3.108 \(\int \csc ^3(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)
Optimal. Leaf size=167 \[ \frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f}-\frac{\sqrt{a} (a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f}+\frac{\sqrt{b} (3 a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 f} \]
[Out]
-(Sqrt[a]*(a + 3*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + (Sqrt[b]*(3*a + b)
*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + (b*Sec[e + f*x]*Sqrt[a - b + b*Sec[e
+ f*x]^2])/f - (Cot[e + f*x]*Csc[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2))/(2*f)
________________________________________________________________________________________
Rubi [A] time = 0.20553, antiderivative size = 167, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used =
{3664, 467, 528, 523, 217, 206, 377, 207} \[ \frac{b \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)-b}}{f}-\frac{\sqrt{a} (a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f}+\frac{\sqrt{b} (3 a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a+b \sec ^2(e+f x)-b}}\right )}{2 f}-\frac{\cot (e+f x) \csc (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{2 f} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
-(Sqrt[a]*(a + 3*b)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + (Sqrt[b]*(3*a + b)
*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + (b*Sec[e + f*x]*Sqrt[a - b + b*Sec[e
+ f*x]^2])/f - (Cot[e + f*x]*Csc[e + f*x]*(a - b + b*Sec[e + f*x]^2)^(3/2))/(2*f)
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]
Rule 467
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)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Rule 528
Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 0]
Rule 523
Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
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 377
Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]
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])
Rubi steps
\begin{align*} \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (a-b+b x^2\right )^{3/2}}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \csc (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a-b+b x^2} \left (a-b+4 b x^2\right )}{-1+x^2} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac{b \sec (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{\operatorname{Subst}\left (\int \frac{2 \left (a^2-b^2\right )+2 b (3 a+b) x^2}{\left (-1+x^2\right ) \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{4 f}\\ &=\frac{b \sec (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{(b (3 a+b)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}+\frac{(a (a+3 b)) \operatorname{Subst}\left (\int \frac{1}{\left (-1+x^2\right ) \sqrt{a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{2 f}\\ &=\frac{b \sec (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{2 f}+\frac{(b (3 a+b)) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac{(a (a+3 b)) \operatorname{Subst}\left (\int \frac{1}{-1+a x^2} \, dx,x,\frac{\sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{2 f}\\ &=-\frac{\sqrt{a} (a+3 b) \tanh ^{-1}\left (\frac{\sqrt{a} \sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac{\sqrt{b} (3 a+b) \tanh ^{-1}\left (\frac{\sqrt{b} \sec (e+f x)}{\sqrt{a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac{b \sec (e+f x) \sqrt{a-b+b \sec ^2(e+f x)}}{f}-\frac{\cot (e+f x) \csc (e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{2 f}\\ \end{align*}
Mathematica [B] time = 6.59162, size = 1022, normalized size = 6.12 \[ \frac{\sqrt{\frac{\cos (2 (e+f x)) a+a+b-b \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\frac{1}{2} b \sec (e+f x)-\frac{1}{2} a \cot (e+f x) \csc (e+f x)\right )}{f}+\frac{\frac{\left (a^2-b^2\right ) \left (2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{b} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )}{\sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}\right )+\sqrt{b} \left (\tanh ^{-1}\left (\frac{-a \tan ^2\left (\frac{1}{2} (e+f x)\right )+2 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a}{\sqrt{a} \sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}\right )+\tanh ^{-1}\left (\frac{2 b+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )}{\sqrt{a} \sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}\right )\right )\right ) (\cos (e+f x)+1) \sqrt{\frac{\cos (2 (e+f x))+1}{(\cos (e+f x)+1)^2}} \sqrt{\frac{a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right ) \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right ) \sqrt{\frac{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}{\left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )^2}}}{4 \sqrt{a} \sqrt{b} \sqrt{a+b+(a-b) \cos (2 (e+f x))} \sqrt{\left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2} \sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}-\frac{\left (a^2+6 b a+b^2\right ) \left (2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{b} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )}{\sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}\right )-\sqrt{b} \left (\tanh ^{-1}\left (\frac{-a \tan ^2\left (\frac{1}{2} (e+f x)\right )+2 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a}{\sqrt{a} \sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}\right )+\tanh ^{-1}\left (\frac{2 b+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )}{\sqrt{a} \sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}\right )\right )\right ) (\cos (e+f x)+1) \sqrt{\frac{\cos (2 (e+f x))+1}{(\cos (e+f x)+1)^2}} \sqrt{\frac{a+b+(a-b) \cos (2 (e+f x))}{\cos (2 (e+f x))+1}} \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right ) \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right ) \sqrt{\frac{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}{\left (\tan ^2\left (\frac{1}{2} (e+f x)\right )+1\right )^2}}}{4 \sqrt{a} \sqrt{b} \sqrt{a+b+(a-b) \cos (2 (e+f x))} \sqrt{\left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2} \sqrt{4 b \tan ^2\left (\frac{1}{2} (e+f x)\right )+a \left (\tan ^2\left (\frac{1}{2} (e+f x)\right )-1\right )^2}}}{2 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
(Sqrt[(a + b + a*Cos[2*(e + f*x)] - b*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(-(a*Cot[e + f*x]*Csc[e + f*x]
)/2 + (b*Sec[e + f*x])/2))/f + (-((a^2 + 6*a*b + b^2)*(2*Sqrt[a]*ArcTanh[(Sqrt[b]*(1 + Tan[(e + f*x)/2]^2))/Sq
rt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]] - Sqrt[b]*(ArcTanh[(a - a*Tan[(e + f*x)/2]^2 + 2*b
*Tan[(e + f*x)/2]^2)/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])] + ArcTanh[(2*b +
a*(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])]))*(1 + Co
s[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f*x])^2]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos
[2*(e + f*x)])]*(-1 + Tan[(e + f*x)/2]^2)*(1 + Tan[(e + f*x)/2]^2)*Sqrt[(4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[
(e + f*x)/2]^2)^2)/(1 + Tan[(e + f*x)/2]^2)^2])/(4*Sqrt[a]*Sqrt[b]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*Sqrt
[(-1 + Tan[(e + f*x)/2]^2)^2]*Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2]) + ((a^2 - b^2)*(2*
Sqrt[a]*ArcTanh[(Sqrt[b]*(1 + Tan[(e + f*x)/2]^2))/Sqrt[4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2
]] + Sqrt[b]*(ArcTanh[(a - a*Tan[(e + f*x)/2]^2 + 2*b*Tan[(e + f*x)/2]^2)/(Sqrt[a]*Sqrt[4*b*Tan[(e + f*x)/2]^2
+ a*(-1 + Tan[(e + f*x)/2]^2)^2])] + ArcTanh[(2*b + a*(-1 + Tan[(e + f*x)/2]^2))/(Sqrt[a]*Sqrt[4*b*Tan[(e + f
*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2])]))*(1 + Cos[e + f*x])*Sqrt[(1 + Cos[2*(e + f*x)])/(1 + Cos[e + f*x]
)^2]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])/(1 + Cos[2*(e + f*x)])]*(-1 + Tan[(e + f*x)/2]^2)*(1 + Tan[(e + f
*x)/2]^2)*Sqrt[(4*b*Tan[(e + f*x)/2]^2 + a*(-1 + Tan[(e + f*x)/2]^2)^2)/(1 + Tan[(e + f*x)/2]^2)^2])/(4*Sqrt[a
]*Sqrt[b]*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*Sqrt[(-1 + Tan[(e + f*x)/2]^2)^2]*Sqrt[4*b*Tan[(e + f*x)/2]^2
+ a*(-1 + Tan[(e + f*x)/2]^2)^2]))/(2*f)
________________________________________________________________________________________
Maple [B] time = 0.192, size = 2904, normalized size = 17.4 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x)
[Out]
-1/8/f/a^(5/2)/b^(1/2)*(cos(f*x+e)-1)^2*(-10*cos(f*x+e)^3*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x
+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^
2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(7/2)*a+10*cos(f*x+e)^3*ln(-4/a^(1/2)*(cos(f*x+e)-1)
*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*
cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(7/2)*a+6*cos(f*x+e)^3*arcta
nh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^
2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(7/2)*b+2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^
(1/2)*a^(7/2)*b^(1/2)*cos(f*x+e)^2+2*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*b^(3/2
)*cos(f*x+e)^2+18*cos(f*x+e)^3*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(co
s(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1
/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(5/2)*a^2+10*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+
e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*
x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(7/2)*cos(f*x+e)^2*a-18*cos(f*x+e)^3*ln(-4/a^(1/2)*(cos(f*x+e)-1)*
