3.557 \(\int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=194 \[ -\frac {3 \left (c^2+6 c d+25 d^2\right ) (c-d) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d^2 (c-9 d) \cos (e+f x)}{4 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {(3 c+13 d) (c-d)^2 \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}} \]

[Out]

-1/16*(c-d)^2*(3*c+13*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))^2/f/(a+a*
sin(f*x+e))^(5/2)-3/32*(c-d)*(c^2+6*c*d+25*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)+1/4*(c-9*d)*d^2*cos(f*x+e)/a^2/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.47, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2765, 2968, 3019, 2751, 2649, 206} \[ -\frac {3 \left (c^2+6 c d+25 d^2\right ) (c-d) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f}+\frac {d^2 (c-9 d) \cos (e+f x)}{4 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {(3 c+13 d) (c-d)^2 \cos (e+f x)}{16 a f (a \sin (e+f x)+a)^{3/2}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(-3*(c - d)*(c^2 + 6*c*d + 25*d^2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqr
t[2]*a^(5/2)*f) - ((c - d)^2*(3*c + 13*d)*Cos[e + f*x])/(16*a*f*(a + a*Sin[e + f*x])^(3/2)) + ((c - 9*d)*d^2*C
os[e + f*x])/(4*a^2*f*Sqrt[a + a*Sin[e + f*x]]) - ((c - d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^2)/(4*f*(a + a*Si
n[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 2751

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

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2968

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

Rule 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {(c+d \sin (e+f x))^3}{(a+a \sin (e+f x))^{5/2}} \, dx &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {(c+d \sin (e+f x)) \left (-\frac {1}{2} a \left (3 c^2+9 c d-4 d^2\right )+\frac {1}{2} a (c-9 d) d \sin (e+f x)\right )}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{2} a c \left (3 c^2+9 c d-4 d^2\right )+\left (\frac {1}{2} a c (c-9 d) d-\frac {1}{2} a d \left (3 c^2+9 c d-4 d^2\right )\right ) \sin (e+f x)+\frac {1}{2} a (c-9 d) d^2 \sin ^2(e+f x)}{(a+a \sin (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {(c-d)^2 (3 c+13 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))^2}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\int \frac {\frac {1}{4} a^2 \left (3 c^3+15 c^2 d+53 c d^2-39 d^3\right )-a^2 (c-9 d) d^2 \sin (e+f x)}{\sqrt {a+a \sin (e+f x)}} \, dx}{8 a^4}\\ &=-\frac {(c-d)^2 (3 c+13 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(c-9 d) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}+\frac {\left (3 (c-d) \left (c^2+6 c d+25 d^2\right )\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2}\\ &=-\frac {(c-d)^2 (3 c+13 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(c-9 d) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}-\frac {\left (3 (c-d) \left (c^2+6 c d+25 d^2\right )\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 {3 (c-d) \left (c^2+6 c d+25 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 {(c-d)^2 (3 c+13 d) \cos (e+f x)}{16 a f (a+a \sin (e+f x))^{3/2}}+\frac {(c-9 d) d^2 \cos (e+f x)}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^2}{4 f (a+a \sin (e+f x))^{5/2}}\\ \end {align*}

