3.319 \(\int \sqrt {a+b \sec (c+d x)} \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=100 \[ \frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^2 d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac {2 \sqrt {a+b \sec (c+d x)}}{d}+\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{d} \]

[Out]

-2/3*a*(a+b*sec(d*x+c))^(3/2)/b^2/d+2/5*(a+b*sec(d*x+c))^(5/2)/b^2/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 = 0.11, antiderivative size = 100, 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} \[ \frac {2 (a+b \sec (c+d x))^{5/2}}{5 b^2 d}-\frac {2 a (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac {2 \sqrt {a+b \sec (c+d x)}}{d}+\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [A]  time = 6.22, size = 194, normalized size = 1.94 \[ \frac {\sqrt {a+b \sec (c+d x)} \left (-\frac {2 \left (2 a^2+15 b^2\right )}{15 b^2}+\frac {2 a \sec (c+d x)}{15 b}+\frac {2}{5} \sec ^2(c+d x)\right )}{d}+\frac {\sin ^2(c+d x) \sqrt {a \cos (c+d x)} \sqrt {a+b \sec (c+d x)} \left (\log \left (\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}+1\right )-\log \left (1-\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}\right )\right )}{d \left (1-\cos ^2(c+d x)\right ) \sqrt {a \cos (c+d x)+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]^3,x]

[Out]

(Sqrt[a + b*Sec[c + d*x]]*((-2*(2*a^2 + 15*b^2))/(15*b^2) + (2*a*Sec[c + d*x])/(15*b) + (2*Sec[c + d*x]^2)/5))
/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*Cos[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 - Cos[
c + d*x]^2))

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fricas [A]  time = 1.45, size = 311, normalized size = 3.11 \[ \left [\frac {15 \, \sqrt {a} b^{2} \cos \left (d x + c\right )^{2} \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} - 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) + 4 \, {\left (a b \cos \left (d x + c\right ) - {\left (2 \, a^{2} + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{30 \, b^{2} d \cos \left (d x + c\right )^{2}}, -\frac {15 \, \sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right )^{2} - 2 \, {\left (a b \cos \left (d x + c\right ) - {\left (2 \, a^{2} + 15 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \, b^{2}\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{15 \, b^{2} d \cos \left (d x + c\right )^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="fricas")

[Out]

[1/30*(15*sqrt(a)*b^2*cos(d*x + c)^2*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*b*cos(d*x + c) - (2*a^2 + 15*b^2
)*cos(d*x + c)^2 + 3*b^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/(b^2*d*cos(d*x + c)^2), -1/15*(15*sqrt(-a)*
b^2*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
)^2 - 2*(a*b*cos(d*x + c) - (2*a^2 + 15*b^2)*cos(d*x + c)^2 + 3*b^2)*sqrt((a*cos(d*x + c) + b)/cos(d*x + c)))/
(b^2*d*cos(d*x + c)^2)]

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giac [B]  time = 2.17, size = 539, normalized size = 5.39 \[ -\frac {2 \, {\left (\frac {15 \, a \arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} + \sqrt {a - b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a}} - \frac {2 \, {\left (15 \, {\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )}^{4} a - 30 \, {\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )}^{3} {\left (a + 2 \, b\right )} \sqrt {a - b} + 20 \, {\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )}^{2} {\left (4 \, a b - 3 \, b^{2}\right )} - 15 \, a^{3} - 10 \, a^{2} b - 35 \, a b^{2} + 12 \, b^{3} + 10 \, {\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )} {\left (3 \, a^{2} - a b + 6 \, b^{2}\right )} \sqrt {a - b}\right )}}{{\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} - \sqrt {a - b}\right )}^{5}}\right )} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}{15 \, 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)^3,x, algorithm="giac")

[Out]

-2/15*(15*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*(15*(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*a - 30*(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*(a + 2*b)*sqrt(a - b) + 20*(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*(4*a*b - 3*b^2) - 1
5*a^3 - 10*a^2*b - 35*a*b^2 + 12*b^3 + 10*(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*a^2 - a*b + 6*b^2)*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))^5)*sgn(cos(d*x + c))/d

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maple [B]  time = 1.72, size = 2342, normalized size = 23.42 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x)

[Out]

-1/60/d*(1+cos(d*x+c))*(-1+cos(d*x+c))^4*(-36*(a-b)^(1/2)*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c
))^2)^(3/2)*b^2+6*(a-b)^(1/2)*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*b^2+6*(a-b)^(1
/2)*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*b^2+54*(a-b)^(1/2)*cos(d*x+c)^4*((b+a*co
s(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*b^3+8*(a-b)^(1/2)*cos(d*x+c)^5*((b+a*cos(d*x+c))*cos(d*x+c)/(1+co
s(d*x+c))^2)^(1/2)*a^3+8*(a-b)^(1/2)*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^3+54*
(a-b)^(1/2)*cos(d*x+c)^3*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*b^3+18*(a-b)^(1/2)*cos(d*x+c)^4*
((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*b^2+30*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^2*b^2-15*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
^3-30*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^2+15*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^3+30*cos(d*x+c)^4*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^3-30*cos(d*x+c)^4*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^3-30*
(a-b)^(1/2)*cos(d*x+c)^2*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*b^2-15*cos(d*x+c)^4*ln(-2*(-1+co
s(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))*b^4+15*
cos(d*x+c)^4*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))*b^4-12*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*(a-b)^(1/2)*b^2+54*(a-b)^(1/2)*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^2+8*(a-b)^(1/2)*cos(d*x+c)^3*((b+a*cos(d*x+
c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^2*b+8*(a-b)^(1/2)*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*
x+c))^2)^(1/2)*a^2*b+54*(a-b)^(1/2)*cos(d*x+c)^4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a*b^2-30
*(a-b)^(1/2)*a^(3/2)*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^2-30*(a-b)^(1/2)*a^(1/2)
*cos(d*x+c)^4*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^3-12*(a-b)^(1/2)*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-4*(a-b)^(1/2)*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c)
)^2)^(3/2)*a*b-4*(a-b)^(1/2)*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-12*(a-b)^(1
/2)*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)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)
*4^(1/2)/((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)/sin(d*x+c)^8/cos(d*x+c)^2/b^2/(a-b)^(1/2)

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maxima [A]  time = 0.42, size = 108, normalized size = 1.08 \[ -\frac {15 \, \sqrt {a} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right ) + 30 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \frac {6 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {5}{2}}}{b^{2}} + \frac {10 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}} a}{b^{2}}}{15 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))^(1/2)*tan(d*x+c)^3,x, algorithm="maxima")

[Out]

-1/15*(15*sqrt(a)*log((sqrt(a + b/cos(d*x + c)) - sqrt(a))/(sqrt(a + b/cos(d*x + c)) + sqrt(a))) + 30*sqrt(a +
b/cos(d*x + c)) - 6*(a + b/cos(d*x + c))^(5/2)/b^2 + 10*(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 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^3\,\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + b/cos(c + d*x))^(1/2),x)

[Out]

int(tan(c + d*x)^3*(a + b/cos(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sec {\left (c + d x \right )}} \tan ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sec(d*x+c))**(1/2)*tan(d*x+c)**3,x)

[Out]

Integral(sqrt(a + b*sec(c + d*x))*tan(c + d*x)**3, x)

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