3.335 \(\int \frac{\tan ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=148 \[ \frac{2 \left (3 a^2-2 b^2\right ) \sqrt{a+b \sec (c+d x)}}{b^4 d}+\frac{2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt{a+b \sec (c+d x)}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{b^4 d} \]

[Out]

(-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + (2*(a^2 - b^2)^2)/(a*b^4*d*Sqrt[a + b*Sec[c + d*x
]]) + (2*(3*a^2 - 2*b^2)*Sqrt[a + b*Sec[c + d*x]])/(b^4*d) - (2*a*(a + b*Sec[c + d*x])^(3/2))/(b^4*d) + (2*(a
+ b*Sec[c + d*x])^(5/2))/(5*b^4*d)

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Rubi [A]  time = 0.171534, antiderivative size = 148, 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} \[ \frac{2 \left (3 a^2-2 b^2\right ) \sqrt{a+b \sec (c+d x)}}{b^4 d}+\frac{2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt{a+b \sec (c+d x)}}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{a^{3/2} d}+\frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{b^4 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^5/(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)^2)/(a*b^4*d*Sqrt[a + b*Sec[c + d*x
]]) + (2*(3*a^2 - 2*b^2)*Sqrt[a + b*Sec[c + d*x]])/(b^4*d) - (2*a*(a + b*Sec[c + d*x])^(3/2))/(b^4*d) + (2*(a
+ b*Sec[c + d*x])^(5/2))/(5*b^4*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 ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (b^2-x^2\right )^2}{x (a+x)^{3/2}} \, dx,x,b \sec (c+d x)\right )}{b^4 d}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{\left (-a^2+b^2+2 a x^2-x^4\right )^2}{x^2 \left (-a+x^2\right )} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{b^4 d}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (3 a^2 \left (1-\frac{2 b^2}{3 a^2}\right )-\frac{\left (a^2-b^2\right )^2}{a x^2}-3 a x^2+x^4-\frac{b^4}{a \left (a-x^2\right )}\right ) \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{b^4 d}\\ &=\frac{2 \left (a^2-b^2\right )^2}{a b^4 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (3 a^2-2 b^2\right ) \sqrt{a+b \sec (c+d x)}}{b^4 d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{b^4 d}+\frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^4 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 )^2}{a b^4 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (3 a^2-2 b^2\right ) \sqrt{a+b \sec (c+d x)}}{b^4 d}-\frac{2 a (a+b \sec (c+d x))^{3/2}}{b^4 d}+\frac{2 (a+b \sec (c+d x))^{5/2}}{5 b^4 d}\\ \end{align*}

Mathematica [A]  time = 6.37868, size = 263, normalized size = 1.78 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b)^2 \left (-\frac{2 \left (b^2-a^2\right )^2}{a^2 b^3 (a \cos (c+d x)+b)}+\frac{2 \left (-20 a^2 b^2+16 a^4+5 b^4\right )}{5 a^2 b^4}-\frac{6 a \sec (c+d x)}{5 b^3}+\frac{2 \sec ^2(c+d x)}{5 b^2}\right )}{d (a+b \sec (c+d x))^{3/2}}-\frac{\tan ^2(c+d x) \sqrt{a \cos (c+d x)} (a \cos (c+d x)+b)^{3/2} \left (\log \left (\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}+1\right )-\log \left (1-\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}\right )\right )}{a^2 d \left (1-\cos ^2(c+d x)\right ) (a+b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^5/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*(16*a^4 - 20*a^2*b^2 + 5*b^4))/(5*a^2*b^4) - (2*(-a^2 + b^2)^2)/(a^
2*b^3*(b + a*Cos[c + d*x])) - (6*a*Sec[c + d*x])/(5*b^3) + (2*Sec[c + d*x]^2)/(5*b^2)))/(d*(a + b*Sec[c + d*x]
)^(3/2)) - (Sqrt[a*Cos[c + d*x]]*(b + a*Cos[c + d*x])^(3/2)*(-Log[1 - Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c +
d*x]]] + Log[1 + Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]])*Tan[c + d*x]^2)/(a^2*d*(1 - Cos[c + d*x]^2)*(
a + b*Sec[c + d*x])^(3/2))

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Maple [B]  time = 0.848, size = 6612, normalized size = 44.7 \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)^5/(a+b*sec(d*x+c))^(3/2),x)

[Out]

result too large to display

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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)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.2865, size = 1098, normalized size = 7.42 \begin{align*} \left [\frac{5 \,{\left (a b^{4} \cos \left (d x + c\right )^{3} + b^{5} \cos \left (d x + c\right )^{2}\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 (2 \, a^{3} b^{2} \cos \left (d x + c\right ) - a^{2} b^{3} -{\left (16 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (4 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{10 \,{\left (a^{3} b^{4} d \cos \left (d x + c\right )^{3} + a^{2} b^{5} d \cos \left (d x + c\right )^{2}\right )}}, \frac{5 \,{\left (a b^{4} \cos \left (d x + c\right )^{3} + b^{5} \cos \left (d x + c\right )^{2}\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 (2 \, a^{3} b^{2} \cos \left (d x + c\right ) - a^{2} b^{3} -{\left (16 \, a^{5} - 20 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 2 \,{\left (4 \, a^{4} b - 5 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{5 \,{\left (a^{3} b^{4} d \cos \left (d x + c\right )^{3} + a^{2} b^{5} d \cos \left (d x + c\right )^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/10*(5*(a*b^4*cos(d*x + c)^3 + b^5*cos(d*x + c)^2)*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*(2*a^3*b^2*
cos(d*x + c) - a^2*b^3 - (16*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos(d*x + c)^3 - 2*(4*a^4*b - 5*a^2*b^3)*cos(d*x + c)
^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^4*d*cos(d*x + c)^3 + a^2*b^5*d*cos(d*x + c)^2), 1/5*(5*(a*
b^4*cos(d*x + c)^3 + b^5*cos(d*x + c)^2)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*co
s(d*x + c)/(2*a*cos(d*x + c) + b)) - 2*(2*a^3*b^2*cos(d*x + c) - a^2*b^3 - (16*a^5 - 20*a^3*b^2 + 5*a*b^4)*cos
(d*x + c)^3 - 2*(4*a^4*b - 5*a^2*b^3)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(a^3*b^4*d*cos(
d*x + c)^3 + a^2*b^5*d*cos(d*x + c)^2)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\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)**5/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(tan(c + d*x)**5/(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 )^{5}}{{\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)^5/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(tan(d*x + c)^5/(b*sec(d*x + c) + a)^(3/2), x)