3.84 \(\int \frac {\csc ^3(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=205 \[ -\frac {(a-6 b) \tanh ^{-1}(\cos (e+f x))}{2 a^4 f}-\frac {b (11 a-12 b) \sec (e+f x)}{8 a^3 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {3 b \sec (e+f x)}{4 a^2 f \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac {\sqrt {b} \left (15 a^2-40 a b+24 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 a^4 f (a-b)^{3/2}}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a+b \sec ^2(e+f x)-b\right )^2} \]

[Out]

-1/2*(a-6*b)*arctanh(cos(f*x+e))/a^4/f-1/2*cot(f*x+e)*csc(f*x+e)/a/f/(a-b+b*sec(f*x+e)^2)^2-3/4*b*sec(f*x+e)/a
^2/f/(a-b+b*sec(f*x+e)^2)^2-1/8*(11*a-12*b)*b*sec(f*x+e)/a^3/(a-b)/f/(a-b+b*sec(f*x+e)^2)-1/8*(15*a^2-40*a*b+2
4*b^2)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/a^4/(a-b)^(3/2)/f

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Rubi [A]  time = 0.29, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3664, 471, 527, 522, 207, 205} \[ -\frac {\sqrt {b} \left (15 a^2-40 a b+24 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 a^4 f (a-b)^{3/2}}-\frac {b (11 a-12 b) \sec (e+f x)}{8 a^3 f (a-b) \left (a+b \sec ^2(e+f x)-b\right )}-\frac {3 b \sec (e+f x)}{4 a^2 f \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac {(a-6 b) \tanh ^{-1}(\cos (e+f x))}{2 a^4 f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a+b \sec ^2(e+f x)-b\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

-(Sqrt[b]*(15*a^2 - 40*a*b + 24*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(8*a^4*(a - b)^(3/2)*f) - ((a
- 6*b)*ArcTanh[Cos[e + f*x]])/(2*a^4*f) - (Cot[e + f*x]*Csc[e + f*x])/(2*a*f*(a - b + b*Sec[e + f*x]^2)^2) -
(3*b*Sec[e + f*x])/(4*a^2*f*(a - b + b*Sec[e + f*x]^2)^2) - ((11*a - 12*b)*b*Sec[e + f*x])/(8*a^3*(a - b)*f*(a
- b + b*Sec[e + f*x]^2))

Rule 205

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3664

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {\csc ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {a-b-5 b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^3} \, dx,x,\sec (e+f x)\right )}{2 a f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {3 b \sec (e+f x)}{4 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^2}+\frac {\operatorname {Subst}\left (\int \frac {2 (2 a-3 b) (a-b)-18 (a-b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{8 a^2 (a-b) f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {3 b \sec (e+f x)}{4 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {(11 a-12 b) b \sec (e+f x)}{8 a^3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {2 (a-b) \left (4 a^2-17 a b+12 b^2\right )-2 (11 a-12 b) (a-b) b x^2}{\left (-1+x^2\right ) \left (a-b+b x^2\right )} \, dx,x,\sec (e+f x)\right )}{16 a^3 (a-b)^2 f}\\ &=-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {3 b \sec (e+f x)}{4 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {(11 a-12 b) b \sec (e+f x)}{8 a^3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}+\frac {(a-6 b) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{2 a^4 f}-\frac {\left (b \left (15 a^2-40 a b+24 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-b+b x^2} \, dx,x,\sec (e+f x)\right )}{8 a^4 (a-b) f}\\ &=-\frac {\sqrt {b} \left (15 a^2-40 a b+24 b^2\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 a^4 (a-b)^{3/2} f}-\frac {(a-6 b) \tanh ^{-1}(\cos (e+f x))}{2 a^4 f}-\frac {\cot (e+f x) \csc (e+f x)}{2 a f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {3 b \sec (e+f x)}{4 a^2 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {(11 a-12 b) b \sec (e+f x)}{8 a^3 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [B]  time = 6.55, size = 414, normalized size = 2.02 \[ \frac {(a-6 b) \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 a^4 f}+\frac {(6 b-a) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 a^4 f}+\frac {8 b^2 \cos (e+f x)-9 a b \cos (e+f x)}{4 a^3 f (a-b) (a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b)}-\frac {\csc ^2\left (\frac {1}{2} (e+f x)\right )}{8 a^3 f}+\frac {\sec ^2\left (\frac {1}{2} (e+f x)\right )}{8 a^3 f}+\frac {b^2 \cos (e+f x)}{a^2 f (a-b) (a \cos (2 (e+f x))+a-b \cos (2 (e+f x))+b)^2}+\frac {\sqrt {b} \sqrt {a-b} \left (15 a^2-40 a b+24 b^2\right ) \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )-\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{8 a^4 f (b-a)^2}+\frac {\sqrt {b} \sqrt {a-b} \left (15 a^2-40 a b+24 b^2\right ) \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (e+f x)\right ) \left (\sqrt {a-b} \cos \left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sin \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {b}}\right )}{8 a^4 f (b-a)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^3/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(Sqrt[a - b]*Sqrt[b]*(15*a^2 - 40*a*b + 24*b^2)*ArcTan[(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] - Sqrt[
a]*Sin[(e + f*x)/2]))/Sqrt[b]])/(8*a^4*(-a + b)^2*f) + (Sqrt[a - b]*Sqrt[b]*(15*a^2 - 40*a*b + 24*b^2)*ArcTan[
(Sec[(e + f*x)/2]*(Sqrt[a - b]*Cos[(e + f*x)/2] + Sqrt[a]*Sin[(e + f*x)/2]))/Sqrt[b]])/(8*a^4*(-a + b)^2*f) +
(b^2*Cos[e + f*x])/(a^2*(a - b)*f*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])^2) + (-9*a*b*Cos[e + f*x]
+ 8*b^2*Cos[e + f*x])/(4*a^3*(a - b)*f*(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])) - Csc[(e + f*x)/2]^2
/(8*a^3*f) + ((-a + 6*b)*Log[Cos[(e + f*x)/2]])/(2*a^4*f) + ((a - 6*b)*Log[Sin[(e + f*x)/2]])/(2*a^4*f) + Sec[
(e + f*x)/2]^2/(8*a^3*f)

