3.4.36 \(\int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [336]
Optimal. Leaf size=88 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {a+b \sec (c+d x)}}{b^2 d} \]
[Out]
2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+2*(a^2-b^2)/a/b^2/d/(a+b*sec(d*x+c))^(1/2)+2*(a+b*sec(d*x+
c))^(1/2)/b^2/d
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Rubi [A]
time = 0.08, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3970, 912,
1275, 212} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {a+b \sec (c+d x)}}{b^2 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]
[Out]
(2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (2*(a^2 - b^2))/(a*b^2*d*Sqrt[a + b*Sec[c + d*x]])
+ (2*Sqrt[a + b*Sec[c + d*x]])/(b^2*d)
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 912
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> With[{q = De
nominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + g*(x^q/e))^n*((c*d^2 + a*e^2)/e^2 - 2*c*
d*(x^q/e^2) + c*(x^(2*q)/e^2))^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[c*d^2 + a*e^2, 0] && IntegersQ[n, p] && FractionQ[m]
Rule 1275
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Rule 3970
Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[-(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[(b^2 - x^2)^((m - 1)/2)*((a + x)^n/x), x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]
Rubi steps
\begin {align*} \int \frac {\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx &=-\frac {\text {Subst}\left (\int \frac {b^2-x^2}{x (a+x)^{3/2}} \, dx,x,b \sec (c+d x)\right )}{b^2 d}\\ &=-\frac {2 \text {Subst}\left (\int \frac {-a^2+b^2+2 a x^2-x^4}{x^2 \left (-a+x^2\right )} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^2 d}\\ &=-\frac {2 \text {Subst}\left (\int \left (-1+\frac {a^2-b^2}{a x^2}-\frac {b^2}{a \left (a-x^2\right )}\right ) \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^2 d}\\ &=\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {a+b \sec (c+d x)}}{b^2 d}+\frac {2 \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{a d}\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {2 \left (a^2-b^2\right )}{a b^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {a+b \sec (c+d x)}}{b^2 d}\\ \end {align*}
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Mathematica [A]
time = 0.99, size = 167, normalized size = 1.90 \begin {gather*} \frac {4 a^2-2 b^2-\frac {b^2 \sqrt {b+a \cos (c+d x)} \log \left (1-\frac {\sqrt {b+a \cos (c+d x)}}{\sqrt {a \cos (c+d x)}}\right )}{\sqrt {a \cos (c+d x)}}+\frac {b^2 \sqrt {b+a \cos (c+d x)} \log \left (1+\frac {\sqrt {b+a \cos (c+d x)}}{\sqrt {a \cos (c+d x)}}\right )}{\sqrt {a \cos (c+d x)}}+2 a b \sec (c+d x)}{a b^2 d \sqrt {a+b \sec (c+d x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Tan[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]
[Out]
(4*a^2 - 2*b^2 - (b^2*Sqrt[b + a*Cos[c + d*x]]*Log[1 - Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]])/Sqrt[a*
Cos[c + d*x]] + (b^2*Sqrt[b + a*Cos[c + d*x]]*Log[1 + Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]])/Sqrt[a*C
os[c + d*x]] + 2*a*b*Sec[c + d*x])/(a*b^2*d*Sqrt[a + b*Sec[c + d*x]])
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(2829\) vs.
\(2(78)=156\).
time = 0.25, size = 2830, normalized size = 32.16
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\(2830\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
[Out]
-1/4/d*(-1+cos(d*x+c))^3*(4*cos(d*x+c)^3*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^4+
4*cos(d*x+c)^2*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^4+12*((b+a*cos(d*x+c))*cos(d
*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^2*(a-b)^(3/2)*a^3-cos(d*x+c)^3*ln(-2*(-1+cos(d*x+c))*(2*((b+a*cos(d*x
+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2
)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^5*b+cos(d*x+c)^3*ln(-2*(-1+cos(
d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos
(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^4*b^2+l
n(-(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(
d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2)
)*cos(d*x+c)^3*a^5*b-ln(-(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a
-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/si
n(d*x+c)^2/(a-b)^(1/2))*cos(d*x+c)^3*a^4*b^2-2*ln(-2*(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*
x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*c
os(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*cos(d*x+c)^2*a^4*b^2+2*cos(d*x+c)^2*ln(-2*(-1+cos(d*x+c))*
(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/
(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^3*b^3+2*cos(d*x
+c)^2*ln(-(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b
+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b
)^(1/2))*a^4*b^2-2*cos(d*x+c)^2*ln(-(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*co
s(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d
*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^3*b^3-4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(3/2)*
a*b^2-ln(-2*(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*(
(b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a
-b)^(1/2))*cos(d*x+c)*a^3*b^3+cos(d*x+c)*ln(-2*(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^
2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x
+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^2*b^4+ln(-(-1+cos(d*x+c))*(2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+c
os(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-
2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*cos(d*x+c)*a^3*b^3-cos(d*x+c)*ln(-(-1+cos(d*x+c))*(2*
