3.7.5 \(\int \frac {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx\) [605]

Optimal. Leaf size=334 \[ \frac {\text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {a \cot (e+f x) \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}+\frac {a \cot (e+f x) \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}} \]

[Out]

arctan((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(3/2)/(a^2+b^2)^(1/4)/f/(sec(f*x+e)^2)^(3/
4)/b^(1/2)-arctanh((sec(f*x+e)^2)^(1/4)*b^(1/2)/(a^2+b^2)^(1/4))*(d*sec(f*x+e))^(3/2)/(a^2+b^2)^(1/4)/f/(sec(f
*x+e)^2)^(3/4)/b^(1/2)-a*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),-b/(a^2+b^2)^(1/2),I)*(d*sec(f*x+e))^(3/2)
*(-tan(f*x+e)^2)^(1/2)/b/f/(sec(f*x+e)^2)^(3/4)/(a^2+b^2)^(1/2)+a*cot(f*x+e)*EllipticPi((sec(f*x+e)^2)^(1/4),b
/(a^2+b^2)^(1/2),I)*(d*sec(f*x+e))^(3/2)*(-tan(f*x+e)^2)^(1/2)/b/f/(sec(f*x+e)^2)^(3/4)/(a^2+b^2)^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 334, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3593, 760, 408, 504, 1227, 551, 455, 65, 304, 211, 214} \begin {gather*} -\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{b f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\text {ArcSin}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{b f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac {(d \sec (e+f x))^{3/2} \text {ArcTan}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{\sqrt {b} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac {(d \sec (e+f x))^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{\sqrt {b} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTan[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(d*Sec[e + f*x])^(3/2))/(Sqrt[b]*(a^2 + b^2)^(1/4)
*f*(Sec[e + f*x]^2)^(3/4)) - (ArcTanh[(Sqrt[b]*(Sec[e + f*x]^2)^(1/4))/(a^2 + b^2)^(1/4)]*(d*Sec[e + f*x])^(3/
2))/(Sqrt[b]*(a^2 + b^2)^(1/4)*f*(Sec[e + f*x]^2)^(3/4)) - (a*Cot[e + f*x]*EllipticPi[-(b/Sqrt[a^2 + b^2]), Ar
cSin[(Sec[e + f*x]^2)^(1/4)], -1]*(d*Sec[e + f*x])^(3/2)*Sqrt[-Tan[e + f*x]^2])/(b*Sqrt[a^2 + b^2]*f*(Sec[e +
f*x]^2)^(3/4)) + (a*Cot[e + f*x]*EllipticPi[b/Sqrt[a^2 + b^2], ArcSin[(Sec[e + f*x]^2)^(1/4)], -1]*(d*Sec[e +
f*x])^(3/2)*Sqrt[-Tan[e + f*x]^2])/(b*Sqrt[a^2 + b^2]*f*(Sec[e + f*x]^2)^(3/4))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

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

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 408

Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[2*(Sqrt[(-b)*(x^2/a)]/x), Subst[I
nt[x^2/(Sqrt[1 - x^4/a]*(b*c - a*d + d*x^4)), x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b
*c - a*d, 0]

Rule 455

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 504

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s
= Denominator[Rt[-a/b, 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), Int[1/
((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 551

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1/(a*Sqr
t[c]*Sqrt[e]*Rt[-d/c, 2]))*EllipticPi[b*(c/(a*d)), ArcSin[Rt[-d/c, 2]*x], c*(f/(d*e))], x] /; FreeQ[{a, b, c,
d, e, f}, x] &&  !GtQ[d/c, 0] && GtQ[c, 0] && GtQ[e, 0] &&  !( !GtQ[f/e, 0] && SimplerSqrtQ[-f/e, -d/c])

Rule 760

Int[1/(((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(1/4)), x_Symbol] :> Dist[d, Int[1/((d^2 - e^2*x^2)*(a + c*x^
2)^(1/4)), x], x] - Dist[e, Int[x/((d^2 - e^2*x^2)*(a + c*x^2)^(1/4)), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ
[c*d^2 + a*e^2, 0]

Rule 1227

Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[Sqrt[-c],
Int[1/((d + e*x^2)*Sqrt[q + c*x^2]*Sqrt[q - c*x^2]), x], x]] /; FreeQ[{a, c, d, e}, x] && GtQ[a, 0] && LtQ[c,
0]