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fricas [B]  time = 0.96, size = 1419, normalized size = 6.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

[1/16*(2*(4*a^4 - 21*a^3*b + 29*a^2*b^2 - 12*a*b^3)*cos(f*x + e)^5 + 2*(17*a^3*b - 40*a^2*b^2 + 24*a*b^3)*cos(
f*x + e)^3 - ((15*a^4 - 70*a^3*b + 119*a^2*b^2 - 88*a*b^3 + 24*b^4)*cos(f*x + e)^6 - (15*a^4 - 100*a^3*b + 229
*a^2*b^2 - 216*a*b^3 + 72*b^4)*cos(f*x + e)^4 - 15*a^2*b^2 + 40*a*b^3 - 24*b^4 - (30*a^3*b - 125*a^2*b^2 + 168
*a*b^3 - 72*b^4)*cos(f*x + e)^2)*sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b))*co
s(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) + 2*(11*a^2*b^2 - 12*a*b^3)*cos(f*x + e) - 4*((a^4 - 9*a^3*b + 2
1*a^2*b^2 - 19*a*b^3 + 6*b^4)*cos(f*x + e)^6 - (a^4 - 11*a^3*b + 37*a^2*b^2 - 45*a*b^3 + 18*b^4)*cos(f*x + e)^
4 - a^2*b^2 + 7*a*b^3 - 6*b^4 - (2*a^3*b - 17*a^2*b^2 + 33*a*b^3 - 18*b^4)*cos(f*x + e)^2)*log(1/2*cos(f*x + e
) + 1/2) + 4*((a^4 - 9*a^3*b + 21*a^2*b^2 - 19*a*b^3 + 6*b^4)*cos(f*x + e)^6 - (a^4 - 11*a^3*b + 37*a^2*b^2 -
45*a*b^3 + 18*b^4)*cos(f*x + e)^4 - a^2*b^2 + 7*a*b^3 - 6*b^4 - (2*a^3*b - 17*a^2*b^2 + 33*a*b^3 - 18*b^4)*cos
(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^7 - 3*a^6*b + 3*a^5*b^2 - a^4*b^3)*f*cos(f*x + e)^6 - (a^7 - 5*
a^6*b + 7*a^5*b^2 - 3*a^4*b^3)*f*cos(f*x + e)^4 - (2*a^6*b - 5*a^5*b^2 + 3*a^4*b^3)*f*cos(f*x + e)^2 - (a^5*b^
2 - a^4*b^3)*f), 1/8*((4*a^4 - 21*a^3*b + 29*a^2*b^2 - 12*a*b^3)*cos(f*x + e)^5 + (17*a^3*b - 40*a^2*b^2 + 24*
a*b^3)*cos(f*x + e)^3 - ((15*a^4 - 70*a^3*b + 119*a^2*b^2 - 88*a*b^3 + 24*b^4)*cos(f*x + e)^6 - (15*a^4 - 100*
a^3*b + 229*a^2*b^2 - 216*a*b^3 + 72*b^4)*cos(f*x + e)^4 - 15*a^2*b^2 + 40*a*b^3 - 24*b^4 - (30*a^3*b - 125*a^
2*b^2 + 168*a*b^3 - 72*b^4)*cos(f*x + e)^2)*sqrt(b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) +
(11*a^2*b^2 - 12*a*b^3)*cos(f*x + e) - 2*((a^4 - 9*a^3*b + 21*a^2*b^2 - 19*a*b^3 + 6*b^4)*cos(f*x + e)^6 - (a^
4 - 11*a^3*b + 37*a^2*b^2 - 45*a*b^3 + 18*b^4)*cos(f*x + e)^4 - a^2*b^2 + 7*a*b^3 - 6*b^4 - (2*a^3*b - 17*a^2*
b^2 + 33*a*b^3 - 18*b^4)*cos(f*x + e)^2)*log(1/2*cos(f*x + e) + 1/2) + 2*((a^4 - 9*a^3*b + 21*a^2*b^2 - 19*a*b
^3 + 6*b^4)*cos(f*x + e)^6 - (a^4 - 11*a^3*b + 37*a^2*b^2 - 45*a*b^3 + 18*b^4)*cos(f*x + e)^4 - a^2*b^2 + 7*a*
b^3 - 6*b^4 - (2*a^3*b - 17*a^2*b^2 + 33*a*b^3 - 18*b^4)*cos(f*x + e)^2)*log(-1/2*cos(f*x + e) + 1/2))/((a^7 -
3*a^6*b + 3*a^5*b^2 - a^4*b^3)*f*cos(f*x + e)^6 - (a^7 - 5*a^6*b + 7*a^5*b^2 - 3*a^4*b^3)*f*cos(f*x + e)^4 -
(2*a^6*b - 5*a^5*b^2 + 3*a^4*b^3)*f*cos(f*x + e)^2 - (a^5*b^2 - a^4*b^3)*f)]