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Mathematica [C]  time = 0.75, size = 400, normalized size = 2.06 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (11 c^3 \sin \left (\frac {1}{2} (e+f x)\right )-3 c^3 \sin \left (\frac {3}{2} (e+f x)\right )-11 c^3 \cos \left (\frac {1}{2} (e+f x)\right )-3 c^3 \cos \left (\frac {3}{2} (e+f x)\right )-9 c^2 d \sin \left (\frac {1}{2} (e+f x)\right )-15 c^2 d \sin \left (\frac {3}{2} (e+f x)\right )+9 c^2 d \cos \left (\frac {1}{2} (e+f x)\right )-15 c^2 d \cos \left (\frac {3}{2} (e+f x)\right )+(6+6 i) (-1)^{3/4} \left (c^3+5 c^2 d+19 c d^2-25 d^3\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )-15 c d^2 \sin \left (\frac {1}{2} (e+f x)\right )+39 c d^2 \sin \left (\frac {3}{2} (e+f x)\right )+15 c d^2 \cos \left (\frac {1}{2} (e+f x)\right )+39 c d^2 \cos \left (\frac {3}{2} (e+f x)\right )+45 d^3 \sin \left (\frac {1}{2} (e+f x)\right )-69 d^3 \sin \left (\frac {3}{2} (e+f x)\right )-16 d^3 \sin \left (\frac {5}{2} (e+f x)\right )-45 d^3 \cos \left (\frac {1}{2} (e+f x)\right )-69 d^3 \cos \left (\frac {3}{2} (e+f x)\right )+16 d^3 \cos \left (\frac {5}{2} (e+f x)\right )\right )}{32 f (a (\sin (e+f x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-11*c^3*Cos[(e + f*x)/2] + 9*c^2*d*Cos[(e + f*x)/2] + 15*c*d^2*Cos[(e
+ f*x)/2] - 45*d^3*Cos[(e + f*x)/2] - 3*c^3*Cos[(3*(e + f*x))/2] - 15*c^2*d*Cos[(3*(e + f*x))/2] + 39*c*d^2*Co
s[(3*(e + f*x))/2] - 69*d^3*Cos[(3*(e + f*x))/2] + 16*d^3*Cos[(5*(e + f*x))/2] + 11*c^3*Sin[(e + f*x)/2] - 9*c
^2*d*Sin[(e + f*x)/2] - 15*c*d^2*Sin[(e + f*x)/2] + 45*d^3*Sin[(e + f*x)/2] + (6 + 6*I)*(-1)^(3/4)*(c^3 + 5*c^
2*d + 19*c*d^2 - 25*d^3)*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 - 3*c^3*Sin[(3*(e + f*x))/2] - 15*c^2*d*Sin[(3*(e + f*x))/2] + 39*c*d^2*Sin[(3*(e + f*x))/2] - 69*d
^3*Sin[(3*(e + f*x))/2] - 16*d^3*Sin[(5*(e + f*x))/2]))/(32*f*(a*(1 + Sin[e + f*x]))^(5/2))

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fricas [B]  time = 0.46, size = 608, normalized size = 3.13 \[ -\frac {3 \, \sqrt {2} {\left ({\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right )^{3} - 4 \, c^{3} - 20 \, c^{2} d - 76 \, c d^{2} + 100 \, d^{3} + 3 \, {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (4 \, c^{3} + 20 \, c^{2} d + 76 \, c d^{2} - 100 \, d^{3} - {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (c^{3} + 5 \, c^{2} d + 19 \, c d^{2} - 25 \, d^{3}\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 (32 \, d^{3} \cos \left (f x + e\right )^{3} - 4 \, c^{3} + 12 \, c^{2} d - 12 \, c d^{2} + 4 \, d^{3} - {\left (3 \, c^{3} + 15 \, c^{2} d - 39 \, c d^{2} + 53 \, d^{3}\right )} \cos \left (f x + e\right )^{2} - {\left (7 \, c^{3} + 3 \, c^{2} d - 27 \, c d^{2} + 81 \, d^{3}\right )} \cos \left (f x + e\right ) - {\left (32 \, d^{3} \cos \left (f x + e\right )^{2} - 4 \, c^{3} + 12 \, c^{2} d - 12 \, c d^{2} + 4 \, d^{3} + {\left (3 \, c^{3} + 15 \, c^{2} d - 39 \, c d^{2} + 85 \, d^{3}\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))^3/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

