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

Optimal. Leaf size=79 \[ \frac {2 (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac {2 a \sqrt {a+b \sec (c+d x)}}{b^2 d}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

[Out]

2/3*(a+b*sec(d*x+c))^(3/2)/b^2/d+2*arctanh((a+b*sec(d*x+c))^(1/2)/a^(1/2))/d/a^(1/2)-2*a*(a+b*sec(d*x+c))^(1/2
)/b^2/d

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Rubi [A]  time = 0.10, antiderivative size = 79, 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} \[ \frac {2 (a+b \sec (c+d x))^{3/2}}{3 b^2 d}-\frac {2 a \sqrt {a+b \sec (c+d x)}}{b^2 d}+\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

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Mathematica [B]  time = 1.23, size = 194, normalized size = 2.46 \[ \frac {\sec (c+d x) (a \cos (c+d x)+b) \left (\frac {2 \sec (c+d x)}{3 b}-\frac {4 a}{3 b^2}\right )}{d \sqrt {a+b \sec (c+d x)}}+\frac {\sin (c+d x) \tan (c+d x) \sqrt {a \cos (c+d x)} \sqrt {a \cos (c+d x)+b} \left (\log \left (\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}+1\right )-\log \left (1-\frac {\sqrt {a \cos (c+d x)+b}}{\sqrt {a \cos (c+d x)}}\right )\right )}{a d \left (1-\cos ^2(c+d x)\right ) \sqrt {a+b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.19, size = 273, normalized size = 3.46 \[ \left [\frac {3 \, \sqrt {a} b^{2} \cos \left (d x + c\right ) \log \left (-8 \, a^{2} \cos \left (d x + c\right )^{2} - 8 \, a b \cos \left (d x + c\right ) - b^{2} - 4 \, {\left (2 \, a \cos \left (d x + c\right )^{2} + b \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}\right ) - 4 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) - a b\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{6 \, a b^{2} d \cos \left (d x + c\right )}, -\frac {3 \, \sqrt {-a} b^{2} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{2 \, a \cos \left (d x + c\right ) + b}\right ) \cos \left (d x + c\right ) + 2 \, {\left (2 \, a^{2} \cos \left (d x + c\right ) - a b\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + b}{\cos \left (d x + c\right )}}}{3 \, a b^{2} d \cos \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*(3*sqrt(a)*b^2*cos(d*x + c)*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^2*cos(d*x + c) - a*b)*sqrt((a*cos(
d*x + c) + b)/cos(d*x + c)))/(a*b^2*d*cos(d*x + c)), -1/3*(3*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^2*cos(d*x + c) - a*b)*sqrt((a
*cos(d*x + c) + b)/cos(d*x + c)))/(a*b^2*d*cos(d*x + c))]

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giac [B]  time = 1.44, size = 288, normalized size = 3.65 \[ \frac {2 \, {\left (\frac {3 \, \arctan \left (-\frac {\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} + \sqrt {a - b}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {2 \, {\left (3 \, {\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}\right )}^{2} - 3 \, a - b\right )}}{{\left (\sqrt {a - b} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b} - \sqrt {a - b}\right )}^{3} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(3*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*(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*ta
n(1/2*d*x + 1/2*c)^2 + a + b))^2 - 3*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))^3*sgn(tan(1/2*d*x + 1/2*c)^
2 - 1)))/d

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maple [B]  time = 1.68, size = 3003, normalized size = 38.01 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/d*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^3*(3*cos(d*x+c)^4*ln(-(-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^5-3*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^5+6*cos(d*x+c)^3*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^2*b^3-3*cos(d*x+c)^3*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^4-3*cos(d*x+c)^3*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^4*b+6*cos(d*x+c)^3*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))*a^3*b^2
-6*cos(d*x+c)^3*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^3+3*cos(d*x+c)^3*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^4+4*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*
(a-b)^(3/2)*a*b-8*cos(d*x+c)^3*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^3-8*cos(d*x+
c)^4*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^3-6*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^4*b+6*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^3*b^2-3*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^2*b^3+6*cos(d*x+c)^4*ln(-2*(-1+cos(d*x+c))*(2*cos(d*x+c)*(a-b)^(1/2)*((b+a*c
os(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+co
s(d*x+c))^2)^(1/2)*(a-b)^(1/2)-b)/sin(d*x+c)^2/(a-b)^(1/2))*a^4*b-6*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*co
s(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^2+3*cos(d*x+c)^4*l
n(-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^3+3*cos(d*x+c)^3*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-6*cos(d*x+c)^3*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^2-8*cos(d*x+c)^2*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos
(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^2*b+6*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+c
os(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-b)^(
3/2)*a^(3/2)*b^2+6*cos(d*x+c)^3*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-b)^(3/2)*a^(1/2)*b^3+12*c
os(d*x+c)^2*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*a*b+4*cos(d*x+c)^3*(a-b)^(3/2)*((
b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)*a*b+12*cos(d*x+c)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c
))^2)^(3/2)*(a-b)^(3/2)*a*b-8*cos(d*x+c)^3*(a-b)^(3/2)*((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(1/2)*a^
2*b)/sin(d*x+c)^6/cos(d*x+c)/((b+a*cos(d*x+c))*cos(d*x+c)/(1+cos(d*x+c))^2)^(3/2)/b^2/a/(a-b)^(3/2)

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maxima [A]  time = 0.43, size = 92, normalized size = 1.16 \[ -\frac {\frac {3 \, \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} - \sqrt {a}}{\sqrt {a + \frac {b}{\cos \left (d x + c\right )}} + \sqrt {a}}\right )}{\sqrt {a}} - \frac {2 \, {\left (a + \frac {b}{\cos \left (d x + c\right )}\right )}^{\frac {3}{2}}}{b^{2}} + \frac {6 \, \sqrt {a + \frac {b}{\cos \left (d x + c\right )}} a}{b^{2}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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 \[ \int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {a + b \sec {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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