3.2.14 \(\int \csc ^4(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\) [114]
Optimal. Leaf size=162 \[ \frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b (3 a+2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}-\frac {(3 a+2 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f} \]
[Out]
1/2*(3*a+2*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))*b^(1/2)/f+1/2*b*(3*a+2*b)*(a+b*tan(f*x+e)^2
)^(1/2)*tan(f*x+e)/a/f-1/3*(3*a+2*b)*cot(f*x+e)*(a+b*tan(f*x+e)^2)^(3/2)/a/f-1/3*cot(f*x+e)^3*(a+b*tan(f*x+e)^
2)^(5/2)/a/f
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Rubi [A]
time = 0.09, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps
used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3744, 464, 283,
201, 223, 212} \begin {gather*} \frac {b (3 a+2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}+\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}-\frac {(3 a+2 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
(Sqrt[b]*(3*a + 2*b)*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*f) + (b*(3*a + 2*b)*Tan[e
+ f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(2*a*f) - ((3*a + 2*b)*Cot[e + f*x]*(a + b*Tan[e + f*x]^2)^(3/2))/(3*a*f) -
(Cot[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(5/2))/(3*a*f)
Rule 201
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^p/(n*p + 1)), x] + Dist[a*n*(p/(n*p + 1)),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])
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 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 283
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 464
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]
Rule 3744
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff^(m + 1)/f), Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)
^(m/2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]
Rubi steps
\begin {align*} \int \csc ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right ) \left (a+b x^2\right )^{3/2}}{x^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}+\frac {(3 a+2 b) \text {Subst}\left (\int \frac {\left (a+b x^2\right )^{3/2}}{x^2} \, dx,x,\tan (e+f x)\right )}{3 a f}\\ &=-\frac {(3 a+2 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}+\frac {(b (3 a+2 b)) \text {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\tan (e+f x)\right )}{a f}\\ &=\frac {b (3 a+2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}-\frac {(3 a+2 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}+\frac {(b (3 a+2 b)) \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {b (3 a+2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}-\frac {(3 a+2 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}+\frac {(b (3 a+2 b)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}\\ &=\frac {\sqrt {b} (3 a+2 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac {b (3 a+2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)}}{2 a f}-\frac {(3 a+2 b) \cot (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 a f}-\frac {\cot ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{5/2}}{3 a f}\\ \end {align*}
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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in
optimal.
time = 1.96, size = 177, normalized size = 1.09 \begin {gather*} \frac {\sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)} \left (-4 (a+2 b) \cot (e+f x)-2 a \cot (e+f x) \csc ^2(e+f x)+\frac {3 \sqrt {2} (3 a+2 b) \cot (e+f x) F\left (\left .\text {ArcSin}\left (\frac {\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt {2}}\right )\right |1\right )}{\sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}+3 b \tan (e+f x)\right )}{6 \sqrt {2} f} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Csc[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]
[Out]
(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*(-4*(a + 2*b)*Cot[e + f*x] - 2*a*Cot[e + f*x]*Csc[e +
f*x]^2 + (3*Sqrt[2]*(3*a + 2*b)*Cot[e + f*x]*EllipticF[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e
+ f*x]^2)/b]/Sqrt[2]], 1])/Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b] + 3*b*Tan[e + f*x]))/(6
*Sqrt[2]*f)
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.40, size = 4594, normalized size = 28.36
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\(4594\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
[Out]
-1/6/f*(4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^4*a*b+3*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/
2)*cos(f*x+e)^2*a*b+12*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x
+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(
f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+
e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-
2*b)/a)^(1/2))*sin(f*x+e)*cos(f*x+e)^2*b^2+9*sin(f*x+e)*cos(f*x+e)^5*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2
)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(
1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I
*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+
8*b^2)/a^2)^(1/2))*a*b-18*sin(f*x+e)*cos(f*x+e)^5*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(
1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b
)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/
2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/(
(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a*b+9*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(
1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b
)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((
2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a
*b+8*b^2)/a^2)^(1/2))*a*b-18*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b
)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(
a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^
(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2
)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a*b-3*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2-9*sin(f*x+e)*
cos(f*x+e)^3*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(co
s(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/
(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^
(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a*b+18*sin(f*x+e)*cos(f*x+e)^3*2^(1/2
)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2
)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(
1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)
+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a*b-9*sin(f*x+
e)*cos(f*x+e)^2*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/
(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-
b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I
*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a*b+18*sin(f*x+e)*cos(f*x+e)^2*2^(
1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(
1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a
)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1
/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a*b-7*((2*I
*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^6*a*b+6*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(
a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2
)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b
)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(
1/2))*sin(f*x+e)*cos(f*x+e)^5*b^2-12*2^(1/2)*((I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+
e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1)/a)^(1/2)*(-2*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f
*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1)/a)^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^
(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2...
