3.6.97 \(\int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx\) [597]
Optimal. Leaf size=178 \[ \frac {2 a \left (a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}} \]
[Out]
2*a*(a^2-6*b^2)*(cos(1/2*arctan(tan(f*x+e)))^2)^(1/2)/cos(1/2*arctan(tan(f*x+e)))*EllipticE(sin(1/2*arctan(tan
(f*x+e))),2^(1/2))*(sec(f*x+e)^2)^(1/4)/f/(d*sec(f*x+e))^(1/2)-2*a*(a^2-6*b^2)*tan(f*x+e)/f/(d*sec(f*x+e))^(1/
2)-2*(b-a*tan(f*x+e))*(a+b*tan(f*x+e))^2/f/(d*sec(f*x+e))^(1/2)-2/3*b*sec(f*x+e)^2*(6*a^2-4*b^2+3*a*b*tan(f*x+
e))/f/(d*sec(f*x+e))^(1/2)
________________________________________________________________________________________
Rubi [A]
time = 0.10, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3593, 753, 794,
233, 202} \begin {gather*} \frac {2 a \left (a^2-6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \text {ArcTan}(\tan (e+f x))\right |2\right )}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(a + b*Tan[e + f*x])^3/Sqrt[d*Sec[e + f*x]],x]
[Out]
(2*a*(a^2 - 6*b^2)*EllipticE[ArcTan[Tan[e + f*x]]/2, 2]*(Sec[e + f*x]^2)^(1/4))/(f*Sqrt[d*Sec[e + f*x]]) - (2*
a*(a^2 - 6*b^2)*Tan[e + f*x])/(f*Sqrt[d*Sec[e + f*x]]) - (2*(b - a*Tan[e + f*x])*(a + b*Tan[e + f*x])^2)/(f*Sq
rt[d*Sec[e + f*x]]) - (2*b*Sec[e + f*x]^2*(2*(3*a^2 - 2*b^2) + 3*a*b*Tan[e + f*x]))/(3*f*Sqrt[d*Sec[e + f*x]])
Rule 202
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]
Rule 233
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]
Rule 753
Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a*e - c*d*x)*((a
+ c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]
Rule 794
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*((a + c*x^2)^(p + 1)/(2*c*(p + 1)*(2*p + 3))), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] && !LeQ[p, -1]
Rule 3593
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[d^(2*
IntPart[m/2])*((d*Sec[e + f*x])^(2*FracPart[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2])), Subst[Int[(a + x)^n*(
1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] &&
!IntegerQ[m/2]
Rubi steps
\begin {align*} \int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx &=\frac {\sqrt [4]{\sec ^2(e+f x)} \text {Subst}\left (\int \frac {(a+x)^3}{\left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{b f \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}+\frac {\left (2 b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {(a+x) \left (\frac {1}{2} \left (4-\frac {a^2}{b^2}\right )-\frac {5 a x}{2 b^2}\right )}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{f \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}+\frac {\left (a \left (6-\frac {a^2}{b^2}\right ) b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{f \sqrt {d \sec (e+f x)}}\\ &=-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}-\frac {\left (a \left (6-\frac {a^2}{b^2}\right ) b \sqrt [4]{\sec ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (1+\frac {x^2}{b^2}\right )^{5/4}} \, dx,x,b \tan (e+f x)\right )}{f \sqrt {d \sec (e+f x)}}\\ &=\frac {2 a \left (a^2-6 b^2\right ) E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right ) \sqrt [4]{\sec ^2(e+f x)}}{f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}}-\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A]
time = 2.06, size = 130, normalized size = 0.73 \begin {gather*} \frac {d \left (6 a \left (a^2-6 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+b \left (-9 a^2+5 b^2+\left (-9 a^2+3 b^2\right ) \cos (2 (e+f x))+9 a b \sin (2 (e+f x))\right )\right ) (a+b \tan (e+f x))^3}{3 f (d \sec (e+f x))^{3/2} (a \cos (e+f x)+b \sin (e+f x))^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(a + b*Tan[e + f*x])^3/Sqrt[d*Sec[e + f*x]],x]
[Out]
(d*(6*a*(a^2 - 6*b^2)*Cos[e + f*x]^(3/2)*EllipticE[(e + f*x)/2, 2] + b*(-9*a^2 + 5*b^2 + (-9*a^2 + 3*b^2)*Cos[
2*(e + f*x)] + 9*a*b*Sin[2*(e + f*x)]))*(a + b*Tan[e + f*x])^3)/(3*f*(d*Sec[e + f*x])^(3/2)*(a*Cos[e + f*x] +
b*Sin[e + f*x])^3)
________________________________________________________________________________________
Maple [C] Result contains complex when optimal does not.
