3.2.9 \(\int \csc ^5(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [109]

Optimal. Leaf size=223 \[ -\frac {3 \left (a^2+6 a b+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{8 \sqrt {a} f}+\frac {3 \sqrt {b} (a+b) \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {3 (a+3 b) \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{8 f}-\frac {3 (a+b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{8 f}-\frac {\cot (e+f x) \csc ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{4 f} \]

[Out]

-1/4*cot(f*x+e)*csc(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(3/2)/f-3/8*(a^2+6*a*b+b^2)*arctanh(sec(f*x+e)*a^(1/2)/(a-b+
b*sec(f*x+e)^2)^(1/2))/f/a^(1/2)+3/2*(a+b)*arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f+3/
8*(a+3*b)*sec(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f-3/8*(a+b)*csc(f*x+e)^2*sec(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)
/f

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Rubi [A]
time = 0.23, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3745, 478, 591, 596, 537, 223, 212, 385, 213} \begin {gather*} -\frac {3 \left (a^2+6 a b+b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{8 \sqrt {a} f}+\frac {3 (a+3 b) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{8 f}-\frac {3 (a+b) \csc ^2(e+f x) \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{8 f}+\frac {3 \sqrt {b} (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 ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(-3*(a^2 + 6*a*b + b^2)*ArcTanh[(Sqrt[a]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(8*Sqrt[a]*f) + (3*Sqr
t[b]*(a + b)*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/(2*f) + (3*(a + 3*b)*Sec[e + f*x]
*Sqrt[a - b + b*Sec[e + f*x]^2])/(8*f) - (3*(a + b)*Csc[e + f*x]^2*Sec[e + f*x]*Sqrt[a - b + b*Sec[e + f*x]^2]
)/(8*f) - (Cot[e + f*x]*Csc[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(3/2))/(4*f)

Rule 212

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

Rule 213

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

Rule 223

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 385

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 478

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 537

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 591

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^q/(a*b*g*n*(p + 1))), x] + Dis
t[1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(
m + 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x]
&& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 596

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

Rule 3745

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

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Mathematica [A]
time = 4.74, size = 415, normalized size = 1.86 \begin {gather*} \frac {\cos (e+f x) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-2 \csc ^2(e+f x) \left (3 a+5 b+2 a \csc ^2(e+f x)\right )+8 b \sec ^2(e+f x)+\frac {3 \left (16 \sqrt {a} \sqrt {b} (a+b) \tanh ^{-1}\left (\frac {-\sqrt {a} \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{2 \sqrt {b}}\right )+\left (a^2+6 a b+b^2\right ) \left (2 \tanh ^{-1}\left (\tan ^2\left (\frac {1}{2} (e+f x)\right )-\frac {\sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}{\sqrt {a}}\right )+\log \left (a-2 b-a \tan ^2\left (\frac {1}{2} (e+f x)\right )+\sqrt {a} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}\right )\right )\right ) \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sqrt {\cos ^2(e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right )}}{\sqrt {a} \sqrt {\left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2} \sqrt {4 b \tan ^2\left (\frac {1}{2} (e+f x)\right )+a \left (-1+\tan ^2\left (\frac {1}{2} (e+f x)\right )\right )^2}}\right )}{16 \sqrt {2} f} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^5*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(Cos[e + f*x]*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-2*Csc[e + f*x]^2*(3*a + 5*b + 2*a*Csc[
e + f*x]^2) + 8*b*Sec[e + f*x]^2 + (3*(16*Sqrt[a]*Sqrt[b]*(a + b)*ArcTanh[(-(Sqrt[a]*(-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])/(2*Sqrt[b])] + (a^2 + 6*a*b + b^2)*(2*ArcTan
h[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[a]] + Log[a - 2*b - a
*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]]))*Sec[(e + f*x)/2]
^2*Sqrt[Cos[e + f*x]^2*Sec[(e + f*x)/2]^4])/(Sqrt[a]*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])))/(16*Sqrt[2]*f)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(6193\) vs. \(2(195)=390\).
time = 0.33, size = 6194, normalized size = 27.78

method	result	size
default	\(\text {Expression too large to display}\)	\(6194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(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)^5, x)

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Fricas [A]
time = 4.12, size = 1449, normalized size = 6.50 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*((a^2 + 6*a*b + b^2)*cos(f*x + e)^5 - 2*(a^2 + 6*a*b + b^2)*cos(f*x + e)^3 + (a^2 + 6*a*b + b^2)*cos(
f*x + e))*sqrt(a)*log(-2*((a - b)*cos(f*x + e)^2 - 2*sqrt(a)*sqrt(((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)) + 12*((a^2 + a*b)*cos(f*x + e)^5 - 2*(a^2 + a*b)*cos(f*x + e)^3 +
(a^2 + 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*(3*(a^2 + 3*a*b)*cos(f*x + e)^4 - (5*a^2 + 13*a*b)*cos
(f*x + e)^2 + 4*a*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x +
e)^3 + a*f*cos(f*x + e)), 1/8*(3*((a^2 + 6*a*b + b^2)*cos(f*x + e)^5 - 2*(a^2 + 6*a*b + b^2)*cos(f*x + e)^3 +
(a^2 + 6*a*b + b^2)*cos(f*x + e))*sqrt(-a)*arctan(sqrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*c
os(f*x + e)/a) + 6*((a^2 + a*b)*cos(f*x + e)^5 - 2*(a^2 + a*b)*cos(f*x + e)^3 + (a^2 + 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) + (3*(a^2 + 3*a*b)*cos(f*x + e)^4 - (5*a^2 + 13*a*b)*cos(f*x + e)^2 + 4*a*b)*sqrt(((a - b
)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x + e)^3 + a*f*cos(f*x + e)), -1/16*(
24*((a^2 + a*b)*cos(f*x + e)^5 - 2*(a^2 + a*b)*cos(f*x + e)^3 + (a^2 + 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) - 3*((a^2 + 6*a*b + b^2)*cos(f*x + e)^5
- 2*(a^2 + 6*a*b + b^2)*cos(f*x + e)^3 + (a^2 + 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*log(-2*((a - b)*cos(f*x +
e)^2 - 2*sqrt(a)*sqrt(((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*(3*(a^2 + 3*a*b)*cos(f*x + e)^4 - (5*a^2 + 13*a*b)*cos(f*x + e)^2 + 4*a*b)*sqrt(((a - b)*cos(f*x + e)^2 +
b)/cos(f*x + e)^2))/(a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x + e)^3 + a*f*cos(f*x + e)), 1/8*(3*((a^2 + 6*a*b + b^
2)*cos(f*x + e)^5 - 2*(a^2 + 6*a*b + b^2)*cos(f*x + e)^3 + (a^2 + 6*a*b + b^2)*cos(f*x + e))*sqrt(-a)*arctan(s
qrt(-a)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/a) - 12*((a^2 + a*b)*cos(f*x + e)^5 - 2
*(a^2 + a*b)*cos(f*x + e)^3 + (a^2 + 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) + (3*(a^2 + 3*a*b)*cos(f*x + e)^4 - (5*a^2 + 13*a*b)*cos(f*x + e)^2 + 4*a
*b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x + e)^3 + a*f*cos(f*
x + e))]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**5*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^5*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(t_

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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