3.336 \(\int \frac{\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=88 \[ \frac{2 \left (a^2-b^2\right )}{a b^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{2 \sqrt{a+b \sec (c+d x)}}{b^2 d} \]
[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)
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Rubi [A] time = 0.123803, antiderivative size = 88, normalized size of antiderivative = 1.,
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 =
{3885, 898, 1261, 206} \[ \frac{2 \left (a^2-b^2\right )}{a b^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{2 \sqrt{a+b \sec (c+d x)}}{b^2 d} \]
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 3885
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]
Rule 898
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 1261
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 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rubi steps
\begin{align*} \int \frac{\tan ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx &=-\frac{\operatorname{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 \operatorname{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 \operatorname{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 \operatorname{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*}
Mathematica [A] time = 1.00991, size = 167, normalized size = 1.9 \[ \frac{4 a^2-\frac{b^2 \sqrt{a \cos (c+d x)+b} \log \left (1-\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}\right )}{\sqrt{a \cos (c+d x)}}+\frac{b^2 \sqrt{a \cos (c+d x)+b} \log \left (\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}+1\right )}{\sqrt{a \cos (c+d x)}}+2 a b \sec (c+d x)-2 b^2}{a b^2 d \sqrt{a+b \sec (c+d x)}} \]
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] time = 0.377, size = 2830, normalized size = 32.2 \begin{align*} \text{output too large to display} \end{align*}
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)
[Out]
-1/4/d/b^2/(a-b)^(3/2)/a^2*(-1+cos(d*x+c))^3*(4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(3/
2)*a^3-2*cos(d*x+c)^2*ln(-2/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)
+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a
-b)^(1/2)-b)/sin(d*x+c)^2)*a^4*b^2+2*cos(d*x+c)^2*ln(-2/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*
x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c
)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^3*b^3-4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2
)^(3/2)*(a-b)^(3/2)*a*b^2+cos(d*x+c)*ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x
+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)
+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^3*b^3-cos(d*x+c)*ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*(
(b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c)
)*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^2*b^4-cos(d*x+c)*ln(-2/(a-b)^(1/2)*(-1+cos
(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d
*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^3*b^3+cos(d*x+c)*l
n(-2/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2
)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)
^2)*a^2*b^4+12*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(3/2)*a^3+cos(d*x+c)^3*
ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/
2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c
)^2)*a^5*b-cos(d*x+c)^3*ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+
c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*
(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^4*b^2-cos(d*x+c)^3*ln(-2/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*
x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c
)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^5*b+cos(d*x+c)^3*ln(-2/(a-b)^(1/2)*(-1+cos(d*x+c))*(2
*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b
+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^4*b^2+2*cos(d*x+c)^2*ln(-1/(a
-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-2*a*c
os(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a^4
*b^2-2*cos(d*x+c)^2*ln(-1/(a-b)^(1/2)*(-1+cos(d*x+c))*(2*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1
)^2)^(1/2)*(a-b)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b
)^(1/2)-b)/sin(d*x+c)^2)*a^3*b^3+4*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^2*b^2+4*
cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(3/2)*a^4+4*cos(d*x+c)^2*((b+a*cos(d*x
+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(3/2)*a^4+12*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)
+1)^2)^(3/2)*(a-b)^(3/2)*a^3+4*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(3/2)*a
^3+8*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(3/2)*a^3*b+4*cos(d*x+c)*((b+a*cos(
d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(3/2)*a^2*b^2+8*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(co
s(d*x+c)+1)^2)^(1/2)*(a-b)^(3/2)*a^3*b-4*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^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)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(3/2)*a*b^2+2*cos(d*x+
c)^3*(a-b)^(3/2)*a^(5/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(
d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^2+4*cos(d*x+c)^2*(a-b)^(3/2)*a^(3
/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+
a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)+2*b)*b^3-12*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*
x+c)+1)^2)^(3/2)*(a-b)^(3/2)*a*b^2+2*cos(d*x+c)*(a-b)^(3/2)*a^(1/2)*ln(4*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+
c)/(cos(d*x+c)+1)^2)^(1/2)*a^(1/2)+4*a*cos(d*x+c)+4*a^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/
2)+2*b)*b^4)*cos(d*x+c)*((b+a*cos(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)/(cos(d*x+c)+1)^2)^(3/2)/sin(d*x+c)^6
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
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]
Exception raised: ValueError
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Fricas [A] time = 2.91271, size = 763, normalized size = 8.67 \begin{align*} \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{align*}
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., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{3}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}
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 [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (d x + c\right )^{3}}{{\left (b \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)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate(tan(d*x + c)^3/(b*sec(d*x + c) + a)^(3/2), x)