time = 0.62, size = 3065, normalized size = 17.22
| | |
| method |
result |
size |
| | |
| default |
\(\text {Expression too large to display}\) |
\(3065\) |
| | |
|
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/6/f*(cos(f*x+e)-1)^2*(12*I*cos(f*x+e)^5*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(co
s(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3-12*I*cos(f*x+e)^5*sin(f*x+
e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos
(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3+48*I*cos(f*x+e)^4*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(
f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3-48*I*cos(f*x+e)^
4*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^
(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3+72*I*cos(f*x+e)^3*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e)
,I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3-72*I*c
os(f*x+e)^3*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f
*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a^3+48*I*cos(f*x+e)^2*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/
sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*
a^3-48*I*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^3+12*I*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f
*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticF(I*(cos(f*x+e)-1)/sin(f*
x+e),I)*a^3-12*I*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+
e)/(cos(f*x+e)+1))^(1/2)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*a^3+4*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b
^3*sin(f*x+e)+24*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a^3-12*cos(f*x+e)^6*(-cos(f*x+e)/(cos(f*x+e
)+1)^2)^(3/2)*a^3+12*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a^3-24*cos(f*x+e)^5*(-cos(f*x+e)/(cos(f
*x+e)+1)^2)^(3/2)*a^3-288*I*cos(f*x+e)^4*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos
(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2+288*I*cos(f*x+e)^4*sin(f*
x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(c
os(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2-432*I*cos(f*x+e)^3*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2+432*I*cos
(f*x+e)^3*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x
+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2-288*I*cos(f*x+e)^2*sin(f*x+e)*EllipticF(I*(cos(f*x+e)-1)
/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)
*a*b^2+288*I*cos(f*x+e)^2*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(
3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2-72*I*cos(f*x+e)^5*sin(f*x+e)*EllipticF(I
*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f
*x+e)+1))^(1/2)*a*b^2+72*I*cos(f*x+e)^5*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(
f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2-72*I*cos(f*x+e)*sin(f*x+e)
*EllipticF(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f
*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2+72*I*cos(f*x+e)*sin(f*x+e)*EllipticE(I*(cos(f*x+e)-1)/sin(f*x+e),I)*(-cos(f*
x+e)/(cos(f*x+e)+1)^2)^(3/2)*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*a*b^2-36*cos(f*x+e)^5*
sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a^2*b+9*cos(f*x+e)^3*sin(f*x+e)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+
e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*a
^2*b-9*cos(f*x+e)^3*sin(f*x+e)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*cos(f
*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*a^2*b-108*cos(f*x+e)^3*sin(f*x+e)*(-cos(f*x+e)/(
cos(f*x+e)+1)^2)^(3/2)*a^2*b-36*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a^2*b-108*cos(f*x
+e)^4*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a^2*b+36*cos(f*x+e)^6*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/
2)*a*b^2+3*cos(f*x+e)^3*sin(f*x+e)*ln(-2*(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-cos(f*x+e)^2+2*c
os(f*x+e)-2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-1)/sin(f*x+e)^2)*b^3+40*cos(f*x+e)^3*sin(f*x+e)*(-cos(f*x+e)/
(cos(f*x+e)+1)^2)^(3/2)*b^3-72*cos(f*x+e)^3*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a*b^2+24*cos(f*x+e)^2*sin(f*x
+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b^3+36*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*a*b^2+12*cos
(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)*b^3+36*cos(f*x+e)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(3/2)
*a*b^2+12*cos(f*x+e)^5*sin(f*x+e)*(-cos(f*x+e)/...
________________________________________________________________________________________
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((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate((b*tan(f*x + e) + a)^3/sqrt(d*sec(f*x + e)), x)
________________________________________________________________________________________
Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 0.13, size = 180, normalized size = 1.01 \begin {gather*} -\frac {3 \, \sqrt {2} {\left (-i \, a^{3} + 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\right ) + 3 \, \sqrt {2} {\left (i \, a^{3} - 6 i \, a b^{2}\right )} \sqrt {d} \cos \left (f x + e\right ) {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\right ) - 2 \, {\left (9 \, a b^{2} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b^{3} - 3 \, {\left (3 \, a^{2} b - b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {d}{\cos \left (f x + e\right )}}}{3 \, d f \cos \left (f x + e\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
-1/3*(3*sqrt(2)*(-I*a^3 + 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, co
s(f*x + e) + I*sin(f*x + e))) + 3*sqrt(2)*(I*a^3 - 6*I*a*b^2)*sqrt(d)*cos(f*x + e)*weierstrassZeta(-4, 0, weie
rstrassPInverse(-4, 0, cos(f*x + e) - I*sin(f*x + e))) - 2*(9*a*b^2*cos(f*x + e)*sin(f*x + e) + b^3 - 3*(3*a^2
*b - b^3)*cos(f*x + e)^2)*sqrt(d/cos(f*x + e)))/(d*f*cos(f*x + e))
________________________________________________________________________________________
Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\sqrt {d \sec {\left (e + f x \right )}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*tan(f*x+e))**3/(d*sec(f*x+e))**(1/2),x)
[Out]
Integral((a + b*tan(e + f*x))**3/sqrt(d*sec(e + f*x)), x)
________________________________________________________________________________________
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((a+b*tan(f*x+e))^3/(d*sec(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate((b*tan(f*x + e) + a)^3/sqrt(d*sec(f*x + e)), x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(1/2),x)
[Out]
int((a + b*tan(e + f*x))^3/(d/cos(e + f*x))^(1/2), x)
________________________________________________________________________________________