3.597 \(\int \frac {(a+b \tan (e+f x))^3}{\sqrt {d \sec (e+f x)}} \, dx\)
Optimal. Leaf size=178 \[ -\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}+\frac {2 a \left (a^2-6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}} \]
[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)
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Rubi [A] time = 0.14, 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 =
{3512, 739, 780, 227, 196} \[ -\frac {2 b \sec ^2(e+f x) \left (2 \left (3 a^2-2 b^2\right )+3 a b \tan (e+f x)\right )}{3 f \sqrt {d \sec (e+f x)}}-\frac {2 a \left (a^2-6 b^2\right ) \tan (e+f x)}{f \sqrt {d \sec (e+f x)}}+\frac {2 a \left (a^2-6 b^2\right ) \sqrt [4]{\sec ^2(e+f x)} E\left (\left .\frac {1}{2} \tan ^{-1}(\tan (e+f x))\right |2\right )}{f \sqrt {d \sec (e+f x)}}-\frac {2 (b-a \tan (e+f x)) (a+b \tan (e+f x))^2}{f \sqrt {d \sec (e+f x)}} \]
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 196
Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]
Rule 227
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 739
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 780
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 3512
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)} \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}
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Mathematica [A] time = 1.92, size = 130, normalized size = 0.73 \[ \frac {d (a+b \tan (e+f x))^3 \left (6 a \left (a^2-6 b^2\right ) \cos ^{\frac {3}{2}}(e+f x) E\left (\left .\frac {1}{2} (e+f x)\right |2\right )+b \left (\left (3 b^2-9 a^2\right ) \cos (2 (e+f x))-9 a^2+9 a b \sin (2 (e+f x))+5 b^2\right )\right )}{3 f (d \sec (e+f x))^{3/2} (a \cos (e+f x)+b \sin (e+f x))^3} \]
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)
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fricas [F] time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{3} \tan \left (f x + e\right )^{3} + 3 \, a b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} b \tan \left (f x + e\right ) + a^{3}\right )} \sqrt {d \sec \left (f x + e\right )}}{d \sec \left (f x + e\right )}, x\right ) \]
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]
integral((b^3*tan(f*x + e)^3 + 3*a*b^2*tan(f*x + e)^2 + 3*a^2*b*tan(f*x + e) + a^3)*sqrt(d*sec(f*x + e))/(d*se
c(f*x + e)), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt {d \sec \left (f x + e\right )}}\,{d x} \]
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)
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maple [C] time = 1.08, size = 3065, normalized size = 17.22 \[ \text {output too large to display} \]
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)
[Out]
1/6/f*(-1+cos(f*x+e))^2*(36*cos(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a*b^2-72*cos(f*x+e)^3*(-cos(f*x+e)
/(1+cos(f*x+e))^2)^(3/2)*a*b^2+24*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*b^3+36*cos(f*x+
e)^2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a*b^2+12*cos(f*x+e)*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*
b^3-72*cos(f*x+e)^4*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a*b^2+40*cos(f*x+e)^3*sin(f*x+e)*(-cos(f*x+e)/(1+cos(
f*x+e))^2)^(3/2)*b^3+36*cos(f*x+e)^6*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a*b^2+12*cos(f*x+e)^5*sin(f*x+e)*(-c
os(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*b^3+3*cos(f*x+e)^3*sin(f*x+e)*ln(-2*(2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*
cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*b^3-3*cos(f*x+e
)^3*sin(f*x+e)*ln(-(2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x
+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*b^3+72*I*cos(f*x+e)^3*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f
*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3-7
2*I*cos(f*x+e)^3*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/
(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3+48*I*cos(f*x+e)^2*sin(f*x+e)*EllipticF(I*(-1+cos(f
*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))
^(1/2)*a^3-48*I*cos(f*x+e)^2*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x
+e)/(1+cos(f*x+e)))^(1/2)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*a^3+12*I*cos(f*x+e)*sin(f*x+e)*EllipticF(I
*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+co
s(f*x+e)))^(1/2)*a^3-12*I*cos(f*x+e)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(
f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3+36*cos(f*x+e)^5*(-cos(f*x+e)/(
1+cos(f*x+e))^2)^(3/2)*a*b^2+36*cos(f*x+e)^4*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*b^3+12*I*cos(f*x+
e)^5*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e
)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3-12*I*cos(f*x+e)^5*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f
*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3+4
8*I*cos(f*x+e)^4*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/
(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a^3-48*I*cos(f*x+e)^4*sin(f*x+e)*EllipticE(I*(-1+cos(f
*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))
^(1/2)*a^3-12*cos(f*x+e)^6*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a^3+4*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*b^3
*sin(f*x+e)+12*cos(f*x+e)^2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a^3+24*cos(f*x+e)^3*(-cos(f*x+e)/(1+cos(f*x+e
))^2)^(3/2)*a^3-24*cos(f*x+e)^5*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a^3-108*cos(f*x+e)^4*sin(f*x+e)*(-cos(f*x
+e)/(1+cos(f*x+e))^2)^(3/2)*a^2*b-36*cos(f*x+e)^5*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a^2*b-9*cos(
f*x+e)^3*sin(f*x+e)*ln(-2*(2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+
cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2)*a^2*b+9*cos(f*x+e)^3*sin(f*x+e)*ln(-(2*(-cos(f*x+e)/(1+cos(
f*x+e))^2)^(1/2)*cos(f*x+e)^2-cos(f*x+e)^2-2*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)+2*cos(f*x+e)-1)/sin(f*x+e)^2
)*a^2*b-108*cos(f*x+e)^3*sin(f*x+e)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a^2*b-36*cos(f*x+e)^2*sin(f*x+e)*(-co
s(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*a^2*b-72*I*cos(f*x+e)*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-
cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2+72*I*cos(f
*x+e)*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+
e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2-72*I*cos(f*x+e)^5*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/si
n(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*
b^2+72*I*cos(f*x+e)^5*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2
)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2-288*I*cos(f*x+e)^4*sin(f*x+e)*EllipticF(I*(
-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(
f*x+e)))^(1/2)*a*b^2+288*I*cos(f*x+e)^4*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+c
os(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2-432*I*cos(f*x+e)^3*sin(f*
x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(
cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2+432*I*cos(f*x+e)^3*sin(f*x+e)*EllipticE(I*(-1+cos(f*x+e))/sin(f*x+e),I)
*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2-288*I*c
os(f*x+e)^2*sin(f*x+e)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+co
s(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*a*b^2+288*I*cos(f*x+e)^2*sin(f*x+e)*EllipticE(I*(-1+cos(f*x
+e))/sin(f*x+e),I)*(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(
1/2)*a*b^2)/(1+cos(f*x+e))/cos(f*x+e)^2/sin(f*x+e)^5/(d/cos(f*x+e))^(1/2)/(-cos(f*x+e)/(1+cos(f*x+e))^2)^(3/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \tan \left (f x + e\right ) + a\right )}^{3}}{\sqrt {d \sec \left (f x + e\right )}}\,{d x} \]
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)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^3}{\sqrt {\frac {d}{\cos \left (e+f\,x\right )}}} \,d x \]
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)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \tan {\left (e + f x \right )}\right )^{3}}{\sqrt {d \sec {\left (e + f x \right )}}}\, dx \]
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)
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