+\frac {a^3 \cos (e+f x)}{16 c^2 f (c-c \sin (e+f x))^{7/2}}-\frac {a^3 \cos (e+f x)}{128 c^3 f (c-c \sin (e+f x))^{5/2}}-\frac {3 a^3 \cos (e+f x)}{512 c^4 f (c-c \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 4.11, size = 435, normalized size = 1.93 \[ \frac {a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (4096 \sin \left (\frac {1}{2} (e+f x)\right )-15 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9-30 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^8-20 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^7-40 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^6+992 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^5+1984 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^4-2688 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^3-5376 \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2+2048 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+(15+15 i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^{10}\right )}{2560 f (c-c \sin (e+f x))^{11/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^(11/2),x]

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(2048*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 2688*(Cos[(e + f*x)/2
] - Sin[(e + f*x)/2])^3 + 992*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 - 20*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2
])^7 - 15*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9 + (15 + 15*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])^10 + 4096*Sin[(e + f*x)/2] - 5376*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 1984*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2] - 40*(Co
s[(e + f*x)/2] - Sin[(e + f*x)/2])^6*Sin[(e + f*x)/2] - 30*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*Sin[(e + f*
x)/2])*(1 + Sin[e + f*x])^3)/(2560*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^(11/2))

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fricas [B]  time = 0.51, size = 600, normalized size = 2.67 \[ \frac {15 \, \sqrt {2} {\left (a^{3} \cos \left (f x + e\right )^{6} - 5 \, a^{3} \cos \left (f x + e\right )^{5} - 18 \, a^{3} \cos \left (f x + e\right )^{4} + 20 \, a^{3} \cos \left (f x + e\right )^{3} + 48 \, a^{3} \cos \left (f x + e\right )^{2} - 16 \, a^{3} \cos \left (f x + e\right ) - 32 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{5} + 6 \, a^{3} \cos \left (f x + e\right )^{4} - 12 \, a^{3} \cos \left (f x + e\right )^{3} - 32 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} \cos \left (f x + e\right ) + 32 \, 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 (15 \, a^{3} \cos \left (f x + e\right )^{5} - 65 \, a^{3} \cos \left (f x + e\right )^{4} + 812 \, a^{3} \cos \left (f x + e\right )^{3} + 1796 \, a^{3} \cos \left (f x + e\right )^{2} - 1144 \, a^{3} \cos \left (f x + e\right ) - 2048 \, a^{3} + {\left (15 \, a^{3} \cos \left (f x + e\right )^{4} + 80 \, a^{3} \cos \left (f x + e\right )^{3} + 892 \, a^{3} \cos \left (f x + e\right )^{2} - 904 \, a^{3} \cos \left (f x + e\right ) - 2048 \, a^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{10240 \, {\left (c^{6} f \cos \left (f x + e\right )^{6} - 5 \, c^{6} f \cos \left (f x + e\right )^{5} - 18 \, c^{6} f \cos \left (f x + e\right )^{4} + 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 48 \, c^{6} f \cos \left (f x + e\right )^{2} - 16 \, c^{6} f \cos \left (f x + e\right ) - 32 \, c^{6} f + {\left (c^{6} f \cos \left (f x + e\right )^{5} + 6 \, c^{6} f \cos \left (f x + e\right )^{4} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} - 32 \, c^{6} f \cos \left (f x + e\right )^{2} + 16 \, c^{6} f \cos \left (f x + e\right ) + 32 \, c^{6} 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))^3/(c-c*sin(f*x+e))^(11/2),x, algorithm="fricas")

[Out]

