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3.191 \(\int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=85 \[ \frac {2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d} \]

[Out]

-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d+2*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*se
c(d*x+c))^(1/2))*2^(1/2)/a^(3/2)/d

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Rubi [A]  time = 0.09, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3887, 481, 203} \[ \frac {2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{a^{3/2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + (2*Sqrt[2]*ArcTan[(Sqrt[a]*Tan[c +
d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(a^(3/2)*d)

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 481

Int[((e_.)*(x_))^(m_.)/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> -Dist[(a*e^n)/(b*c -
a*d), Int[(e*x)^(m - n)/(a + b*x^n), x], x] + Dist[(c*e^n)/(b*c - a*d), Int[(e*x)^(m - n)/(c + d*x^n), x], x]
/; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LeQ[n, m, 2*n - 1]

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
 n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rubi steps

\begin {align*} \int \frac {\tan ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {2 \operatorname {Subst}\left (\int \frac {x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a d}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a d}\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {2 \sqrt {2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}\\ \end {align*}

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Mathematica [C]  time = 23.84, size = 4739, normalized size = 55.75 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-32*Cos[(c + d*x)/4]^2*Cos[(c + d*x)/2]^2*(-2/Sqrt[Sec[c + d*x]] + 2*Sqrt[Sec[c + d*x]])*Sec[c + d*x]^2*((((2
 + Sqrt[2])*EllipticF[ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4]
)]], 1/2] - (2 + Sqrt[2])*EllipticPi[-(1/Sqrt[2]), ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (
1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] + (-2 + Sqrt[2])*EllipticPi[1/Sqrt[2], ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2]
 + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2])*Sqrt[(-1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1
+ Sqrt[2] + Tan[(c + d*x)/4])]*Sqrt[(1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1 +
Sqrt[2] + Tan[(c + d*x)/4])^2*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])])/4 + El
lipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(c + d*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]*Sqrt[3 + 2*Sqrt[2] - T
an[(c + d*x)/4]^2]*Sqrt[(-3 + 2*Sqrt[2])*(-3 + 2*Sqrt[2] + Tan[(c + d*x)/4]^2)]))/(d*(a*(1 + Sec[c + d*x]))^(3
/2)*(16*Cos[(c + d*x)/4]*Sqrt[Sec[c + d*x]]*Sin[(c + d*x)/4]*((((2 + Sqrt[2])*EllipticF[ArcSin[2^(1/4)*Sqrt[(1
 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] - (2 + Sqrt[2])*EllipticPi[-(1/Sqr
t[2]), ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] + (-2
 + Sqrt[2])*EllipticPi[1/Sqrt[2], ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[
(c + d*x)/4])]], 1/2])*Sqrt[(-1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*Sqrt[(1 - Sqr
t[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1 + Sqrt[2] + Tan[(c + d*x)/4])^2*Sqrt[(1 + Sqrt
[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])])/4 + EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(c + d*x
)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]*Sqrt[3 + 2*Sqrt[2] - Tan[(c + d*x)/4]^2]*Sqrt[(-3 + 2*Sqrt[2])*(-3
 + 2*Sqrt[2] + Tan[(c + d*x)/4]^2)]) - 16*Cos[(c + d*x)/4]^2*Sec[c + d*x]^(3/2)*Sin[c + d*x]*((((2 + Sqrt[2])*
EllipticF[ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] -
(2 + Sqrt[2])*EllipticPi[-(1/Sqrt[2]), ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])
*Tan[(c + d*x)/4])]], 1/2] + (-2 + Sqrt[2])*EllipticPi[1/Sqrt[2], ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c +
d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2])*Sqrt[(-1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] +
Tan[(c + d*x)/4])]*Sqrt[(1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1 + Sqrt[2] + Ta
n[(c + d*x)/4])^2*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])])/4 + EllipticPi[-3
+ 2*Sqrt[2], ArcSin[Tan[(c + d*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]*Sqrt[3 + 2*Sqrt[2] - Tan[(c + d*x)
/4]^2]*Sqrt[(-3 + 2*Sqrt[2])*(-3 + 2*Sqrt[2] + Tan[(c + d*x)/4]^2)]) - 32*Cos[(c + d*x)/4]^2*Sqrt[Sec[c + d*x]
]*((((2 + Sqrt[2])*EllipticF[ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c +
d*x)/4])]], 1/2] - (2 + Sqrt[2])*EllipticPi[-(1/Sqrt[2]), ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])
/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] + (-2 + Sqrt[2])*EllipticPi[1/Sqrt[2], ArcSin[2^(1/4)*Sqrt[(1 +
Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2])*Sec[(c + d*x)/4]^2*Sqrt[(-1 - Sqrt[2
] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*Sqrt[(1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] +
 Tan[(c + d*x)/4])]*(-1 + Sqrt[2] + Tan[(c + d*x)/4])*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Ta
