∫ 1____________ (a+a sin(e+fx))5∕2(c+d sin(e+fx)) dx

3.561 \(\int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx\)

Optimal. Leaf size=218 \[ -\frac {\left (3 c^2-14 c d+43 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f (c-d)^3}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{a^{5/2} f (c-d)^3 \sqrt {c+d}}-\frac {(3 c-11 d) \cos (e+f x)}{16 a f (c-d)^2 (a \sin (e+f x)+a)^{3/2}}-\frac {\cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}} \]

[Out]

-1/4*cos(f*x+e)/(c-d)/f/(a+a*sin(f*x+e))^(5/2)-1/16*(3*c-11*d)*cos(f*x+e)/a/(c-d)^2/f/(a+a*sin(f*x+e))^(3/2)-1

/32*(3*c^2-14*c*d+43*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)/(c-d)^3/f*2^(

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

(1/2)

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Rubi [A] time = 0.74, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2766, 2978, 2985, 2649, 206, 2773, 208} \[ -\frac {\left (3 c^2-14 c d+43 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{16 \sqrt {2} a^{5/2} f (c-d)^3}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{a^{5/2} f (c-d)^3 \sqrt {c+d}}-\frac {(3 c-11 d) \cos (e+f x)}{16 a f (c-d)^2 (a \sin (e+f x)+a)^{3/2}}-\frac {\cos (e+f x)}{4 f (c-d) (a \sin (e+f x)+a)^{5/2}} \]

Antiderivative was successfully veriﬁed.

[In]

Int[1/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x])),x]

[Out]

-((3*c^2 - 14*c*d + 43*d^2)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(16*Sqrt[2]*a^

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

)/(a^(5/2)*(c - d)^3*Sqrt[c + d]*f) - Cos[e + f*x]/(4*(c - d)*f*(a + a*Sin[e + f*x])^(5/2)) - ((3*c - 11*d)*Co

s[e + f*x])/(16*a*(c - d)^2*f*(a + a*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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,

b}, x] && NegQ[a/b]

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 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim

p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis

t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*

(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,

0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ

[m] && EqQ[c, 0]))

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*

b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*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]

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*

x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +

1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*

(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2

- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,

0])

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_

.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(

B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,

A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin {align*} \int \frac {1}{(a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))} \, dx &=-\frac {\cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{2} a (3 c-8 d)-\frac {3}{2} a d \sin (e+f x)}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))} \, dx}{4 a^2 (c-d)}\\ &=-\frac {\cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c-11 d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac {\int \frac {\frac {1}{4} a^2 \left (3 c^2-11 c d+32 d^2\right )+\frac {1}{4} a^2 (3 c-11 d) d \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{8 a^4 (c-d)^2}\\ &=-\frac {\cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c-11 d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}-\frac {d^3 \int \frac {\sqrt {a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{a^3 (c-d)^3}+\frac {\left (3 c^2-14 c d+43 d^2\right ) \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx}{32 a^2 (c-d)^3}\\ &=-\frac {\cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c-11 d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}+\frac {\left (2 d^3\right ) \operatorname {Subst}\left (\int \frac {1}{a c+a d-d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{a^2 (c-d)^3 f}-\frac {\left (3 c^2-14 c d+43 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 (c-d)^3 f}\\ &=-\frac {\left (3 c^2-14 c d+43 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} (c-d)^3 f}+\frac {2 d^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a+a \sin (e+f x)}}\right )}{a^{5/2} (c-d)^3 \sqrt {c+d} f}-\frac {\cos (e+f x)}{4 (c-d) f (a+a \sin (e+f x))^{5/2}}-\frac {(3 c-11 d) \cos (e+f x)}{16 a (c-d)^2 f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] time = 3.29, size = 501, normalized size = 2.30 \[ \frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (\frac {(1+i) (-1)^{3/4} \left (3 c^2-14 c d+43 d^2\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )}{(c-d)^3}+\frac {8 d^{5/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}-\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(c-d)^3 \sqrt {c+d}}+\frac {8 d^{5/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}+\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )-\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right )\right )+e+f x\right )}{(d-c)^3 \sqrt {c+d}}+\frac {8 \sin \left (\frac {1}{2} (e+f x)\right )}{c-d}+\frac {(11 d-3 c) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3}{(c-d)^2}+\frac {2 (3 c-11 d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^2}{(c-d)^2}-\frac {4 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{c-d}\right )}{16 f (a (\sin (e+f x)+1))^{5/2}} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/((a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x])),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((8*Sin[(e + f*x)/2])/(c - d) - (4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]

