3.5.21 \(\int \frac {\tan ^6(e+f x)}{(a+b \sec ^2(e+f x))^{3/2}} \, dx\) [421]
Optimal. Leaf size=172 \[ -\frac {\text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {(3 a+5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f} \]
[Out]
-arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/a^(3/2)/f-1/2*(3*a+5*b)*arctanh(b^(1/2)*tan(f*x+e)/(a+b
+b*tan(f*x+e)^2)^(1/2))/b^(5/2)/f+1/2*(3*a+2*b)*(a+b+b*tan(f*x+e)^2)^(1/2)*tan(f*x+e)/a/b^2/f-(a+b)*tan(f*x+e)
^3/a/b/f/(a+b+b*tan(f*x+e)^2)^(1/2)
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Rubi [A]
time = 0.23, antiderivative size = 172, 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 = {4226, 2000,
481, 596, 537, 223, 212, 385, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{a^{3/2} f}-\frac {(3 a+5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{2 b^{5/2} f}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{2 a b^2 f}-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b \tan ^2(e+f x)+b}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
-(ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]]/(a^(3/2)*f)) - ((3*a + 5*b)*ArcTanh[(Sqrt[b]*T
an[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/(2*b^(5/2)*f) - ((a + b)*Tan[e + f*x]^3)/(a*b*f*Sqrt[a + b + b*T
an[e + f*x]^2]) + ((3*a + 2*b)*Tan[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/(2*a*b^2*f)
Rule 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
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 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 481
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 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] && GtQ[m - n + 1, n] && 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 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 2000
Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&& !BinomialMatchQ[{u, v}, x]
Rule 4226
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[(d*ff*x)^m*((a + b*(1 + ff^2*x^2)^(n/2))^p/(1 + ff^2
*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])
Rubi steps
\begin {align*} \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b+b x^2\right )^{3/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (a+b)+(3 a+2 b) x^2\right )}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a b f}\\ &=-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f}-\frac {\text {Subst}\left (\int \frac {(a+b) (3 a+2 b)+a (3 a+5 b) x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 a b^2 f}\\ &=-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f}-\frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{a f}-\frac {(3 a+5 b) \text {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 b^2 f}\\ &=-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f}-\frac {\text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a f}-\frac {(3 a+5 b) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^2 f}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{a^{3/2} f}-\frac {(3 a+5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{2 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{a b f \sqrt {a+b+b \tan ^2(e+f x)}}+\frac {(3 a+2 b) \tan (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{2 a b^2 f}\\ \end {align*}
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Mathematica [A]
time = 10.10, size = 247, normalized size = 1.44 \begin {gather*} -\frac {\left (\frac {2 b^2 \text {ArcTan}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {a}}+\frac {a (3 a+5 b) \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b-a \sin ^2(e+f x)}}\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 e+2 f x))^{3/2} \sec ^3(e+f x)}{4 \sqrt {2} a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {(a+2 b+a \cos (2 (e+f x))) \left (3 a^2+6 a b+2 b^2+\left (3 a^2+4 a b+2 b^2\right ) \cos (2 (e+f x))\right ) \sec ^4(e+f x) \tan (e+f x)}{8 a b^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2)^(3/2),x]
[Out]
-1/4*(((2*b^2*ArcTan[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[a] + (a*(3*a + 5*b)*ArcTanh[
(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b - a*Sin[e + f*x]^2]])/Sqrt[b])*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2)*Sec[e +
f*x]^3)/(Sqrt[2]*a*b^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) + ((a + 2*b + a*Cos[2*(e + f*x)])*(3*a^2 + 6*a*b + 2*b^
2 + (3*a^2 + 4*a*b + 2*b^2)*Cos[2*(e + f*x)])*Sec[e + f*x]^4*Tan[e + f*x])/(8*a*b^2*f*(a + b*Sec[e + f*x]^2)^(
3/2))
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.20, size = 3860, normalized size = 22.44
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\(\text {Expression too large to display}\) |
\(3860\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
[Out]
-1/4/f*(8*EllipticPi((cos(f*x+e)-1)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)
+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*sin(f*x+e)*cos(f
*x+e)^2*2^(1/2)*b^(7/2)*((I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e))/(a+b))
^(1/2)*(-2*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+cos(f*x+e))/(a+b))^(1/2)+7*arcta
nh(1/4*(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/((b+a*cos(f*x+e)^2)/(1+cos(f*x+
e))^2)^(1/2)*b^(1/2))*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*((2*I*a^(1/2)*b^(1/2
)+a-b)/(a+b))^(1/2)*a*b^2-7*arctanh(1/8*(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/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*4^(1/2))*sin(f*x+e)*cos(f*x+e)^2*((b+a*cos(f*x+e)^2)/(1+
cos(f*x+e))^2)^(1/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b^2-4*EllipticF((cos(f*x+e)-1)*((2*I*a^(1/2)*b^
(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*s
in(f*x+e)*cos(f*x+e)^2*2^(1/2)*b^(7/2)*((I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos
(f*x+e))/(a+b))^(1/2)*(-2*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+cos(f*x+e))/(a+b)
)^(1/2)+7*arctanh(1/4*(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/((b+a*cos(f*x+e)
^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2))*sin(f*x+e)*cos(f*x+e)^4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*((2*I
*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-7*arctanh(1/8*(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/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*4^(1/2))*sin(f*x+e)*cos(f*x+e)^4*((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2*b-8*cos(f*x+e)^3*b^(5/2)*((2*I