((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*(a-b)^(1/2)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+
cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^2*b^4+4*((b+a*cos(
d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(3/2)*a^2*b^2+4*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(1+
cos(d*x+c))^2)^(3/2)*(a-b)^(3/2)*a^3+12*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^
(3/2)*a^3+2*ln(4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*a^(1/2)+4*a^(1/2)*((b+a*cos(d
*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*cos(d*x+c)*(a-b)^(3/2)*a^(1/2)*b^4+8*cos(d*x+c)*
(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^3*b-4*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*
x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(3/2)*a*b^2-12*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3
/2)*(a-b)^(3/2)*a*b^2+4*cos(d*x+c)*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^2*b^2+2*
ln(4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*a^(1/2)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d
*x+c)/(1+cos(d*x+c))^2)^(1/2)+4*a*cos(d*x+c)+2*b)*cos(d*x+c)^3*(a-b)^(3/2)*a^(5/2)*b^2+4*ln(4*((b+a*cos(d*x+c)
)*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)*a^(1/2)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^
2)^(1/2)+4*a*cos(d*x+c)+2*b)*cos(d*x+c)^2*(a-b)^(3/2)*a^(3/2)*b^3-12*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)
/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(3/2)*a*b^2+8*cos(d*x+c)^2*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+
c))^2)^(1/2)*a^3*b+4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(3/2)*a^3)*cos(d*x+c)*((b+a*co
s(d*x+c))/cos(d*x+c))^(1/2)*4^(1/2)/(b+a*cos(d*x+c))/((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)/sin(
d*x+c)^6/b^2/(a-b)^(3/2)/a^2
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Maxima [A]
time = 0.50, size = 110, normalized size = 1.25 \begin {gather*} -\frac {\frac {\log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {2}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a} - \frac {2 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}}}{b^{2}} - \frac {2 \, a}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} b^{2}}}{d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
-(log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a)))/a^(3/2) + 2/(sqrt(a + b/cos(d
*x + c))*a) - 2*sqrt(a + b/cos(d*x + c))/b^2 - 2*a/(sqrt(a + b/cos(d*x + c))*b^2))/d
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Fricas [A]
time = 4.58, size = 317, normalized size = 3.60 \begin {gather*} \left [\frac {{\left (a b^{2} \cos \left (d x + c\right ) + b^{3}\right )} \sqrt {a} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} - 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \, {\left (a^{2} b + {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{2 \, {\left (a^{3} b^{2} d \cos \left (d x + c\right ) + a^{2} b^{3} d\right )}}, -\frac {{\left (a b^{2} \cos \left (d x + c\right ) + b^{3}\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) - 2 \, {\left (a^{2} b + {\left (2 \, a^{3} - a b^{2}\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{a^{3} b^{2} d \cos \left (d x + c\right ) + a^{2} b^{3} d}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
[1/2*((a*b^2*cos(d*x + c) + b^3)*sqrt(a)*log(-8*a^2*cos(d*x + c)^2 - 8*a*b*cos(d*x + c) - b^2 - 4*(2*a*cos(d*x
+ c)^2 + b*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))) + 4*(a^2*b + (2*a^3 - a*b^2)*cos(d*
x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^2*d*cos(d*x + c) + a^2*b^3*d), -((a*b^2*cos(d*x + c) +
b^3)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*cos(d*x + c)/(2*a*cos(d*x + c) + b))
- 2*(a^2*b + (2*a^3 - a*b^2)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^2*d*cos(d*x + c) +
a^2*b^3*d)]
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)**3/(a+b*sec(d*x+c))**(3/2),x)
[Out]
Integral(tan(c + d*x)**3/(a + b*sec(c + d*x))**(3/2), x)
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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 258 vs.
\(2 (78) = 156\).
time = 1.22, size = 258, normalized size = 2.93 \begin {gather*} -\frac {2 \, {\left (\frac {\frac {{\left (2 \, a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - a^{2} b \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - a b^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} b^{2}} - \frac {2 \, a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + a^{2} b \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - a b^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{a^{2} b^{2}}}{\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}} + \frac {\arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} + \sqrt {a - b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )}}{d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
[Out]
-2*(((2*a^3*sgn(cos(d*x + c)) - a^2*b*sgn(cos(d*x + c)) - a*b^2*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/(a^2
*b^2) - (2*a^3*sgn(cos(d*x + c)) + a^2*b*sgn(cos(d*x + c)) - a*b^2*sgn(cos(d*x + c)))/(a^2*b^2))/sqrt(a*tan(1/
2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2 + a + b) + arctan(-1/2*(sqrt(a - b)*t
an(1/2*d*x + 1/2*c)^2 - sqrt(a*tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*a*tan(1/2*d*x + 1/2*c)^2
+ a + b) + sqrt(a - b))/sqrt(-a))/(sqrt(-a)*a*sgn(cos(d*x + c))))/d
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^3}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(c + d*x)^3/(a + b/cos(c + d*x))^(3/2),x)
[Out]
int(tan(c + d*x)^3/(a + b/cos(c + d*x))^(3/2), x)
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