1/10240*(15*sqrt(2)*(a^3*cos(f*x + e)^6 - 5*a^3*cos(f*x + e)^5 - 18*a^3*cos(f*x + e)^4 + 20*a^3*cos(f*x + e)^3
+ 48*a^3*cos(f*x + e)^2 - 16*a^3*cos(f*x + e) - 32*a^3 + (a^3*cos(f*x + e)^5 + 6*a^3*cos(f*x + e)^4 - 12*a^3*
cos(f*x + e)^3 - 32*a^3*cos(f*x + e)^2 + 16*a^3*cos(f*x + e) + 32*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*(15*a^3*cos(f*x + e)^5 - 65*a^3*cos(f*x + e)^4 + 812*a^3*cos(f*x + e)^3 + 1796*a^3*cos(f*x + e)^2 - 1144*
a^3*cos(f*x + e) - 2048*a^3 + (15*a^3*cos(f*x + e)^4 + 80*a^3*cos(f*x + e)^3 + 892*a^3*cos(f*x + e)^2 - 904*a^
3*cos(f*x + e) - 2048*a^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c))/(c^6*f*cos(f*x + e)^6 - 5*c^6*f*cos(f*x +
e)^5 - 18*c^6*f*cos(f*x + e)^4 + 20*c^6*f*cos(f*x + e)^3 + 48*c^6*f*cos(f*x + e)^2 - 16*c^6*f*cos(f*x + e) - 3
2*c^6*f + (c^6*f*cos(f*x + e)^5 + 6*c^6*f*cos(f*x + e)^4 - 12*c^6*f*cos(f*x + e)^3 - 32*c^6*f*cos(f*x + e)^2 +
16*c^6*f*cos(f*x + e) + 32*c^6*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))^3/(c-c*sin(f*x+e))^(11/2),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*(1/5120*(-5135*a^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*
tan((f*x+exp(1))/2)^2+c))^19-15645*a^3*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^
18-150755*a^3*c*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^17-270725*a^3*sqrt(c)*c*(-sqrt(
c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^16-385756*a^3*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c
*tan((f*x+exp(1))/2)^2+c))^15+333260*a^3*sqrt(c)*c^2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^
2+c))^14+983140*a^3*c^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^13+258460*a^3*sqrt(c)*c
^3*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^12-1312610*a^3*c^4*(-sqrt(c)*tan((f*x+exp(1)
)/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^11-597862*a^3*sqrt(c)*c^4*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+
exp(1))/2)^2+c))^10+1413590*a^3*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^9+283610*a^
3*sqrt(c)*c^5*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^8-1071020*a^3*c^6*(-sqrt(c)*tan((
f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^7+237340*a^3*sqrt(c)*c^6*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*
tan((f*x+exp(1))/2)^2+c))^6+268532*a^3*c^7*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^5-35
2180*a^3*sqrt(c)*c^7*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^4+96505*a^3*c^8*(-sqrt(c)*
tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^3+1205*a^3*c^9*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((
f*x+exp(1))/2)^2+c))-38245*a^3*sqrt(c)*c^8*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2-31
7*a^3*sqrt(c)*c^9)/c^5/(-(-sqrt(c)*tan((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+exp(1))/2)^2+c))+c)^10/sign(tan((f*x+exp(1))/2)-1)-3/1024*a^3*atan((-sqrt(
c)*tan((f*x+exp(1))/2)+sqrt(c)+sqrt(c*tan((f*x+exp(1))/2)^2+c))/sqrt(2)/sqrt(-c))/sqrt(2)/c^5/sqrt(-c)/sign(ta
n((f*x+exp(1))/2)-1))

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maple [A]  time = 1.12, size = 353, normalized size = 1.57 \[ \frac {a^{3} \left (15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{5}\left (f x +e \right )\right ) c^{7}-30 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {9}{2}} c^{\frac {5}{2}}+280 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {7}{2}} c^{\frac {7}{2}}+1024 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}} c^{\frac {9}{2}}-1120 \left (c \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} c^{\frac {11}{2}}+480 \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, c^{\frac {13}{2}}-75 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{4}\left (f x +e \right )\right ) c^{7}+150 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{3}\left (f x +e \right )\right ) c^{7}-150 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \left (\sin ^{2}\left (f x +e \right )\right ) c^{7}+75 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) \sin \left (f x +e \right ) c^{7}-15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c^{7}\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}}{5120 c^{\frac {25}{2}} \left (\sin \left (f x +e \right )-1\right )^{4} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(11/2),x)

[Out]

1/5120*a^3*(15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*sin(f*x+e)^5*c^7-30*(c*(1+sin(f*x
+e)))^(9/2)*c^(5/2)+280*(c*(1+sin(f*x+e)))^(7/2)*c^(7/2)+1024*(c*(1+sin(f*x+e)))^(5/2)*c^(9/2)-1120*(c*(1+sin(
f*x+e)))^(3/2)*c^(11/2)+480*(c*(1+sin(f*x+e)))^(1/2)*c^(13/2)-75*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^7+150*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^7-150*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^7+75*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^7-15*2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^
(1/2)/c^(1/2))*c^7)*(c*(1+sin(f*x+e)))^(1/2)/c^(25/2)/(sin(f*x+e)-1)^4/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 \[ \int \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}{{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^(11/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^3/(-c*sin(f*x + e) + c)^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3}{{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(11/2),x)

[Out]

int((a + a*sin(e + f*x))^3/(c - c*sin(e + f*x))^(11/2), 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))**3/(c-c*sin(f*x+e))**(11/2),x)

[Out]

Timed out

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