-1/64*(3*sqrt(2)*((c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e)^3 - 4*c^3 - 20*c^2*d - 76*c*d^2 + 100*d^3 +
3*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e)^2 - 2*(c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e) - (
4*c^3 + 20*c^2*d + 76*c*d^2 - 100*d^3 - (c^3 + 5*c^2*d + 19*c*d^2 - 25*d^3)*cos(f*x + e)^2 + 2*(c^3 + 5*c^2*d
+ 19*c*d^2 - 25*d^3)*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*(32*d^3*cos(f*x + e)^3 - 4*c^3 +
12*c^2*d - 12*c*d^2 + 4*d^3 - (3*c^3 + 15*c^2*d - 39*c*d^2 + 53*d^3)*cos(f*x + e)^2 - (7*c^3 + 3*c^2*d - 27*c
*d^2 + 81*d^3)*cos(f*x + e) - (32*d^3*cos(f*x + e)^2 - 4*c^3 + 12*c^2*d - 12*c*d^2 + 4*d^3 + (3*c^3 + 15*c^2*d
- 39*c*d^2 + 85*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*co
s(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))^3/(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)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)2/f*(2*sqrt(a*tan((f*x+exp(1))/2)^2+a)*(-1/2*d^3/a^2/sign(tan((f*x+exp(1))/2)+1)+1/2
*d^3*tan((f*x+exp(1))/2)/a^2/sign(tan((f*x+exp(1))/2)+1))/(a*tan((f*x+exp(1))/2)^2+a)+2*(1/32*(-29*c^3*(-sqrt(
a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7-43*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x
+exp(1))/2)^2+a))^7+57*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+15*c^2*d*(-sqrt(
a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^7+75*sqrt(a)*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*
tan((f*x+exp(1))/2)^2+a))^6+237*sqrt(a)*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-3
99*sqrt(a)*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6+87*sqrt(a)*c^2*d*(-sqrt(a)*t
an((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^6-55*a*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+e
xp(1))/2)^2+a))^5-161*a*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5+267*a*c*d^2*(-sqr
t(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^5-51*a*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*ta
n((f*x+exp(1))/2)^2+a))^5-91*sqrt(a)*a*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-22
1*sqrt(a)*a*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4+351*sqrt(a)*a*c*d^2*(-sqrt(a)
*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^4-39*sqrt(a)*a*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(
a*tan((f*x+exp(1))/2)^2+a))^4+a^2*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3-25*a^2*
d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3+51*a^2*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2
)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^3-27*a^2*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a
))^3+27*a^3*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+93*a^3*d^3*(-sqrt(a)*tan((f*x+e
xp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+65*sqrt(a)*a^2*c^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+ex
p(1))/2)^2+a))^2+103*sqrt(a)*a^2*d^3*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))^2-159*a^3*
c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))/2)^2+a))+39*a^3*c^2*d*(-sqrt(a)*tan((f*x+exp(1))/2
)+sqrt(a*tan((f*x+exp(1))/2)^2+a))-141*sqrt(a)*a^2*c*d^2*(-sqrt(a)*tan((f*x+exp(1))/2)+sqrt(a*tan((f*x+exp(1))
/2)^2+a))^2-27*sqrt(a)*a^2*c^2*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^3+17*sqrt(a)*a^3*d^3-27*sqrt(a)*a^3*c*d^2+3*sqrt(a)*a^3*c^2*d)/a^2/(-(-sqrt(a)*tan((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+exp(1))/2)^2+a))+a)^4/sign
(tan((f*x+exp(1))/2)+1)+1/32*(3*c^3-75*d^3+57*c*d^2+15*c^2*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 = 1.32, size = 688, normalized size = 3.55 \[ -\frac {\left (2 \sin \left (f x +e \right ) \left (64 d^{3} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}+3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{3}+15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d +57 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}-75 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}\right )+\left (-3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{3}-15 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d -57 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}+75 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}-64 d^{3} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}\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^{3}+30 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c^{2} d +114 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} c \,d^{2}-150 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} d^{3}-6 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c^{3}-30 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c^{2} d +78 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \sqrt {a}\, c \,d^{2}-42 \left (a -a \sin \left (f x +e \right )\right )^{\frac {3}{2}} d^{3} \sqrt {a}+20 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c^{3}+36 \sqrt {a -a \sin \left (f x +e \right )}\, a^{\frac {3}{2}} c^{2} d -132 c \,d^{2} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}+204 d^{3} a^{\frac {3}{2}} \sqrt {a -a \sin \left (f x +e \right )}\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))^3/(a+a*sin(f*x+e))^(5/2),x)

[Out]

-1/32*(2*sin(f*x+e)*(64*d^3*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+3*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2
)/a^(1/2))*a^2*c^3+15*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2*d+57*2^(1/2)*arctanh
(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c*d^2-75*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a
^(1/2))*a^2*d^3)+(-3*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^3-15*2^(1/2)*arctanh(1/
2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2*d-57*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1
/2))*a^2*c*d^2+75*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3-64*d^3*a^(3/2)*(a-a*sin(
f*x+e))^(1/2))*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^3+30*2^(1/2)*a
rctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*c^2*d+114*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^
(1/2)/a^(1/2))*a^2*c*d^2-150*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*d^3-6*(a-a*sin(f*
x+e))^(3/2)*a^(1/2)*c^3-30*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c^2*d+78*(a-a*sin(f*x+e))^(3/2)*a^(1/2)*c*d^2-42*(a-
a*sin(f*x+e))^(3/2)*d^3*a^(1/2)+20*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^3+36*(a-a*sin(f*x+e))^(1/2)*a^(3/2)*c^2*d-
132*c*d^2*a^(3/2)*(a-a*sin(f*x+e))^(1/2)+204*d^3*a^(3/2)*(a-a*sin(f*x+e))^(1/2))*(-a*(sin(f*x+e)-1))^(1/2)/a^(
9/2)/(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 \[ \int \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{3}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^3/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^3/(a*sin(f*x + e) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^3}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*sin(e + f*x))^3/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int((c + d*sin(e + f*x))^3/(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))**3/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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