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]

-2/3*a*(a^2-2*b^2)*(a+b*sec(d*x+c))^(3/2)/b^4/d+2/5*(3*a^2-2*b^2)*(a+b*sec(d*x+c))^(5/2)/b^4/d-6/7*a*(a+b*sec(
d*x+c))^(7/2)/b^4/d+2/9*(a+b*sec(d*x+c))^(9/2)/b^4/d-2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d+2*(a+
b*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.17, antiderivative size = 169, 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, 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 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 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 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 \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*}

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Mathematica [A]  time = 6.31, size = 254, normalized size = 1.50 \[ \frac {\sqrt {a+b \sec (c+d x)} \left (-\frac {4 \left (a^2+21 b^2\right ) \sec ^2(c+d x)}{105 b^2}-\frac {4 a \left (21 b^2-4 a^2\right ) \sec (c+d x)}{315 b^3}+\frac {2 \left (-16 a^4+84 a^2 b^2+315 b^4\right )}{315 b^4}+\frac {2 a \sec ^3(c+d x)}{63 b}+\frac {2}{9} \sec ^4(c+d x)\right )}{d}-\frac {\sin ^2(c+d x) \sqrt {a \cos (c+d x)} \sqrt {a+b \sec (c+d x)} \left (\log \left (\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}+1\right )-\log \left (1-\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}\right )\right )}{d \left (1-\cos ^2(c+d x)\right ) \sqrt {a \cos (c+d x)+b}} \]

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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fricas [A]  time = 1.64, size = 425, normalized size = 2.51 \[ \left [\frac {315 \, \sqrt {a} b^{4} \cos \left (d x + c\right )^{4} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} + 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \, {\left (5 \, a b^{3} \cos \left (d x + c\right ) - {\left (16 \, a^{4} - 84 \, a^{2} b^{2} - 315 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 35 \, b^{4} + 2 \, {\left (4 \, a^{3} b - 21 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{2} b^{2} + 21 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{630 \, b^{4} d \cos \left (d x + c\right )^{4}}, \frac {315 \, \sqrt {-a} b^{4} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right )^{4} + 2 \, {\left (5 \, a b^{3} \cos \left (d x + c\right ) - {\left (16 \, a^{4} - 84 \, a^{2} b^{2} - 315 \, b^{4}\right )} \cos \left (d x + c\right )^{4} + 35 \, b^{4} + 2 \, {\left (4 \, a^{3} b - 21 \, a b^{3}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{2} b^{2} + 21 \, b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{315 \, b^{4} d \cos \left (d x + c\right )^{4}}\right ] \]

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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giac [B]  time = 3.76, size = 966, normalized size = 5.72 \[ \text {result too large to display} \]

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]

2/315*(315*a*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) - 2*(315*(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))^8*a - 3150*(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))^7*sqrt(a - b)*a + 210*(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*(39*a^2 - 5*a*b - 32*b
^2) - 630*(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*(9*a^2 + 15*a*b - 16*b^2)*sqrt(a - b) - 252*(25*a^3 - 37*a^2*b + 80*a*b^2
- 72*b^3)*(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 + 945*a^5 + 3864*a^4*b + 2562*a^3*b^2 + 2448*a^2*b^3 - 1083*a*b^4 + 224*b^
5 + 42*(255*a^3 + 2*a^2*b + 655*a*b^2 - 288*b^3)*(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*sqrt(a - b) - 18*(175*a^4 - 483*a^3
*b + 1113*a^2*b^2 - 773*a*b^3 + 448*b^4)*(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 - 18*(105*a^4 + 637*a^3*b + 203*a^2*b^2 + 4
47*a*b^3 - 112*b^4)*(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 - 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))^9)*sgn(cos(d*x
+ c))/d

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maple [B]  time = 2.05, size = 3268, normalized size = 19.34 \[ \text {output too large to display} \]

