3.393 \(\int \tan ^3(x) (a+b \tan ^4(x))^{3/2} \, dx\)

Optimal. Leaf size=148 \[ \frac {\left (3 a^2+12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{16 \sqrt {b}}-\frac {1}{24} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}-\frac {1}{16} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}+\frac {1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right ) \]

[Out]

1/2*(a+b)^(3/2)*arctanh((a-b*tan(x)^2)/(a+b)^(1/2)/(a+b*tan(x)^4)^(1/2))+1/16*(3*a^2+12*a*b+8*b^2)*arctanh(b^(
1/2)*tan(x)^2/(a+b*tan(x)^4)^(1/2))/b^(1/2)-1/16*(a+b*tan(x)^4)^(1/2)*(8*a+8*b-(3*a+4*b)*tan(x)^2)-1/24*(4-3*t
an(x)^2)*(a+b*tan(x)^4)^(3/2)

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Rubi [A]  time = 0.31, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3670, 1252, 815, 844, 217, 206, 725} \[ \frac {\left (3 a^2+12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{16 \sqrt {b}}-\frac {1}{24} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}-\frac {1}{16} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}+\frac {1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^3*(a + b*Tan[x]^4)^(3/2),x]

[Out]

((3*a^2 + 12*a*b + 8*b^2)*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]])/(16*Sqrt[b]) + ((a + b)^(3/2)*ArcT
anh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])])/2 - ((8*(a + b) - (3*a + 4*b)*Tan[x]^2)*Sqrt[a + b*T
an[x]^4])/16 - ((4 - 3*Tan[x]^2)*(a + b*Tan[x]^4)^(3/2))/24

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 0]

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \tan ^3(x) \left (a+b \tan ^4(x)\right )^{3/2} \, dx &=\operatorname {Subst}\left (\int \frac {x^3 \left (a+b x^4\right )^{3/2}}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x \left (a+b x^2\right )^{3/2}}{1+x} \, dx,x,\tan ^2(x)\right )\\ &=-\frac {1}{24} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}+\frac {\operatorname {Subst}\left (\int \frac {(-a b+b (3 a+4 b) x) \sqrt {a+b x^2}}{1+x} \, dx,x,\tan ^2(x)\right )}{8 b}\\ &=-\frac {1}{16} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}-\frac {1}{24} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}+\frac {\operatorname {Subst}\left (\int \frac {-a b^2 (5 a+4 b)+b^2 \left (3 a^2+12 a b+8 b^2\right ) x}{(1+x) \sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right )}{16 b^2}\\ &=-\frac {1}{16} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}-\frac {1}{24} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}-\frac {1}{2} (a+b)^2 \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right )+\frac {1}{16} \left (3 a^2+12 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right )\\ &=-\frac {1}{16} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}-\frac {1}{24} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}+\frac {1}{2} (a+b)^2 \operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\frac {a-b \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )+\frac {1}{16} \left (3 a^2+12 a b+8 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )\\ &=\frac {\left (3 a^2+12 a b+8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )}{16 \sqrt {b}}+\frac {1}{2} (a+b)^{3/2} \tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )-\frac {1}{16} \left (8 (a+b)-(3 a+4 b) \tan ^2(x)\right ) \sqrt {a+b \tan ^4(x)}-\frac {1}{24} \left (4-3 \tan ^2(x)\right ) \left (a+b \tan ^4(x)\right )^{3/2}\\ \end {align*}

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Mathematica [B]  time = 6.08, size = 324, normalized size = 2.19 \[ \frac {1}{2} a \tan ^2(x) \sqrt {a+b \tan ^4(x)} \left (\frac {b \tan ^4(x)}{a}+1\right )^2 \left (\frac {1}{4} \left (\frac {1}{\frac {b \tan ^4(x)}{a}+1}+\frac {3}{2 \left (\frac {b \tan ^4(x)}{a}+1\right )^2}\right )+\frac {3 \sqrt {a} \cot ^2(x) \sinh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a}}\right )}{8 \sqrt {b} \left (\frac {b \tan ^4(x)}{a}+1\right )^{5/2}}\right )+\frac {1}{2} \left (-\frac {1}{3} \left (a+b \tan ^4(x)\right )^{3/2}-(a+b) \left (\sqrt {a+b \tan ^4(x)}-\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )-\sqrt {a+b} \tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )\right )+b \tan ^2(x) \sqrt {a+b \tan ^4(x)} \left (\frac {b \tan ^4(x)}{a}+1\right ) \left (\frac {1}{2 \left (\frac {b \tan ^4(x)}{a}+1\right )}+\frac {\sqrt {a} \cot ^2(x) \sinh ^{-1}\left (\frac {\sqrt {b} \tan ^2(x)}{\sqrt {a}}\right )}{2 \sqrt {b} \left (\frac {b \tan ^4(x)}{a}+1\right )^{3/2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^3*(a + b*Tan[x]^4)^(3/2),x]