))/(c - d) + (2*(3*c - 11*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2)/(c - d)^2 + ((-3*c + 11

*d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)/(c - d)^2 + ((1 + I)*(-1)^(3/4)*(3*c^2 - 14*c*d + 43*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)/(c - d)^3 + (8*d^(5/

2)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] + Sqrt[d]*Cos[(e + f*x)/2] - S

qrt[d]*Sin[(e + f*x)/2])])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)/((c - d)^3*Sqrt[c + d]) + (8*d^(5/2)*(e +

f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] - Sqrt[d]*Cos[(e + f*x)/2] + Sqrt[d]*S

in[(e + f*x)/2])])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)/((-c + d)^3*Sqrt[c + d])))/(16*f*(a*(1 + Sin[e + f

*x]))^(5/2))

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fricas [B] time = 1.06, size = 2015, normalized size = 9.24 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[-1/64*(sqrt(2)*((3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e)^2 - 12*c^

2 + 56*c*d - 172*d^2 - 2*(3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e) + ((3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e)^2 -

12*c^2 + 56*c*d - 172*d^2 - 2*(3*c^2 - 14*c*d + 43*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*co

s(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)) +

32*(a*d^2*cos(f*x + e)^3 + 3*a*d^2*cos(f*x + e)^2 - 2*a*d^2*cos(f*x + e) - 4*a*d^2 + (a*d^2*cos(f*x + e)^2 -

2*a*d^2*cos(f*x + e) - 4*a*d^2)*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*co

s(f*x + e)^2 - c^2 - 2*c*d - d^2 - 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)

*cos(f*x + e) + (c^2 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d

/(a*c + a*d)) - (c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*d^

2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 +

d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) - 4*((3*c^2

- 14*c*d + 11*d^2)*cos(f*x + e)^2 + 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 22*c*d + 15*d^2)*cos(f*x + e) - (4*c^2 -

8*c*d + 4*d^2 - (3*c^2 - 14*c*d + 11*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^3*c^3 - 3*

a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(

f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a

^3*c*d^2 - a^3*d^3)*f + ((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3

*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f)*sin(f*x

+ e)), -1/64*(sqrt(2)*((3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e)^3 + 3*(3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e)^2 -

12*c^2 + 56*c*d - 172*d^2 - 2*(3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e) + ((3*c^2 - 14*c*d + 43*d^2)*cos(f*x + e

)^2 - 12*c^2 + 56*c*d - 172*d^2 - 2*(3*c^2 - 14*c*d + 43*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)) - 64*(a*d^2*cos(f*x + e)^3 + 3*a*d^2*cos(f*x + e)^2 - 2*a*d^2*cos(f*x + e) - 4*a*d^2 + (a*d^2*cos(f*x + e

)^2 - 2*a*d^2*cos(f*x + e) - 4*a*d^2)*sin(f*x + e))*sqrt(-d/(a*c + a*d))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(

d*sin(f*x + e) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*cos(f*x + e))) - 4*((3*c^2 - 14*c*d + 11*d^2)*cos(f*x + e)^2

+ 4*c^2 - 8*c*d + 4*d^2 + (7*c^2 - 22*c*d + 15*d^2)*cos(f*x + e) - (4*c^2 - 8*c*d + 4*d^2 - (3*c^2 - 14*c*d +

11*d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3

)*f*cos(f*x + e)^3 + 3*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c

^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f + ((a^3*c^3

- 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f

*cos(f*x + e) - 4*(a^3*c^3 - 3*a^3*c^2*d + 3*a^3*c*d^2 - a^3*d^3)*f)*sin(f*x + e))]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(5/2)/(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*p

i/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)Unab

le 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*p

i/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)Unab

le 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*p

i/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)Unab

le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to ch

eck sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warnin

g, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(

cos((f*t_nostep+exp(1))/2-pi/4))]Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sig

n: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Discontinuities at zeroes of cos((f*t_nostep+exp(1))/2-pi/4) were not c

heckedUnable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi

/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>

(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nost

ep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/