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*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*(630*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))*cos(d*x+c)^6*
b^6-280*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1/2)*b^4-630*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))*cos(d*x+c)^6*b^6+630*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))*
cos(d*x+c)^7*a*b^5-1197*cos(d*x+c)^6*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*b^4+1260
*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))*cos(d*x+c)^6*a*b^5-1260*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+co
s(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))*cos(d*x+c)^6*a*b^5+2625*cos(d*x+c)^4*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)
/(1+cos(d*x+c))^2)^(3/2)*b^4+168*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)^(1/2)
*b^4-2121*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*b^5-2121*cos(d*x+c)^5*
((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*b^5-189*cos(d*x+c)^5*(a-b)^(1/2)*((b+a*cos(d*
x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*b^4+2744*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)^(1/2)*b^4-840*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1/2)*b^4-399*
cos(d*x+c)^7*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*b^4+128*cos(d*x+c)^6*((b+a*cos(d
*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a^5+128*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos
(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a^5+1260*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))*cos(d*x+c)^7*a^2*b^4-630*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))*cos(d*x+c)^7*a*b^5-1260*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))*cos(d*x+c)^
7*a^2*b^4-648*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a^2*b^3+128*cos(d*
x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a^4*b-648*cos(d*x+c)^6*((b+a*cos(d*x+c
))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a^3*b^2-2121*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(1+co
s(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a*b^4-2121*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a
-b)^(1/2)*a*b^4-648*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a^3*b^2+128*
cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^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)/(1+cos(d*x+c))^2)^(1/2)*(a-b)^(1/2)*a^2*b^3+1260*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)*a^(3/2)*cos(d*x+c)^7*(a-b)^(1/2)*b^4+1260*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)*a^(1/2)*c
os(d*x+c)^6*(a-b)^(1/2)*b^5-72*cos(d*x+c)^6*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^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)/(1+cos(d*x+c))^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)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1/2)*a*b^3+120*cos(d*x+c)^4*((b+a*cos(d*x+c))*c
os(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1/2)*a^2*b^2-64*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)^(1/2)*a^3*b+216*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)^(1/
2)*a*b^3+48*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)^(1/2)*a^2*b^2-40*cos(d*x+c
)*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*a*b^3-24*cos(d*x+c)^7*((b+a*cos(d*x+c))*cos
(d*x+c)/(1+cos(d*x+c))^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)/(1+cos(d*x+c)
)^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)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1/2)
*a*b^3-24*cos(d*x+c)^5*(a-b)^(1/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*a^2*b^2-192*cos(d*x+c)
^4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1/2)*a^3*b+968*cos(d*x+c)^4*((b+a*cos(d*x+c))*c
os(d*x+c)/(1+cos(d*x+c))^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)/(1+cos(d*x+c
))^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)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1
/2)*a*b^3)/sin(d*x+c)^8/cos(d*x+c)^4/((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)/(a-b)^(1/2)/b^4

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maxima [A]  time = 0.43, size = 191, normalized size = 1.13 \[ \frac {315 \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 630 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \frac {70 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {9}{2}}}{b^{4}} - \frac {270 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {7}{2}} a}{b^{4}} + \frac {378 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}} a^{2}}{b^{4}} - \frac {210 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a^{3}}{b^{4}} - \frac {252 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{b^{2}} + \frac {420 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}{b^{2}}}{315 \, d} \]

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]

1/315*(315*sqrt(a)*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a))) + 630*sqrt(a
+ b/cos(d*x + c)) + 70*(a + b/cos(d*x + c))^(9/2)/b^4 - 270*(a + b/cos(d*x + c))^(7/2)*a/b^4 + 378*(a + b/cos
(d*x + c))^(5/2)*a^2/b^4 - 210*(a + b/cos(d*x + c))^(3/2)*a^3/b^4 - 252*(a + b/cos(d*x + c))^(5/2)/b^2 + 420*(
a + b/cos(d*x + c))^(3/2)*a/b^2)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^5\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]

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 \[ \int \sqrt {a + b \sec {\left (c + d x \right )}} \tan ^{5}{\left (c + d x \right )}\, dx \]

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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