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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(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)2/f*((-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+12*(1-cos(f
*x+exp(1)))/(1+cos(f*x+exp(1)))*b-a)*1/16/a^4/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1)))+(9*((1-cos(f*x+exp(1)))/
(1+cos(f*x+exp(1))))^3*a^3*b-32*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^3*a^2*b^2+24*((1-cos(f*x+exp(1)))/(1
+cos(f*x+exp(1))))^3*a*b^3-27*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a^3*b+102*((1-cos(f*x+exp(1)))/(1+co
s(f*x+exp(1))))^2*a^2*b^2-152*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a*b^3+80*((1-cos(f*x+exp(1)))/(1+cos
(f*x+exp(1))))^2*b^4+27*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a^3*b-80*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)
))*a^2*b^2+56*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a*b^3-9*a^3*b+10*a^2*b^2)/(8*a^5-8*a^4*b)/(((1-cos(f*x+e
xp(1)))/(1+cos(f*x+exp(1))))^2*a-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+4*(1-cos(f*x+exp(1)))/(1+cos(f*x+
exp(1)))*b+a)^2+(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*1/16/a^3+(a-6*b)*1/8/a^4*ln(abs(1-cos(f*x+exp(1)))/abs
(1+cos(f*x+exp(1))))+(-15*a^2*b+40*a*b^2-24*b^3)*1/4/(4*a^5-4*a^4*b)/sqrt(-b^2+a*b)*atan((-a*cos(f*x+exp(1))+b
*cos(f*x+exp(1))+b)/(sqrt(-b^2+a*b)*cos(f*x+exp(1))+sqrt(-b^2+a*b))))