*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a+4*cos(f*x+e)^4*b^(5/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a-8*cos(f*
x+e)^3*b^(3/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2+8*cos(f*x+e)^4*b^(3/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(
a+b))^(1/2)*a^2-8*cos(f*x+e)^5*b^(3/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a^2-4*cos(f*x+e)^5*b^(5/2)*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a+8*cos(f*x+e)^2*b^(5/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a-2*cos(f
*x+e)*b^(5/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a+8*cos(f*x+e)^2*b^(3/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b
))^(1/2)*a^2+6*cos(f*x+e)^4*a^3*b^(1/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)-6*cos(f*x+e)^5*a^3*b^(1/2)*((2
*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)-10*EllipticF((cos(f*x+e)-1)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f
*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*sin(f*x+e)*cos(f*x+e)^4*2^(1/2
)*b^(3/2)*((I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*
cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+cos(f*x+e))/(a+b))^(1/2)*a^2+12*EllipticPi((co
s(f*x+e)-1)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2
)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*a^3*b^(1/
2)*((I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*cos(f*x
+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+cos(f*x+e))/(a+b))^(1/2)-4*EllipticF((cos(f*x+e)-1)*(
(2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a
+b)^2)^(1/2))*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*b^(5/2)*((I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+cos(f*x
+e)*a+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+co
s(f*x+e))/(a+b))^(1/2)*a+4*cos(f*x+e)^2*b^(7/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)-4*cos(f*x+e)^3*b^(7/2)
*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)+2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^(5/2)*a-6*EllipticF((cos(
f*x+e)-1)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*
a*b-b^2)/(a+b)^2)^(1/2))*sin(f*x+e)*cos(f*x+e)^4*2^(1/2)*a^3*b^(1/2)*((I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*
b^(1/2)+cos(f*x+e)*a+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-cos(f*
x+e)*a-b)/(1+cos(f*x+e))/(a+b))^(1/2)+20*EllipticPi((cos(f*x+e)-1)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin
(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/
(a+b))^(1/2))*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*b^(5/2)*((I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+cos(f*x
+e)*a+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)-cos(f*x+e)*a-b)/(1+co
s(f*x+e))/(a+b))^(1/2)*a-10*EllipticF((cos(f*x+e)-1)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I
*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*sin(f*x+e)*cos(f*x+e)^2*2^(1/2)*b^(5/2)*((
I*cos(f*x+e)*a^(1/2)*b^(1/2)-I*a^(1/2)*b^(1/2)+...
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 422 vs.
\(2 (160) = 320\).
time = 8.49, size = 2001, normalized size = 11.63 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
[-1/8*((a*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f
*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 -
32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x +
e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - ((3*a^4 + 5*a^3*b)*cos(f*x + e)^3 + (3*a^3*b + 5*a
^2*b^2)*cos(f*x + e))*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)^2 - 4*((a -
b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e) + 8*b^
2)/cos(f*x + e)^4) - 4*(a^2*b^2 + (3*a^3*b + 4*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/
cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f*x + e)^3 + a^2*b^4*f*cos(f*x + e)), -1/8*(2*((3*a^4 + 5*a^3*b)*
cos(f*x + e)^3 + (3*a^3*b + 5*a^2*b^2)*cos(f*x + e))*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*
x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) + (a*b
^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(-a)*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 +
32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7
*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 - 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5
*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos
(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 4*(a^2*b^2 + (3*a^3*b + 4*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f*x + e)^3 + a^2*b^4*f*cos(f*x + e)),
1/8*(2*(a*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos
(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*co
s(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) + ((3*a^4 + 5*a^3*b)*cos(f*x + e
)^3 + (3*a^3*b + 5*a^2*b^2)*cos(f*x + e))*sqrt(b)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(
f*x + e)^2 - 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)
*sin(f*x + e) + 8*b^2)/cos(f*x + e)^4) + 4*(a^2*b^2 + (3*a^3*b + 4*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*
cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f*x + e)^3 + a^2*b^4*f*cos(f*x + e)), 1/4*((a
*b^3*cos(f*x + e)^3 + b^4*cos(f*x + e))*sqrt(a)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^
3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)
^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e))) - ((3*a^4 + 5*a^3*b)*cos(f*x + e)^3 + (3*a
^3*b + 5*a^2*b^2)*cos(f*x + e))*sqrt(-b)*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt
((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e))) + 2*(a^2*b^2 + (3*a^3*b + 4
*a^2*b^2 + 2*a*b^3)*cos(f*x + e)^2)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))/(a^3*b^3*f*cos(f
*x + e)^3 + a^2*b^4*f*cos(f*x + e))]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)**6/(a+b*sec(f*x+e)**2)**(3/2),x)
[Out]
Integral(tan(e + f*x)**6/(a + b*sec(e + f*x)**2)**(3/2), x)
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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(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate(tan(f*x + e)^6/(b*sec(f*x + e)^2 + a)^(3/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^6}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2)^(3/2),x)
[Out]
int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2)^(3/2), x)
________________________________________________________________________________________