3.326 \(\int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\)
Optimal. Leaf size=148 \[ \frac {2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}+\frac {2 (a+b \sec (c+d x))^{7/2}}{7 b^4 d}-\frac {6 a (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]
[Out]
2/3*(3*a^2-2*b^2)*(a+b*sec(d*x+c))^(3/2)/b^4/d-6/5*a*(a+b*sec(d*x+c))^(5/2)/b^4/d+2/7*(a+b*sec(d*x+c))^(7/2)/b
^4/d-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)-2*a*(a^2-2*b^2)*(a+b*sec(d*x+c))^(1/2)/b^4/d
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Rubi [A] time = 0.14, antiderivative size = 148, 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 =
{3885, 898, 1153, 207} \[ \frac {2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}+\frac {2 (a+b \sec (c+d x))^{7/2}}{7 b^4 d}-\frac {6 a (a+b \sec (c+d x))^{5/2}}{5 b^4 d}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]
Antiderivative was successfully verified.
[In]
Int[Tan[c + d*x]^5/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(-2*ArcTanh[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d) - (2*a*(a^2 - 2*b^2)*Sqrt[a + b*Sec[c + d*x]])/(b^4
*d) + (2*(3*a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(3/2))/(3*b^4*d) - (6*a*(a + b*Sec[c + d*x])^(5/2))/(5*b^4*d) +
(2*(a + b*Sec[c + d*x])^(7/2))/(7*b^4*d)
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])
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 1153
Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]
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]
Rubi steps
\begin {align*} \int \frac {\tan ^5(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (b^2-x^2\right )^2}{x \sqrt {a+x}} \, 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}{-a+x^2} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^4 d}\\ &=\frac {2 \operatorname {Subst}\left (\int \left (-a^3+2 a b^2+\left (3 a^2-2 b^2\right ) x^2-3 a x^4+x^6+\frac {b^4}{-a+x^2}\right ) \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{b^4 d}\\ &=-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}+\frac {2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}-\frac {6 a (a+b \sec (c+d x))^{5/2}}{5 b^4 d}+\frac {2 (a+b \sec (c+d x))^{7/2}}{7 b^4 d}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{-a+x^2} \, dx,x,\sqrt {a+b \sec (c+d x)}\right )}{d}\\ &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}-\frac {2 a \left (a^2-2 b^2\right ) \sqrt {a+b \sec (c+d x)}}{b^4 d}+\frac {2 \left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{3/2}}{3 b^4 d}-\frac {6 a (a+b \sec (c+d x))^{5/2}}{5 b^4 d}+\frac {2 (a+b \sec (c+d x))^{7/2}}{7 b^4 d}\\ \end {align*}
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Mathematica [A] time = 6.32, size = 248, normalized size = 1.68 \[ \frac {\sec (c+d x) (a \cos (c+d x)+b) \left (\frac {8 a \left (35 b^2-12 a^2\right )}{105 b^4}-\frac {4 \left (35 b^2-12 a^2\right ) \sec (c+d x)}{105 b^3}-\frac {12 a \sec ^2(c+d x)}{35 b^2}+\frac {2 \sec ^3(c+d x)}{7 b}\right )}{d \sqrt {a+b \sec (c+d x)}}-\frac {\sin (c+d x) \tan (c+d x) \sqrt {a \cos (c+d x)} \sqrt {a \cos (c+d x)+b} \left (\log \left (\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}+1\right )-\log \left (1-\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}\right )\right )}{a d \left (1-\cos ^2(c+d x)\right ) \sqrt {a+b \sec (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[c + d*x]^5/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
((b + a*Cos[c + d*x])*Sec[c + d*x]*((8*a*(-12*a^2 + 35*b^2))/(105*b^4) - (4*(-12*a^2 + 35*b^2)*Sec[c + d*x])/(
105*b^3) - (12*a*Sec[c + d*x]^2)/(35*b^2) + (2*Sec[c + d*x]^3)/(7*b)))/(d*Sqrt[a + b*Sec[c + d*x]]) - (Sqrt[a*
Cos[c + d*x]]*Sqrt[b + 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*Cos[c + d*x]]/Sqrt[a*Cos[c + d*x]]])*Sin[c + d*x]*Tan[c + d*x])/(a*d*(1 - Cos[c + d*x]^2)*Sqrt[a + b*Se
c[c + d*x]])
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fricas [A] time = 2.01, size = 381, normalized size = 2.57 \[ \left [\frac {105 \, \sqrt {a} b^{4} \cos \left (d x + c\right )^{3} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) - 4 \, {\left (18 \, a^{2} b^{2} \cos \left (d x + c\right ) - 15 \, a b^{3} + 4 \, {\left (12 \, a^{4} - 35 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (12 \, a^{3} b - 35 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{210 \, a b^{4} d \cos \left (d x + c\right )^{3}}, \frac {105 \, \sqrt {-a} b^{4} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right )^{3} - 2 \, {\left (18 \, a^{2} b^{2} \cos \left (d x + c\right ) - 15 \, a b^{3} + 4 \, {\left (12 \, a^{4} - 35 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} - 2 \, {\left (12 \, a^{3} b - 35 \, a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{105 \, a b^{4} d \cos \left (d x + c\right )^{3}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
[1/210*(105*sqrt(a)*b^4*cos(d*x + c)^3*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*(18*a^2*b^2*cos(d*x + c) - 15*a*b