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maple [B]  time = 0.82, size = 435, normalized size = 2.12 \[ -\frac {9 b \left (\cos ^{3}\left (f x +e \right )\right )}{8 f \,a^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}+\frac {b^{2} \left (\cos ^{3}\left (f x +e \right )\right )}{f \,a^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2}}-\frac {7 b^{2} \cos \left (f x +e \right )}{8 f \,a^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2} \left (a -b \right )}+\frac {b^{3} \cos \left (f x +e \right )}{f \,a^{3} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )^{2} \left (a -b \right )}+\frac {15 b \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{8 f \,a^{2} \left (a -b \right ) \sqrt {\left (a -b \right ) b}}-\frac {5 b^{2} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{3} \left (a -b \right ) \sqrt {\left (a -b \right ) b}}+\frac {3 b^{3} \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {\left (a -b \right ) b}}\right )}{f \,a^{4} \left (a -b \right ) \sqrt {\left (a -b \right ) b}}+\frac {1}{4 f \,a^{3} \left (-1+\cos \left (f x +e \right )\right )}+\frac {\ln \left (-1+\cos \left (f x +e \right )\right )}{4 f \,a^{3}}-\frac {3 \ln \left (-1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{4}}+\frac {1}{4 f \,a^{3} \left (1+\cos \left (f x +e \right )\right )}-\frac {\ln \left (1+\cos \left (f x +e \right )\right )}{4 f \,a^{3}}+\frac {3 \ln \left (1+\cos \left (f x +e \right )\right ) b}{2 f \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x)

[Out]

-9/8/f*b/a^2/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*cos(f*x+e)^3+1/f*b^2/a^3/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2*
cos(f*x+e)^3-7/8/f*b^2/a^2/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2/(a-b)*cos(f*x+e)+1/f*b^3/a^3/(a*cos(f*x+e)^2-co
s(f*x+e)^2*b+b)^2/(a-b)*cos(f*x+e)+15/8/f*b/a^2/(a-b)/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))
-5/f*b^2/a^3/(a-b)/((a-b)*b)^(1/2)*arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))+3/f*b^3/a^4/(a-b)/((a-b)*b)^(1/2)*
arctan((a-b)*cos(f*x+e)/((a-b)*b)^(1/2))+1/4/f/a^3/(-1+cos(f*x+e))+1/4/f/a^3*ln(-1+cos(f*x+e))-3/2/f/a^4*ln(-1
+cos(f*x+e))*b+1/4/f/a^3/(1+cos(f*x+e))-1/4/f/a^3*ln(1+cos(f*x+e))+3/2/f/a^4*ln(1+cos(f*x+e))*b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*tan(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for mor
e details)Is b-a positive or negative?

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mupad [B]  time = 13.22, size = 1652, normalized size = 8.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^3),x)

[Out]