n[(c + d*x)/4])])/8 + ((-3 + 2*Sqrt[2])*EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(c + d*x)/4]/Sqrt[3 - 2*Sqrt[2]]
], 17 - 12*Sqrt[2]]*Sec[(c + d*x)/4]^2*Tan[(c + d*x)/4]*Sqrt[3 + 2*Sqrt[2] - Tan[(c + d*x)/4]^2])/(4*Sqrt[(-3
+ 2*Sqrt[2])*(-3 + 2*Sqrt[2] + Tan[(c + d*x)/4]^2)]) - (EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(c + d*x)/4]/Sqr
t[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]]*Sec[(c + d*x)/4]^2*Tan[(c + d*x)/4]*Sqrt[(-3 + 2*Sqrt[2])*(-3 + 2*Sqrt[2]
+ Tan[(c + d*x)/4]^2)])/(4*Sqrt[3 + 2*Sqrt[2] - Tan[(c + d*x)/4]^2]) + (Sec[(c + d*x)/4]^2*Sqrt[3 + 2*Sqrt[2]
- Tan[(c + d*x)/4]^2]*Sqrt[(-3 + 2*Sqrt[2])*(-3 + 2*Sqrt[2] + Tan[(c + d*x)/4]^2)])/(4*Sqrt[3 - 2*Sqrt[2]]*Sqr
t[1 - Tan[(c + d*x)/4]^2/(3 - 2*Sqrt[2])]*Sqrt[1 - ((17 - 12*Sqrt[2])*Tan[(c + d*x)/4]^2)/(3 - 2*Sqrt[2])]*(1
- ((-3 + 2*Sqrt[2])*Tan[(c + d*x)/4]^2)/(3 - 2*Sqrt[2]))) + (((2 + Sqrt[2])*EllipticF[ArcSin[2^(1/4)*Sqrt[(1 +
 Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] - (2 + Sqrt[2])*EllipticPi[-(1/Sqrt[
2]), ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] + (-2 +
 Sqrt[2])*EllipticPi[1/Sqrt[2], ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c
 + d*x)/4])]], 1/2])*Sqrt[(1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1 + Sqrt[2] +
Tan[(c + d*x)/4])^2*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1/4*(Sec[(c + d
*x)/4]^2*(-1 - Sqrt[2] + Tan[(c + d*x)/4]))/(-1 + Sqrt[2] + Tan[(c + d*x)/4])^2 + Sec[(c + d*x)/4]^2/(4*(-1 +
Sqrt[2] + Tan[(c + d*x)/4]))))/(8*Sqrt[(-1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]) +
 (((2 + Sqrt[2])*EllipticF[ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*
x)/4])]], 1/2] - (2 + Sqrt[2])*EllipticPi[-(1/Sqrt[2]), ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(
1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] + (-2 + Sqrt[2])*EllipticPi[1/Sqrt[2], ArcSin[2^(1/4)*Sqrt[(1 + Sq
rt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2])*Sqrt[(-1 - Sqrt[2] + Tan[(c + d*x)/4])
/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1 + Sqrt[2] + Tan[(c + d*x)/4])^2*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/
(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1/4*(Sec[(c + d*x)/4]^2*(1 - Sqrt[2] + Tan[(c + d*x)/4]))/(-1 + Sqrt[2] +
 Tan[(c + d*x)/4])^2 + Sec[(c + d*x)/4]^2/(4*(-1 + Sqrt[2] + Tan[(c + d*x)/4]))))/(8*Sqrt[(1 - Sqrt[2] + Tan[(
c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]) + (((2 + Sqrt[2])*EllipticF[ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2]
+ Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] - (2 + Sqrt[2])*EllipticPi[-(1/Sqrt[2]), ArcS
in[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]], 1/2] + (-2 + Sqrt[2])
*EllipticPi[1/Sqrt[2], ArcSin[2^(1/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4
])]], 1/2])*Sqrt[(-1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*Sqrt[(1 - Sqrt[2] + Tan[
(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*(-1 + Sqrt[2] + Tan[(c + d*x)/4])^2*(Sec[(c + d*x)/4]^2/(4*(-
1 + Sqrt[2] + Tan[(c + d*x)/4])) - (Sec[(c + d*x)/4]^2*(1 + Sqrt[2] + Tan[(c + d*x)/4]))/(4*(-1 + Sqrt[2] + Ta
n[(c + d*x)/4])^2)))/(8*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]) + (Sqrt[(-1
- Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sqrt[2] + Tan[(c + d*x)/4])]*Sqrt[(1 - Sqrt[2] + Tan[(c + d*x)/4])/(-1 + S
qrt[2] + Tan[(c + d*x)/4])]*(-1 + Sqrt[2] + Tan[(c + d*x)/4])^2*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(-1 + Sq
rt[2] + Tan[(c + d*x)/4])]*(((2 + Sqrt[2])*(-1/4*((1 + Sqrt[2])*Sec[(c + d*x)/4]^2*(1 + Sqrt[2] + Tan[(c + d*x
)/4]))/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])^2 + Sec[(c + d*x)/4]^2/(4*(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4]))))/
(2^(3/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]*Sqrt[1 - (1 + Sqrt[2] + T
an[(c + d*x)/4])/(Sqrt[2]*(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4]))]*Sqrt[1 - (Sqrt[2]*(1 + Sqrt[2] + Tan[(c + d*x
)/4]))/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]) + ((-2 + Sqrt[2])*(-1/4*((1 + Sqrt[2])*Sec[(c + d*x)/4]^2*(1 + S
qrt[2] + Tan[(c + d*x)/4]))/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])^2 + Sec[(c + d*x)/4]^2/(4*(1 + (1 + Sqrt[2])*
Tan[(c + d*x)/4]))))/(2^(3/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]*(1 -
 (1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4]))*Sqrt[1 - (1 + Sqrt[2] + Tan[(c + d*x)/
4])/(Sqrt[2]*(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4]))]*Sqrt[1 - (Sqrt[2]*(1 + Sqrt[2] + Tan[(c + d*x)/4]))/(1 + (
1 + Sqrt[2])*Tan[(c + d*x)/4])]) - ((2 + Sqrt[2])*(-1/4*((1 + Sqrt[2])*Sec[(c + d*x)/4]^2*(1 + Sqrt[2] + Tan[(
c + d*x)/4]))/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])^2 + Sec[(c + d*x)/4]^2/(4*(1 + (1 + Sqrt[2])*Tan[(c + d*x)/
4]))))/(2^(3/4)*Sqrt[(1 + Sqrt[2] + Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4])]*(1 + (1 + Sqrt[2]
+ Tan[(c + d*x)/4])/(1 + (1 + Sqrt[2])*Tan[(c + d*x)/4]))*Sqrt[1 - (1 + Sqrt[2] + Tan[(c + d*x)/4])/(Sqrt[2]*(
1 + (1 + Sqrt[2])*Tan[(c + d*x)/4]))]*Sqrt[1 - (Sqrt[2]*(1 + Sqrt[2] + Tan[(c + d*x)/4]))/(1 + (1 + Sqrt[2])*T
an[(c + d*x)/4])])))/4)))