^3 + 4*(12*a^4 - 35*a^2*b^2)*cos(d*x + c)^3 - 2*(12*a^3*b - 35*a*b^3)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b
)/cos(d*x + c)))/(a*b^4*d*cos(d*x + c)^3), 1/105*(105*sqrt(-a)*b^4*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))*cos(d*x + c)^3 - 2*(18*a^2*b^2*cos(d*x + c) - 15*a*b^3 + 4
*(12*a^4 - 35*a^2*b^2)*cos(d*x + c)^3 - 2*(12*a^3*b - 35*a*b^3)*cos(d*x + c)^2)*sqrt((a*cos(d*x + c) + b)/cos(
d*x + c)))/(a*b^4*d*cos(d*x + c)^3)]
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giac [B] time = 6.08, size = 722, normalized size = 4.88 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
-2/105*(105*arctan(-1/2*(sqrt(a - b)*tan(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)*sgn(tan(1/2*d*x + 1/2*c)^2 -
1)) - 2*(105*(sqrt(a - b)*tan(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))^6 - 840*(sqrt(a - b)*tan(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))^5*sqrt(a - b) + 35*(sqrt(a - b)*tan(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
))^4*(27*a - 23*b) + 280*(sqrt(a - b)*tan(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))^3*(3*a + 4*b)*sqrt(a - b) - 21*(sqrt(a - b)*tan(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))^2*(65*
a^2 - 2*a*b - 15*b^2) + 315*a^3 + 707*a^2*b - 7*a*b^2 - 55*b^3 - 56*(sqrt(a - b)*tan(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))*(19*a*b + 5*b^2)*s
qrt(a - b))/((sqrt(a - b)*tan(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))^7*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d
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maple [B] time = 1.75, size = 4997, normalized size = 33.76 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x)
[Out]
-1/420/d*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(1+cos(d*x+c))*(-1+cos(d*x+c))^4*(192*(a-b)^(3/2)*((b+a*c
os(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^6*a^5+192*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1
+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^5*a^5-210*cos(d*x+c)^6*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*
cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+c
os(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^6*b-105*cos(d*x+c)^6*ln(-2*(-1+cos(d*x+c))*(2*c
os(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a
*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^5*b^2+420*cos(d*x+c
)^6*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(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)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b
)^(1/2))*a^4*b^3-315*cos(d*x+c)^6*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)
/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a
-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^3*b^4+105*cos(d*x+c)^6*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2
)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x
+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^2*b^5+210*cos(d*x+c)^6*ln(-(-1+cos(d*x+
c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)
+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^6*b+105*cos
(d*x+c)^6*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2
*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/
(a-b)^(1/2))*a^5*b^2-420*cos(d*x+c)^6*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+
c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*
(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^4*b^3+315*cos(d*x+c)^6*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2
)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x
+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^3*b^4-105*cos(d*x+c)^6*ln(-(-1+cos(d*x+
c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)
+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^2*b^5+105*c
os(d*x+c)^5*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/
2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c
)^2/(a-b)^(1/2))*a^6*b-210*cos(d*x+c)^5*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(
d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1
/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^5*b^2-105*cos(d*x+c)^5*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b
)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*c
os(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^4*b^3+420*cos(d*x+c)^5*ln(-2*(-1+
cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*co
s(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^3*b
^4-315*cos(d*x+c)^5*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c)
)^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/s
in(d*x+c)^2/(a-b)^(1/2))*a^2*b^5+105*cos(d*x+c)^5*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*
x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x
+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a*b^6-105*cos(d*x+c)^5*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c
)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x