[Out]

(-1/3*(a + b*Tan[x]^4)^(3/2) - (a + b)*(-(Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[x]^2)/Sqrt[a + b*Tan[x]^4]]) - Sqrt[a +
b]*ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])] + Sqrt[a + b*Tan[x]^4]) + b*Tan[x]^2*Sqrt[a +
b*Tan[x]^4]*(1 + (b*Tan[x]^4)/a)*((Sqrt[a]*ArcSinh[(Sqrt[b]*Tan[x]^2)/Sqrt[a]]*Cot[x]^2)/(2*Sqrt[b]*(1 + (b*T
an[x]^4)/a)^(3/2)) + 1/(2*(1 + (b*Tan[x]^4)/a))))/2 + (a*Tan[x]^2*Sqrt[a + b*Tan[x]^4]*(1 + (b*Tan[x]^4)/a)^2*
((3*Sqrt[a]*ArcSinh[(Sqrt[b]*Tan[x]^2)/Sqrt[a]]*Cot[x]^2)/(8*Sqrt[b]*(1 + (b*Tan[x]^4)/a)^(5/2)) + (3/(2*(1 +
(b*Tan[x]^4)/a)^2) + (1 + (b*Tan[x]^4)/a)^(-1))/4))/2

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fricas [A]  time = 0.87, size = 758, normalized size = 5.12 \[ \left [\frac {3 \, {\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \sqrt {b} \log \left (-2 \, b \tan \relax (x)^{4} - 2 \, \sqrt {b \tan \relax (x)^{4} + a} \sqrt {b} \tan \relax (x)^{2} - a\right ) + 24 \, {\left (a b + b^{2}\right )} \sqrt {a + b} \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \relax (x)^{4} - 2 \, a b \tan \relax (x)^{2} - 2 \, \sqrt {b \tan \relax (x)^{4} + a} {\left (b \tan \relax (x)^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \relax (x)^{4} + 2 \, \tan \relax (x)^{2} + 1}\right ) + 2 \, {\left (6 \, b^{2} \tan \relax (x)^{6} - 8 \, b^{2} \tan \relax (x)^{4} + 3 \, {\left (5 \, a b + 4 \, b^{2}\right )} \tan \relax (x)^{2} - 32 \, a b - 24 \, b^{2}\right )} \sqrt {b \tan \relax (x)^{4} + a}}{96 \, b}, -\frac {3 \, {\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \relax (x)^{4} + a} \sqrt {-b}}{b \tan \relax (x)^{2}}\right ) - 12 \, {\left (a b + b^{2}\right )} \sqrt {a + b} \log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \relax (x)^{4} - 2 \, a b \tan \relax (x)^{2} - 2 \, \sqrt {b \tan \relax (x)^{4} + a} {\left (b \tan \relax (x)^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \relax (x)^{4} + 2 \, \tan \relax (x)^{2} + 1}\right ) - {\left (6 \, b^{2} \tan \relax (x)^{6} - 8 \, b^{2} \tan \relax (x)^{4} + 3 \, {\left (5 \, a b + 4 \, b^{2}\right )} \tan \relax (x)^{2} - 32 \, a b - 24 \, b^{2}\right )} \sqrt {b \tan \relax (x)^{4} + a}}{48 \, b}, \frac {48 \, {\left (a b + b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {b \tan \relax (x)^{4} + a} {\left (b \tan \relax (x)^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \relax (x)^{4} + a^{2} + a b}\right ) + 3 \, {\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \sqrt {b} \log \left (-2 \, b \tan \relax (x)^{4} - 2 \, \sqrt {b \tan \relax (x)^{4} + a} \sqrt {b} \tan \relax (x)^{2} - a\right ) + 2 \, {\left (6 \, b^{2} \tan \relax (x)^{6} - 8 \, b^{2} \tan \relax (x)^{4} + 3 \, {\left (5 \, a b + 4 \, b^{2}\right )} \tan \relax (x)^{2} - 32 \, a b - 24 \, b^{2}\right )} \sqrt {b \tan \relax (x)^{4} + a}}{96 \, b}, \frac {24 \, {\left (a b + b^{2}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {b \tan \relax (x)^{4} + a} {\left (b \tan \relax (x)^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \relax (x)^{4} + a^{2} + a b}\right ) - 3 \, {\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b \tan \relax (x)^{4} + a} \sqrt {-b}}{b \tan \relax (x)^{2}}\right ) + {\left (6 \, b^{2} \tan \relax (x)^{6} - 8 \, b^{2} \tan \relax (x)^{4} + 3 \, {\left (5 \, a b + 4 \, b^{2}\right )} \tan \relax (x)^{2} - 32 \, a b - 24 \, b^{2}\right )} \sqrt {b \tan \relax (x)^{4} + a}}{48 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3*(a+b*tan(x)^4)^(3/2),x, algorithm="fricas")

