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

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

[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 = 0.29, 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 = {3512, 746, 399, 490, 1213, 537, 444, 63, 298, 205, 208} \[ \frac {(d \sec (e+f x))^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{\sqrt {b} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac {(d \sec (e+f x))^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt [4]{\sec ^2(e+f x)}}{\sqrt [4]{a^2+b^2}}\right )}{\sqrt {b} f \sqrt [4]{a^2+b^2} \sec ^2(e+f x)^{3/4}}-\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (-\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{b f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}}+\frac {a \sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^{3/2} \Pi \left (\frac {b}{\sqrt {a^2+b^2}};\left .\sin ^{-1}\left (\sqrt [4]{\sec ^2(e+f x)}\right )\right |-1\right )}{b f \sqrt {a^2+b^2} \sec ^2(e+f x)^{3/4}} \]

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 63

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 205

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

Rule 208

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

Rule 298

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 399

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 444

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 490

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), In
t[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 537

Int[1/(((a_) + (b_.)*(x_)^2)*Sqrt[(c_) + (d_.)*(x_)^2]*Sqrt[(e_) + (f_.)*(x_)^2]), x_Symbol] :> Simp[(1*Ellipt
icPi[(b*c)/(a*d), ArcSin[Rt[-(d/c), 2]*x], (c*f)/(d*e)])/(a*Sqrt[c]*Sqrt[e]*Rt[-(d/c), 2]), 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 746

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 1213

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 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 {(d \sec (e+f x))^{3/2}}{a+b \tan (e+f x)} \, dx &=\frac {(d \sec (e+f x))^{3/2} \operatorname {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} \operatorname {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 ) \operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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} \operatorname {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} \operatorname {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 ) \operatorname {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 ) \operatorname {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]  time = 21.94, size = 6301, normalized size = 18.87 \[ \text {Result too large to show} \]

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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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

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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maple [B]  time = 1.36, size = 3737, normalized size = 11.19 \[ \text {output too large to display} \]

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)

[Out]

1/2/f*(1+cos(f*x+e))^2*(-4*I*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(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)*(-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)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(a^2+b^2)^(1/2)*a*b^3-4*I*b*a^3*(1/(1+cos(f*x+e)))^(1/2
)*(cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*EllipticPi(I*(-1+cos(f*x+e))/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*b*a^3*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)
))^(1/2)*EllipticPi(I*(-1+cos(f*x+e))/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*(1/(1+cos(f*x+e)))^(1/2)*(cos(f*x+e)/(1+cos(f*x+e)))^(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)*(-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)*EllipticF(I*(-1+cos(f*x+e))/sin(f*x+e),I)*(a^2+b^2)^(3/2)*a*b+(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*l
n(-(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*((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)^(3/2)*a^2-(-cos(f*x+e)/(1+cos(f*x+e))
^2)^(1/2)*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*((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-(-cos(f*x+e)
/(1+cos(f*x+e))^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)*(-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)*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^2)^(3/2)*a^2
+(-cos(f*x+e)/(1+cos(f*x+e))^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)*
(-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)*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
^2)^(1/2)*a^2*b^2+(-cos(f*x+e)/(1+cos(f*x+e))^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)^(3/2
)*a^2-(-cos(f*x+e)/(1+cos(f*x+e))^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)^(1/2)*a^2*b^2+(-
cos(f*x+e)/(1+cos(f*x+e))^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)*arc
tanh(1/2*(-1+cos(f*x+e))*(cos(f*x+e)*(a^2+b^2)^(1/2)*b-cos(f*x+e)*a^2-cos(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/si
n(f*x+e)^2/(-cos(f*x+e)/(1+cos(f*x+e))^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^2)^(3/2)*b^2-(-cos(f*x+e)/(1+cos(f*x+e))^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*(-1+cos(f*x+e))*(cos(f*x+e)*(a^2+b^2)^(1/2)*b-cos(f*x+e)*a^2-co
s(f*x+e)*b^2-b*(a^2+b^2)^(1/2)+b^2)/sin(f*x+e)^2/(-cos(f*x+e)/(1+cos(f*x+e))^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^2)^(1/2)*b^4-(-cos(f*x+e)/(1+cos(f*x+e))^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*(-1+cos(f*x+e))*(cos(f*x+
e)*(a^2+b^2)^(1/2)*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)/(1+cos(f*x
+e))^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)
/(1+cos(f*x+e))^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*(
-1+cos(f*x+e))*(cos(f*x+e)*(a^2+b^2)^(1/2)*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)/(1+cos(f*x+e))^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-(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*arctanh(1/2*(-1+cos(f*x+e))*(cos(f*x+e)*(a^2+b^2)^(1/2)*b+c
os(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)/(1+cos(f*x+e))^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)*(a^2+b^2)^(3/2)*b^2+(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*arctanh(1/2*(-1+cos(f
*x+e))*(cos(f*x+e)*(a^2+b^2)^(1/2)*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)/(1+cos(f*x+e))^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)*(a^2+b^2)^(1/2)*b^4-(-cos(f*x+e)/(1+cos(
f*x+e))^2)^(1/2)*arctanh(1/2*(-1+cos(f*x+e))*(cos(f*x+e)*(a^2+b^2)^(1/2)*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)/(1+cos(f*x+e))^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)*a^
4*b-(-cos(f*x+e)/(1+cos(f*x+e))^2)^(1/2)*arctanh(1/2*(-1+cos(f*x+e))*(cos(f*x+e)*(a^2+b^2)^(1/2)*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)/(1+cos(f*x+e))^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)*a^2*b^3)*(-1+cos(f*x+e))*(d/cos(f*x+e))^(3/2)*cos(f*x+e)/sin(f*x+e)^2/b/a^2/(-b+(a^2+b^2
)^(1/2))/(a^2+b^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)/(b+(a^2+b^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)

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

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

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

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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