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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[(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)

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Rubi [A]  time = 0.0861983, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, 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 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]

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 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])

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*}

Mathematica [C]  time = 23.4534, size = 4752, normalized size = 55.91 \[ \text{Result too large to show} \]

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] -
Tan[(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/Sq
rt[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 - Sq
rt[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 + Sqr
t[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] +
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)]) - 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 - Sqr
t[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] +
 Tan[(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
]/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])*(-3 + 2*Sqr
t[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*Sqr
t[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]
]*Sqrt[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])*T
an[(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])]*(-(Sec[(c +
d*x)/4]^2*(-1 - Sqrt[2] + Tan[(c + d*x)/4]))/(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
 + 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])]*(-(Sec[(c + d*x)/4]^2*(1 - Sqrt[2] + Tan[(c + d*x)/4]))/(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 + Sqr
t[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 + Sqr
t[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]
 + Tan[(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 + 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])]*(((2 + Sqrt[2])*(-((1 + Sqrt[2])*Sec[(c + d*x)/4]^2*(1 + Sqrt[2] + Tan[(c + d*
x)/4]))/(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
] + 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 + Sqrt[2])*Sec[(c + d*x)/4]^2*(1 +
Sqrt[2] + Tan[(c + d*x)/4]))/(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 + Sqrt[2])*Sec[(c + d*x)/4]^2*(1 + Sqrt[2] + Tan[
(c + d*x)/4]))/(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 + Sqr
t[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])])))/4)))

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Maple [B]  time = 0.127, size = 142, normalized size = 1.7 \begin{align*}{\frac{1}{d{a}^{2}}\sqrt{{\frac{a \left ( \cos \left ( dx+c \right ) +1 \right ) }{\cos \left ( dx+c \right ) }}}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) +2\,\ln \left ( -{\frac{1}{\sin \left ( dx+c \right ) } \left ( -\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) } \right ) \right ) } \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{2}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [A]  time = 2.0362, size = 784, normalized size = 9.22 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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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Giac [B]  time = 12.5451, size = 302, normalized size = 3.55 \begin{align*} \frac{\frac{\sqrt{2} \log \left ({\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{\log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} - a{\left (2 \, \sqrt{2} + 3\right )} \right |}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{\log \left ({\left |{\left (\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}\right )}^{2} + a{\left (2 \, \sqrt{2} - 3\right )} \right |}\right )}{\sqrt{-a} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}}{d} \end{align*}

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

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