(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*c
os(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(5/2)*a^2-10*ln(-4/a^(1/2)*(c
os(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos
(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(7/2)*cos(f*x+e)
^2*a+2*cos(f*x+e)^3*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin
(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(5/2)*b^2-6*arctanh(1/8*b^(1/2)*4^(1/2
)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(
cos(f*x+e)+1)^2)^(1/2))*a^(7/2)*cos(f*x+e)^2*b-3*cos(f*x+e)^3*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos
(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x
+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(3/2)*a^3-18*cos(f*x+e)^2*ln(-2/a^(1/2)*(cos(f*x
+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e
)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(5/2)*a^2+18*cos(f*x+e
)^2*ln(-4/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2
)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2
)*b^(5/2)*a^2-2*arctanh(1/8*b^(1/2)*4^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x
+e)^2/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2))*a^(5/2)*cos(f*x+e)^2*b^2-3*cos(f*x+e)^3*ln(-
4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a
*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*b^(3/2)*a^3-2*((a*cos(f*x+e
)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(5/2)*b^(3/2)-cos(f*x+e)^3*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f
*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x
+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(1/2)*a^4+3*cos(f*x+e)^2*ln(-2/a^(1
/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)-cos(f*x+e)*a
+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/sin(f*x+e)^2)*b^(3/2)*a^3-
cos(f*x+e)^3*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+e)*a
-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*b^(1/2)*a^
4+3*cos(f*x+e)^2*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+cos(f*x+
e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+b)/(cos(f*x+e)-1))*b^(3/2
)*a^3+cos(f*x+e)^2*ln(-2/a^(1/2)*(cos(f*x+e)-1)*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^
2)^(1/2)*a^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+
b)/sin(f*x+e)^2)*b^(1/2)*a^4+cos(f*x+e)^2*ln(-4*(cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^
2)^(1/2)*a^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(1/2)*a^(1/2)+
b)/(cos(f*x+e)-1))*b^(1/2)*a^4)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)*4^(1/2)/((a*
cos(f*x+e)^2-cos(f*x+e)^2*b+b)/(cos(f*x+e)+1)^2)^(3/2)/sin(f*x+e)^6
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^3, x)
________________________________________________________________________________________
Fricas [A] time = 7.95492, size = 2515, normalized size = 15.06 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/4*(((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*s
qrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) + ((3*a + b)*cos(
f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x +
e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((a + b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*c
os(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), -1/4*(2*((3*a + b)*cos(f*x + e)^3 - (
3*a + b)*cos(f*x + e))*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)
/b) - ((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*s
qrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + a + b)/(cos(f*x + e)^2 - 1)) - 2*((a + b)*cos(
f*x + e)^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e)), 1/4*(2
*((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b
)/cos(f*x + e)^2)*cos(f*x + e)/a) + ((3*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt(b)*log(-((a - b)*
cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^
2) + 2*((a + b)*cos(f*x + e)^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*c
os(f*x + e)), 1/2*(((a + 3*b)*cos(f*x + e)^3 - (a + 3*b)*cos(f*x + e))*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*
cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) - ((3*a + b)*cos(f*x + e)^3 - (3*a + b)*cos(f*x + e))*sqrt
(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + ((a + b)*cos(f*x + e)
^2 - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(f*cos(f*x + e)^3 - f*cos(f*x + e))]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)**3*(a+b*tan(f*x+e)**2)**(3/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \csc \left (f x + e\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e)^2 + a)^(3/2)*csc(f*x + e)^3, x)