+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^6*b+210*cos(d*x+c)^5*ln(-(-
1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*
cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^5
*b^2+105*cos(d*x+c)^5*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c)
)^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/s
in(d*x+c)^2/(a-b)^(1/2))*a^4*b^3-420*cos(d*x+c)^5*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+
c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c
))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^3*b^4+315*cos(d*x+c)^5*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c
)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x
+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^2*b^5-105*cos(d*x+c)^5*ln(-
(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+
b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a
*b^6-60*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(3/2)*a*b^3+105*cos(d*x+c)^6*ln(-2*(-1+cos(
d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)-2*a*cos(d*x+c)+b*cos(d*
x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^7-105*c
os(d*x+c)^6*ln(-(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)
-2*a*cos(d*x+c)+b*cos(d*x+c)+2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^
2/(a-b)^(1/2))*a^7-524*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^5*a^3*b^2-5
24*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^6*a^3*b^2+192*(a-b)^(3/2)*((b+a
*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^4*a^4*b-524*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c
)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^4*a^2*b^3+210*(a-b)^(3/2)*a^(3/2)*cos(d*x+c)^6*ln(4*a^(1/2)*cos(d*x+c)*((
b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(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)*b^4+210*(a-b)^(3/2)*a^(1/2)*cos(d*x+c)^5*ln(4*a^(1/2)*cos(d*x+c)*((b+a*cos(d*x+c))*cos
(d*x+c)/(1+cos(d*x+c))^2)^(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)*b^5-108*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^5*a^2*b^2-288*(a-b)^(
3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^4*a^3*b+840*(a-b)^(3/2)*((b+a*cos(d*x+c))
*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^4*a*b^3+180*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x
+c))^2)^(3/2)*cos(d*x+c)^3*a^2*b^2-96*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x
+c)^2*a^3*b+100*(a-b)^(3/2)*((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+72*(a-b)^(
3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)*a^2*b^2+192*(a-b)^(3/2)*((b+a*cos(d*x+c))
*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*cos(d*x+c)^5*a^4*b-524*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x
+c))^2)^(1/2)*cos(d*x+c)^5*a^2*b^3-36*(a-b)^(3/2)*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^
(3/2)*a^2*b^2-96*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^5*a^3*b+280*(a-b)
^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^5*a*b^3-36*(a-b)^(3/2)*((b+a*cos(d*x+c)
)*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^4*a^2*b^2-288*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(
d*x+c))^2)^(3/2)*cos(d*x+c)^3*a^3*b+780*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d
*x+c)^3*a*b^3+216*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)^2*a^2*b^2-180*(a
-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*cos(d*x+c)*a*b^3)/cos(d*x+c)^3/sin(d*x+c)^8/((b
+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)/b^4/a/(a-b)^(3/2)
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maxima [A] time = 0.43, size = 175, normalized size = 1.18 \[ \frac {\frac {105 \, \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a}} + \frac {30 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}}}{b^{4}} - \frac {126 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} a}{b^{4}} + \frac {210 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a^{2}}{b^{4}} - \frac {210 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a^{3}}{b^{4}} - \frac {140 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{b^{2}} + \frac {420 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a}{b^{2}}}{105 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)^5/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
1/105*(105*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a)))/sqrt(a) + 30*(a + b/
cos(d*x + c))^(7/2)/b^4 - 126*(a + b/cos(d*x + c))^(5/2)*a/b^4 + 210*(a + b/cos(d*x + c))^(3/2)*a^2/b^4 - 210*
sqrt(a + b/cos(d*x + c))*a^3/b^4 - 140*(a + b/cos(d*x + c))^(3/2)/b^2 + 420*sqrt(a + b/cos(d*x + c))*a/b^2)/d
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^5}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(c + d*x)^5/(a + b/cos(c + d*x))^(1/2),x)
[Out]
int(tan(c + d*x)^5/(a + b/cos(c + d*x))^(1/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(d*x+c)**5/(a+b*sec(d*x+c))**(1/2),x)
[Out]
Integral(tan(c + d*x)**5/sqrt(a + b*sec(c + d*x)), x)
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