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 {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx &=\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{(a+x) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {x}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}}+\frac {\left (a (d \sec (e+f x))^{3/2}\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-x^2\right ) \sqrt [4]{1+\frac {x^2}{b^2}}} \, dx,x,b \tan (e+f x)\right )}{b f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{\left (a^2-x\right ) \sqrt [4]{1+\frac {x}{b^2}}} \, dx,x,b^2 \tan ^2(e+f x)\right )}{2 b f \sec ^2(e+f x)^{3/4}}+\frac {\left (2 a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1-x^4} \left (1+\frac {a^2}{b^2}-x^4\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b^2 f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {\left (2 b (d \sec (e+f x))^{3/2}\right ) \text {Subst}\left (\int \frac {x^2}{a^2+b^2-b^2 x^4} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sec ^2(e+f x)^{3/4}}+\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b^2}-b x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}-\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\left (\sqrt {a^2+b^2}+b x^2\right ) \sqrt {1-x^4}} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\\ &=-\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}-b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sec ^2(e+f x)^{3/4}}+\frac {(d \sec (e+f x))^{3/2} \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}+b x^2} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{f \sec ^2(e+f x)^{3/4}}+\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a^2+b^2}-b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}-\frac {\left (a \cot (e+f x) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+x^2} \left (\sqrt {a^2+b^2}+b x^2\right )} \, dx,x,\sqrt [4]{\sec ^2(e+f x)}\right )}{b f \sec ^2(e+f x)^{3/4}}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right ) (d \sec (e+f x))^{3/2}}{\sqrt {b} \sqrt [4]{a^2+b^2} f \sec ^2(e+f x)^{3/4}}-\frac {a \cot (e+f x) \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}+\frac {a \cot (e+f x) \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right ) (d \sec (e+f x))^{3/2} \sqrt {-\tan ^2(e+f x)}}{b \sqrt {a^2+b^2} f \sec ^2(e+f x)^{3/4}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 42.93, size = 6301, normalized size = 18.87 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3736 vs. \(2 (282 ) = 564\).
time = 0.80, size = 3737, normalized size = 11.19

method	result	size
default	\(\text {Expression too large to display}\)	\(3737\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(cos(f*x+e)+1)^2*(-4*I*b*a^3*(1/(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticPi(I*(co
s(f*x+e)-1)/sin(f*x+e),-1/(-b+(a^2+b^2)^(1/2))^2*a^2,I)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+
2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(a^2+b^2)^(1/2)+4*I
*(a^2+b^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)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2
+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a*b-4*I*(a^2+b^2)^(1/2)*(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)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+
2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a*b^3-4*I*b*a^3*(1/
(cos(f*x+e)+1))^(1/2)*(cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*EllipticPi(I*(cos(f*x+e)-1)/sin(f*x+e),-1/(b+(a^2+b^2)
^(1/2))^2*a^2,I)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*
a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*(a^2+b^2)^(1/2)-(a^2+b^2)^(3/2)*(-cos(f*x+e)/(cos(f*x+e)+1
)^2)^(1/2)*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)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1
/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^2+(a^2+b^2)^(1/2)*(-cos(f*x+e)/
(cos(f*x+e)+1)^2)^(1/2)*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)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2
*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^2*b^2+(a^2+b^2)^(3
/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*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)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2
)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^2-(
a^2+b^2)^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(-(2*cos(f*x+e)^2*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)-c
os(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)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^
2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^
(1/2)*a^2*b^2+(a^2+b^2)^(3/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(
1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^
2-(a^2+b^2)^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*ln(2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a
^2*b+2*b^3)/a^4)^(1/2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*a^2*b^2+(a^2+b
^2)^(3/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^
4)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1
/2)+b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^
2*b-2*b^3)/a^4)^(1/2)/a^2)*b^2-(a^2+b^2)^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*
(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b-cos(f*x
+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)
^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*b^4-(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*(b*((
a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)
*cos(f*x+e)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)
^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*a^4*b-(-cos(f*x+e)/(cos(f
*x+e)+1)^2)^(1/2)*(b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)*arctanh(1/2*(cos(f*x
+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/sin(f*x+e)^2/(-cos(f
*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)/a^2)*a^
2*b^3-(a^2+b^2)^(3/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*arctanh(1/2*(cos(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x
+e)*b+cos(f*x+e)*a^2+cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)-b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(
b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^2)^(1/2)*b^2+2*a^2*b+2*b^3)/a^4)^(1/2)/a^2)*(-b*((a^2+b^2)^(1/2)*a^2+2*(a^2+b^
2)^(1/2)*b^2-2*a^2*b-2*b^3)/a^4)^(1/2)*b^2+(a^2+b^2)^(1/2)*(-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)*arctanh(1/2*(c
os(f*x+e)-1)*((a^2+b^2)^(1/2)*cos(f*x+e)*b+cos(f*x+e)*a^2+cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)-b^2)/sin(f*x+e)^2/(
-cos(f*x+e)/(cos(f*x+e)+1)^2)^(1/2)/(b*((a^2+b^...

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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((d*sec(f*x+e))^(3/2)/(a+b*tan(f*x+e)),x, algorithm="maxima")

[Out]

integrate((d*sec(f*x + e))^(3/2)/(b*tan(f*x + e) + a), x)

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Fricas [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((d*sec(f*x+e))^(3/2)/(a+b*tan(f*x+e)),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   catdef: division by zero

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {3}{2}}}{a + b \tan {\left (e + f x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sec(f*x+e))**(3/2)/(a+b*tan(f*x+e)),x)

[Out]

Integral((d*sec(e + f*x))**(3/2)/(a + b*tan(e + f*x)), 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((d*sec(f*x+e))^(3/2)/(a+b*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(3/2)/(b*tan(f*x + e) + a), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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