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Maxima [A]
time = 0.28, size = 187, normalized size = 1.15 \begin {gather*} \frac {9 \, a \sqrt {b} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) + 6 \, b^{\frac {3}{2}} \operatorname {arsinh}\left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right ) + 9 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b \tan \left (f x + e\right ) + \frac {6 \, \sqrt {b \tan \left (f x + e\right )^{2} + a} b^{2} \tan \left (f x + e\right )}{a} - \frac {6 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{\tan \left (f x + e\right )} - \frac {4 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} b}{a \tan \left (f x + e\right )} - \frac {2 \, {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}}{a \tan \left (f x + e\right )^{3}}}{6 \, f} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
1/6*(9*a*sqrt(b)*arcsinh(b*tan(f*x + e)/sqrt(a*b)) + 6*b^(3/2)*arcsinh(b*tan(f*x + e)/sqrt(a*b)) + 9*sqrt(b*ta
n(f*x + e)^2 + a)*b*tan(f*x + e) + 6*sqrt(b*tan(f*x + e)^2 + a)*b^2*tan(f*x + e)/a - 6*(b*tan(f*x + e)^2 + a)^
(3/2)/tan(f*x + e) - 4*(b*tan(f*x + e)^2 + a)^(3/2)*b/(a*tan(f*x + e)) - 2*(b*tan(f*x + e)^2 + a)^(5/2)/(a*tan
(f*x + e)^3))/f
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Fricas [A]
time = 2.74, size = 531, normalized size = 3.28 \begin {gather*} \left [\frac {3 \, {\left ({\left (3 \, a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 2 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {b} \log \left (\frac {{\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 8 \, {\left (a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right ) + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) \sin \left (f x + e\right ) - 4 \, {\left ({\left (4 \, a + 11 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{24 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}, -\frac {3 \, {\left ({\left (3 \, a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a + 2 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \arctan \left (\frac {{\left ({\left (a - 2 \, b\right )} \cos \left (f x + e\right )^{3} + 2 \, b \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, {\left ({\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 2 \, {\left ({\left (4 \, a + 11 \, b\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} + 3 \, b\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{12 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[1/24*(3*((3*a + 2*b)*cos(f*x + e)^3 - (3*a + 2*b)*cos(f*x + e))*sqrt(b)*log(((a^2 - 8*a*b + 8*b^2)*cos(f*x +
e)^4 + 8*(a*b - 2*b^2)*cos(f*x + e)^2 + 4*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt(((a - b)*
cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 4*((4*a + 11*b)*cos(f
*x + e)^4 - 2*(3*a + 7*b)*cos(f*x + e)^2 + 3*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))*sin(f*x + e)), -1/12*(3*((3*a + 2*b)*cos(f*x + e)^3 - (3*a + 2*b)*cos(f*x + e))*sqrt
(-b)*arctan(1/2*((a - 2*b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f
*x + e)^2)/(((a*b - b^2)*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*sin(f*x + e) + 2*((4*a + 11*b)*cos(f*x + e)^4 -
2*(3*a + 7*b)*cos(f*x + e)^2 + 3*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))*sin(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)**4*(a+b*tan(f*x+e)**2)**(3/2),x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 3004 deep
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)^4*(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)^4, x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\sin \left (e+f\,x\right )}^4} \,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)^4,x)
[Out]
int((a + b*tan(e + f*x)^2)^(3/2)/sin(e + f*x)^4, x)
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