tan(e/2 + (f*x)/2)^2/(8*a^3*f) - (a^2/2 + (tan(e/2 + (f*x)/2)^4*(96*a*b^2 - 38*a^2*b + 3*a^3 - 64*b^3))/(a - b
) + (tan(e/2 + (f*x)/2)^8*(64*a*b^2 - 19*a^2*b + a^3 - 48*b^3))/(2*(a - b)) - (tan(e/2 + (f*x)/2)^2*(14*a*b^2
- 15*a^2*b + 2*a^3))/(a - b) - (tan(e/2 + (f*x)/2)^6*(2*a^4 - 33*a^3*b - 152*a*b^3 + 80*b^4 + 106*a^2*b^2))/(a
*(a - b)))/(f*(4*a^5*tan(e/2 + (f*x)/2)^2 + 4*a^5*tan(e/2 + (f*x)/2)^10 + tan(e/2 + (f*x)/2)^6*(24*a^5 - 64*a^
4*b + 64*a^3*b^2) + tan(e/2 + (f*x)/2)^4*(32*a^4*b - 16*a^5) + tan(e/2 + (f*x)/2)^8*(32*a^4*b - 16*a^5))) + (l
og(tan(e/2 + (f*x)/2))*(a - 6*b))/(2*a^4*f) + (b^(1/2)*atan(((tan(e/2 + (f*x)/2)^2*((((b^(3/2)*(15*a^2 - 40*a*
b + 24*b^2)^3*(128*a^16 - 3712*a^15*b + 6144*a^10*b^6 - 27648*a^11*b^5 + 49408*a^12*b^4 - 43904*a^13*b^3 + 195
84*a^14*b^2))/(32768*a^12*(a - b)^(9/2)*(3*a^10*b - a^11 + a^8*b^3 - 3*a^9*b^2)) + (b^(1/2)*(15*a^2 - 40*a*b +
24*b^2)*(360*a^9*b - 13824*a^2*b^8 + 66816*a^3*b^7 - 132864*a^4*b^6 + 139776*a^5*b^5 - 83240*a^6*b^4 + 27836*
a^7*b^3 - 4860*a^8*b^2))/(128*a^4*(a - b)^(3/2)*(3*a^10*b - a^11 + a^8*b^3 - 3*a^9*b^2)))*(63*a^6 - 1013*a^5*b
- 9600*a*b^5 + 2304*b^6 + 15792*a^2*b^4 - 12888*a^3*b^3 + 5342*a^4*b^2))/(2*a^5*(a - b)^(9/2)*(5760*a*b^4 - 7
35*a^4*b + 64*a^5 - 1728*b^5 - 6960*a^2*b^3 + 3600*a^3*b^2)) - (((6912*a*b^6 - 1728*b^7 - 10800*a^2*b^5 + 8240
*a^3*b^4 - 3075*a^4*b^3 + 450*a^5*b^2)/(8*(3*a^10*b - a^11 + a^8*b^3 - 3*a^9*b^2)) + (b*(15*a^2 - 40*a*b + 24*
b^2)^2*(1936*a^12*b - 64*a^13 + 18432*a^6*b^7 - 86016*a^7*b^6 + 161664*a^8*b^5 - 155008*a^9*b^4 + 78736*a^10*b
^3 - 19680*a^11*b^2))/(2048*a^8*(a - b)^3*(3*a^10*b - a^11 + a^8*b^3 - 3*a^9*b^2)))*(3072*a*b^4 - 145*a^4*b +
4*a^5 - 1152*b^5 - 2856*a^2*b^3 + 1080*a^3*b^2))/(a^5*b^(1/2)*(a - b)^3*(5760*a*b^4 - 735*a^4*b + 64*a^5 - 172
8*b^5 - 6960*a^2*b^3 + 3600*a^3*b^2))) + (((b^(3/2)*(15*a^2 - 40*a*b + 24*b^2)^3*(128*a^16 - 768*a^15*b + 512*
a^12*b^4 - 1536*a^13*b^3 + 1664*a^14*b^2))/(32768*a^12*(a - b)^(9/2)*(a^11 - 2*a^10*b + a^9*b^2)) + (b^(1/2)*(
15*a^2 - 40*a*b + 24*b^2)*(240*a^9*b - 5760*a^4*b^6 + 19200*a^5*b^5 - 23776*a^6*b^4 + 13344*a^7*b^3 - 3250*a^8
*b^2))/(128*a^4*(a - b)^(3/2)*(a^11 - 2*a^10*b + a^9*b^2)))*(63*a^6 - 1013*a^5*b - 9600*a*b^5 + 2304*b^6 + 157
92*a^2*b^4 - 12888*a^3*b^3 + 5342*a^4*b^2))/(2*a^5*(a - b)^(9/2)*(5760*a*b^4 - 735*a^4*b + 64*a^5 - 1728*b^5 -
6960*a^2*b^3 + 3600*a^3*b^2)) - (((12096*a*b^6 - 3456*b^7 - 15840*a^2*b^5 + 9520*a^3*b^4 - 2550*a^4*b^3 + 225
*a^5*b^2)/(8*(a^11 - 2*a^10*b + a^9*b^2)) + (b*(15*a^2 - 40*a*b + 24*b^2)^2*(1248*a^12*b - 64*a^13 + 3072*a^8*
b^5 - 9728*a^9*b^4 + 11328*a^10*b^3 - 5856*a^11*b^2))/(2048*a^8*(a - b)^3*(a^11 - 2*a^10*b + a^9*b^2)))*(3072*
a*b^4 - 145*a^4*b + 4*a^5 - 1152*b^5 - 2856*a^2*b^3 + 1080*a^3*b^2))/(a^5*b^(1/2)*(a - b)^3*(5760*a*b^4 - 735*
a^4*b + 64*a^5 - 1728*b^5 - 6960*a^2*b^3 + 3600*a^3*b^2)))*(256*a^13*(a - b)^(9/2) - 768*a^12*b*(a - b)^(9/2)
- 256*a^10*b^3*(a - b)^(9/2) + 768*a^11*b^2*(a - b)^(9/2)))/(225*a^4*b - 1920*a*b^4 + 576*b^5 + 2320*a^2*b^3 -
1200*a^3*b^2))*(15*a^2 - 40*a*b + 24*b^2))/(8*a^4*f*(a - b)^(3/2))

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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(csc(f*x+e)**3/(a+b*tan(f*x+e)**2)**3,x)

[Out]

Timed out

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