3.318 \(\int \sqrt{a+b \sec (c+d x)} \tan ^5(c+d x) \, dx\)
Optimal. Leaf size=169 \[ \frac{2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac{2 a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}+\frac{2 (a+b \sec (c+d x))^{9/2}}{9 b^4 d}-\frac{6 a (a+b \sec (c+d x))^{7/2}}{7 b^4 d}+\frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d} \]
[Out]
(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/d + (2*Sqrt[a + b*Sec[c + d*x]])/d - (2*a*(a^2 - 2*b^2)
*(a + b*Sec[c + d*x])^(3/2))/(3*b^4*d) + (2*(3*a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(5/2))/(5*b^4*d) - (6*a*(a +
b*Sec[c + d*x])^(7/2))/(7*b^4*d) + (2*(a + b*Sec[c + d*x])^(9/2))/(9*b^4*d)
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Rubi [A] time = 0.17029, antiderivative size = 169, 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, 207} \[ \frac{2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac{2 a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}+\frac{2 (a+b \sec (c+d x))^{9/2}}{9 b^4 d}-\frac{6 a (a+b \sec (c+d x))^{7/2}}{7 b^4 d}+\frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]^5,x]
[Out]
(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/d + (2*Sqrt[a + b*Sec[c + d*x]])/d - (2*a*(a^2 - 2*b^2)
*(a + b*Sec[c + d*x])^(3/2))/(3*b^4*d) + (2*(3*a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(5/2))/(5*b^4*d) - (6*a*(a +
b*Sec[c + d*x])^(7/2))/(7*b^4*d) + (2*(a + b*Sec[c + d*x])^(9/2))/(9*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 207
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \sqrt{a+b \sec (c+d x)} \tan ^5(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sqrt{a+x} \left (b^2-x^2\right )^2}{x} \, dx,x,b \sec (c+d x)\right )}{b^4 d}\\ &=\frac{2 \operatorname{Subst}\left (\int \frac{x^2 \left (-a^2+b^2+2 a x^2-x^4\right )^2}{-a+x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{b^4 d}\\ &=\frac{2 \operatorname{Subst}\left (\int \left (b^4-a \left (a^2-2 b^2\right ) x^2+\left (3 a^2-2 b^2\right ) x^4-3 a x^6+x^8+\frac{a b^4}{-a+x^2}\right ) \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{b^4 d}\\ &=\frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}+\frac{2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac{6 a (a+b \sec (c+d x))^{7/2}}{7 b^4 d}+\frac{2 (a+b \sec (c+d x))^{9/2}}{9 b^4 d}+\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{-a+x^2} \, dx,x,\sqrt{a+b \sec (c+d x)}\right )}{d}\\ &=-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 \sqrt{a+b \sec (c+d x)}}{d}-\frac{2 a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}+\frac{2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac{6 a (a+b \sec (c+d x))^{7/2}}{7 b^4 d}+\frac{2 (a+b \sec (c+d x))^{9/2}}{9 b^4 d}\\ \end{align*}
Mathematica [A] time = 6.2912, size = 254, normalized size = 1.5 \[ \frac{\sqrt{a+b \sec (c+d x)} \left (-\frac{4 \left (a^2+21 b^2\right ) \sec ^2(c+d x)}{105 b^2}-\frac{4 a \left (21 b^2-4 a^2\right ) \sec (c+d x)}{315 b^3}+\frac{2 \left (84 a^2 b^2-16 a^4+315 b^4\right )}{315 b^4}+\frac{2 a \sec ^3(c+d x)}{63 b}+\frac{2}{9} \sec ^4(c+d x)\right )}{d}-\frac{\sin ^2(c+d x) \sqrt{a \cos (c+d x)} \sqrt{a+b \sec (c+d x)} \left (\log \left (\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}+1\right )-\log \left (1-\frac{\sqrt{a \cos (c+d x)+b}}{\sqrt{a \cos (c+d x)}}\right )\right )}{d \left (1-\cos ^2(c+d x)\right ) \sqrt{a \cos (c+d x)+b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]^5,x]
[Out]
(Sqrt[a + b*Sec[c + d*x]]*((2*(-16*a^4 + 84*a^2*b^2 + 315*b^4))/(315*b^4) - (4*a*(-4*a^2 + 21*b^2)*Sec[c + d*x
])/(315*b^3) - (4*(a^2 + 21*b^2)*Sec[c + d*x]^2)/(105*b^2) + (2*a*Sec[c + d*x]^3)/(63*b) + (2*Sec[c + d*x]^4)/
9))/d - (Sqrt[a*Cos[c + d*x]]*(-Log[1 - Sqrt[b + a*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]] + Log[1 + Sqrt[b + a*Co
s[c + d*x]]/Sqrt[a*Cos[c + d*x]]])*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x]^2)/(d*Sqrt[b + a*Cos[c + d*x]]*(1 - C
os[c + d*x]^2))
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Maple [B] time = 0.818, size = 3268, normalized size = 19.3 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^5,x)
[Out]
-1/2520/d/(a-b)^(1/2)/b^4*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(cos(d*x+c)+1)*(-1+cos(d*x+c))^4*(968*co
s(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a*b^3+144*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)^(1/2)*a^2*b^2-120*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(c
os(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a*b^3-24*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(
a-b)^(1/2)*a^2*b^2-64*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a^3*b+336*
cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a*b^3+1260*cos(d*x+c)^7*(a-b)^(1
/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^4+1260*cos(d*x+c)^6*(a-b)^(1/2)*a^(1/2)*ln(4*c
os(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^5-72*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)
^(3/2)*(a-b)^(1/2)*a^2*b^2-192*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a
^3*b+1008*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a*b^3+120*cos(d*x+c)^4
*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a^2*b^2-64*cos(d*x+c)^3*((b+a*cos(d*x+c))*co