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fricas [A]  time = 0.63, size = 295, normalized size = 3.47 \[ \left [\frac {\sqrt {2} a \sqrt {-\frac {1}{a}} \log \left (-\frac {2 \, \sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {-\frac {1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, \cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right )}{a^{2} d}, -\frac {2 \, {\left (\sqrt {2} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) - \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right )\right )}}{a^{2} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[(sqrt(2)*a*sqrt(-1/a)*log(-(2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x
 + c) - 3*cos(d*x + c)^2 - 2*cos(d*x + c) + 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - sqrt(-a)*log((2*a*cos(
d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a
)/(cos(d*x + c) + 1)))/(a^2*d), -2*(sqrt(2)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos
(d*x + c)/(sqrt(a)*sin(d*x + c))) - sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(
a)*sin(d*x + c))))/(a^2*d)]

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giac [A]  time = 2.97, size = 73, normalized size = 0.86 \[ -\frac {\sqrt {2} {\left (\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-a + \frac {a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}}{2 \, \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {2 \, \arctan \left (\frac {\sqrt {-a + \frac {a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

-sqrt(2)*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-a + a/tan(1/2*d*x + 1/2*c)^2)/sqrt(a))/a^(3/2) - 2*arctan(sqrt(-a +
 a/tan(1/2*d*x + 1/2*c)^2)/sqrt(a))/a^(3/2))/d

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maple [B]  time = 0.90, size = 142, normalized size = 1.67 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+2 \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^2/(a+a*sec(d*x+c))^(3/2),x)

[Out]

1/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(2^(1/2)*arctanh(1/2*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))+2*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)+cos(d*x+c)-1)/sin(d*x+c)))/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^2/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^2/(a*sec(d*x + c) + a)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^2}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2/(a + a/cos(c + d*x))^(3/2),x)

[Out]

int(tan(c + d*x)^2/(a + a/cos(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**2/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Integral(tan(c + d*x)**2/(a*(sec(c + d*x) + 1))**(3/2), x)

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