3.558 \(\int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=147 \[ -\frac {\left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

-3/16*(c-d)*(c+3*d)*cos(f*x+e)/a/f/(a+a*sin(f*x+e))^(3/2)-1/4*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))/f/(a+a*sin(f*x
+e))^(5/2)-1/32*(3*c^2+10*c*d+19*d^2)*arctanh(1/2*cos(f*x+e)*a^(1/2)*2^(1/2)/(a+a*sin(f*x+e))^(1/2))/a^(5/2)/f
*2^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2760, 2750, 2649, 206} \[ -\frac {\left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a \sin (e+f x)+a)^{5/2}}-\frac {3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^
(5/2)*f) - (3*(c - d)*(c + 3*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])^(3/2)) - ((c - d)*Cos[e + f*x]*(c +
d*Sin[e + f*x]))/(4*f*(a + a*Sin[e + f*x])^(5/2))

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 2750

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

Rule 2760

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

Rubi steps

\begin {align*} \int \frac {(c+d \sin (e+f x))^2}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{2} a \left (3 c^2+7 c d-2 d^2\right )-\frac {1}{2} a d (c+7 d) \sin (e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (3 c^2+10 c d+19 d^2\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac {3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\left (3 c^2+10 c d+19 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{16 a^2 f}\\ &=-\frac {\left (3 c^2+10 c d+19 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{16 \sqrt {2} a^{5/2} f}-\frac {3 (c-d) (c+3 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.54, size = 252, normalized size = 1.71 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (2 \left (3 c^2+10 c d-13 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2+(1+i) (-1)^{3/4} \left (3 c^2+10 c d+19 d^2\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )+8 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-(c-d) (3 c+13 d) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3-4 (c-d)^2 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{16 f (a (\sin (e+f x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(8*(c - d)^2*Sin[(e + f*x)/2] - 4*(c - d)^2*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2]) + 2*(3*c^2 + 10*c*d - 13*d^2)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (c - d)*(3
*c + 13*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (1 + I)*(-1)^(3/4)*(3*c^2 + 10*c*d + 19*d^2)*ArcTanh[(1/2
+ I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4))/(16*f*(a*(1 + Sin[e + f*
x]))^(5/2))

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fricas [B]  time = 0.47, size = 492, normalized size = 3.35 \[ \frac {\sqrt {2} {\left ({\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} - 40 \, c d - 76 \, d^{2} - 2 \, {\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right ) + {\left ({\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )^{2} - 12 \, c^{2} - 40 \, c d - 76 \, d^{2} - 2 \, {\left (3 \, c^{2} + 10 \, c d + 19 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )} + 3 \, a \cos \left (f x + e\right ) - {\left (a \cos \left (f x + e\right ) - 2 \, a\right )} \sin \left (f x + e\right ) + 2 \, a}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 4 \, {\left ({\left (3 \, c^{2} + 10 \, c d - 13 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, c^{2} - 8 \, c d + 4 \, d^{2} + {\left (7 \, c^{2} + 2 \, c d - 9 \, d^{2}\right )} \cos \left (f x + e\right ) - {\left (4 \, c^{2} - 8 \, c d + 4 \, d^{2} - {\left (3 \, c^{2} + 10 \, c d - 13 \, d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{64 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(sqrt(2)*((3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^3 + 3*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*c^2
- 40*c*d - 76*d^2 - 2*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e) + ((3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e)^2 - 12*
c^2 - 40*c*d - 76*d^2 - 2*(3*c^2 + 10*c*d + 19*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2
- 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*
x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(
(3*c^2 + 10*c*d - 13*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 + 2*c*d - 9*d^2)*cos(f*x + e) - (4*c
^2 - 8*c*d + 4*d^2 - (3*c^2 + 10*c*d - 13*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f*co
s(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f
*x + e) - 4*a^3*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((c+d*sin(f*x+e))^2/(a+a*sin(f*x+e))^(5/2),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*(1/32*(-29*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan(
(f*x+exp(1))/2)^2+a))^7+19*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+10*c*d*(-sqrt(
a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+75*sqrt(a)*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*
tan((f*x+exp(1))/2)^2+a))^6-133*sqrt(a)*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+5
8*sqrt(a)*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-55*a*c^2*(-sqrt(a)*tan((f*x+exp
(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5+89*a*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2
+a))^5-34*a*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-91*sqrt(a)*a*c^2*(-sqrt(a)*ta
n((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4+117*sqrt(a)*a*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*ta
n((f*x+exp(1))/2)^2+a))^4-26*sqrt(a)*a*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4+a^
2*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+17*a^2*d^2*(-sqrt(a)*tan((f*x+exp(1))/2
)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3-18*a^2*c*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))
^3+27*a^3*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-53*a^3*d^2*(-sqrt(a)*tan((f*x+exp
(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+65*sqrt(a)*a^2*c^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(
1))/2)^2+a))^2-47*sqrt(a)*a^2*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+26*a^3*c*d*
(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-18*sqrt(a)*a^2*c*d*(-sqrt(a)*tan((f*x+exp(1))/2
)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+7*sqrt(a)*a^3*c^2-9*sqrt(a)*a^3*d^2+2*sqrt(a)*a^3*c*d)/a^2/(-(-sqrt(a)*ta
n((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2+2*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+e
xp(1))/2)^2+a))+a)^4/sign(tan((f*x+exp(1))/2)+1)+1/32*(3*c^2+19*d^2+10*c*d)*atan((-sqrt(a)*tan((f*x+exp(1))/2)
-sqrt(a)+sqrt(a*tan((f*x+exp(1))/2)^2+a))/sqrt(2)/sqrt(-a))/sqrt(2)/a^2/sqrt(-a)/sign(tan((f*x+exp(1))/2)+1))

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maple [B]  time = 1.43, size = 378, normalized size = 2.57 \[ \frac {\left (-2 \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{2} \left (3 c^{2}+10 c d +19 d^{2}\right )+\arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) \sqrt {2}\, a^{2} \left (3 c^{2}+10 c d +19 d^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-6 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2}-20 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c d -38 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{2}+6 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c^{2}+20 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c d -26 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, d^{2}-20 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c^{2}-24 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c d +44 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} d^{2}\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}}{32 a^{\frac {9}{2}} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32/a^(9/2)*(-2*sin(f*x+e)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*(3*c^2+10*c*d+19*d
^2)+arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^2*(3*c^2+10*c*d+19*d^2)*cos(f*x+e)^2-6*2^(1/
2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2-20*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2
^(1/2)/a^(1/2))*a^2*c*d-38*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^2+6*(a-a*sin(f*x+
e))^(3/2)*a^(1/2)*c^2+20*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d-26*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*d^2-20*(a-a*sin(
f*x+e))^(1/2)*a^(3/2)*c^2-24*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c*d+44*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*d^2)*(-a*(si
n(f*x+e)-1))^(1/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*sin(f*x + e) + c)^2/(a*sin(f*x + e) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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