s(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a^3*b+216*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)^(1/2)*a*b^3+48*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)^(1/2)*
a^2*b^2-40*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a*b^3+128*cos(d*x+c)^6*
((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^4*b-648*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos(
d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^3*b^2-2121*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+
1)^2)^(1/2)*(a-b)^(1/2)*a*b^4-648*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2
)*a^3*b^2-648*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^2*b^3-2121*cos(d
*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a*b^4+128*cos(d*x+c)^5*((b+a*cos(d*x+
c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^4*b-648*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*
x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^2*b^3-24*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b
)^(1/2)*a^2*b^2-192*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*a^3*b-630*co
s(d*x+c)^6*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)*b^6+630*cos(d*x+c)^6*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)*b^6-280*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(
3/2)*b^4+168*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)^(1/2)*b^4+1260*cos(d*x+c)
^7*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^4-630*cos(d*x+c)^7*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*b^5-1260*cos(d*x+c)^7*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^2*b^4+630*cos(d*x+c)^7*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*b^5+2744*co
s(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^4-840*cos(d*x+c)*((b+a*cos(d*x+c
))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^4-189*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c
)+1)^2)^(3/2)*(a-b)^(1/2)*b^4+128*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2
)*a^5-2121*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*b^5+128*cos(d*x+c)^7*
((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*a^5-2121*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d
*x+c)/(cos(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)*b^5-399*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^
(3/2)*(a-b)^(1/2)*b^4-1197*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^4+1
260*cos(d*x+c)^6*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*b^5-1260*cos(d*x+c)^6*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)/(c
os(d*x+c)+1)^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2)*a*b^5+2625*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d
*x+c)+1)^2)^(3/2)*(a-b)^(1/2)*b^4)/sin(d*x+c)^8/cos(d*x+c)^4/((b+a*cos(d*x+c))*cos(d*x+c)/(cos(d*x+c)+1)^2)^(3
/2)
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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((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^5,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 3.7358, size = 1048, normalized size = 6.2 \begin{align*} \left [\frac{315 \, \sqrt{a} b^{4} \cos \left (d x + c\right )^{4} \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 (5 \, a b^{3} \cos \left (d x + c\right ) -{\left (16 \, a^{4} - 84 \, a^{2} b^{2} - 315 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 35 \, b^{4} + 2 \,{\left (4 \, a^{3} b - 21 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 6 \,{\left (a^{2} b^{2} + 21 \, b^{4}\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 )}}}{630 \, b^{4} d \cos \left (d x + c\right )^{4}}, \frac{315 \, \sqrt{-a} b^{4} \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 ) \cos \left (d x + c\right )^{4} + 2 \,{\left (5 \, a b^{3} \cos \left (d x + c\right ) -{\left (16 \, a^{4} - 84 \, a^{2} b^{2} - 315 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 35 \, b^{4} + 2 \,{\left (4 \, a^{3} b - 21 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 6 \,{\left (a^{2} b^{2} + 21 \, b^{4}\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 )}}}{315 \, b^{4} d \cos \left (d x + c\right )^{4}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^5,x, algorithm="fricas")
[Out]
[1/630*(315*sqrt(a)*b^4*cos(d*x + c)^4*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*(5*a*b^3*cos(d*x + c) - (16*a^4 -
84*a^2*b^2 - 315*b^4)*cos(d*x + c)^4 + 35*b^4 + 2*(4*a^3*b - 21*a*b^3)*cos(d*x + c)^3 - 6*(a^2*b^2 + 21*b^4)*
cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(b^4*d*cos(d*x + c)^4), 1/315*(315*sqrt(-a)*b^4*arcta
n(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))*cos(d*x + c)^4 + 2*(
5*a*b^3*cos(d*x + c) - (16*a^4 - 84*a^2*b^2 - 315*b^4)*cos(d*x + c)^4 + 35*b^4 + 2*(4*a^3*b - 21*a*b^3)*cos(d*
x + c)^3 - 6*(a^2*b^2 + 21*b^4)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(b^4*d*cos(d*x + c)^4
)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec{\left (c + d x \right )}} \tan ^{5}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))**(1/2)*tan(d*x+c)**5,x)
[Out]
Integral(sqrt(a + b*sec(c + d*x))*tan(c + d*x)**5, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (d x + c\right ) + a} \tan \left (d x + c\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^5,x, algorithm="giac")
[Out]
integrate(sqrt(b*sec(d*x + c) + a)*tan(d*x + c)^5, x)