[Out]

[1/96*(3*(3*a^2 + 12*a*b + 8*b^2)*sqrt(b)*log(-2*b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*sqrt(b)*tan(x)^2 - a) + 2
4*(a*b + b^2)*sqrt(a + b)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 -
a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) + 2*(6*b^2*tan(x)^6 - 8*b^2*tan(x)^4 + 3*(5*a*b + 4
*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sqrt(b*tan(x)^4 + a))/b, -1/48*(3*(3*a^2 + 12*a*b + 8*b^2)*sqrt(-b)*arctan(s
qrt(b*tan(x)^4 + a)*sqrt(-b)/(b*tan(x)^2)) - 12*(a*b + b^2)*sqrt(a + b)*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*ta
n(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) - (6*
b^2*tan(x)^6 - 8*b^2*tan(x)^4 + 3*(5*a*b + 4*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sqrt(b*tan(x)^4 + a))/b, 1/96*(4
8*(a*b + b^2)*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2)*tan(x)^4 + a
^2 + a*b)) + 3*(3*a^2 + 12*a*b + 8*b^2)*sqrt(b)*log(-2*b*tan(x)^4 - 2*sqrt(b*tan(x)^4 + a)*sqrt(b)*tan(x)^2 -
a) + 2*(6*b^2*tan(x)^6 - 8*b^2*tan(x)^4 + 3*(5*a*b + 4*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sqrt(b*tan(x)^4 + a))/
b, 1/48*(24*(a*b + b^2)*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2)*ta
n(x)^4 + a^2 + a*b)) - 3*(3*a^2 + 12*a*b + 8*b^2)*sqrt(-b)*arctan(sqrt(b*tan(x)^4 + a)*sqrt(-b)/(b*tan(x)^2))
+ (6*b^2*tan(x)^6 - 8*b^2*tan(x)^4 + 3*(5*a*b + 4*b^2)*tan(x)^2 - 32*a*b - 24*b^2)*sqrt(b*tan(x)^4 + a))/b]

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giac [A]  time = 0.60, size = 176, normalized size = 1.19 \[ \frac {1}{48} \, \sqrt {b \tan \relax (x)^{4} + a} {\left ({\left (2 \, {\left (3 \, b \tan \relax (x)^{2} - 4 \, b\right )} \tan \relax (x)^{2} + \frac {3 \, {\left (5 \, a b^{2} + 4 \, b^{3}\right )}}{b^{2}}\right )} \tan \relax (x)^{2} - \frac {8 \, {\left (4 \, a b^{2} + 3 \, b^{3}\right )}}{b^{2}}\right )} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \arctan \left (-\frac {\sqrt {b} \tan \relax (x)^{2} - \sqrt {b \tan \relax (x)^{4} + a} + \sqrt {b}}{\sqrt {-a - b}}\right )}{\sqrt {-a - b}} - \frac {{\left (3 \, a^{2} \sqrt {b} + 12 \, a b^{\frac {3}{2}} + 8 \, b^{\frac {5}{2}}\right )} \log \left ({\left | -\sqrt {b} \tan \relax (x)^{2} + \sqrt {b \tan \relax (x)^{4} + a} \right |}\right )}{16 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3*(a+b*tan(x)^4)^(3/2),x, algorithm="giac")

[Out]

1/48*sqrt(b*tan(x)^4 + a)*((2*(3*b*tan(x)^2 - 4*b)*tan(x)^2 + 3*(5*a*b^2 + 4*b^3)/b^2)*tan(x)^2 - 8*(4*a*b^2 +
3*b^3)/b^2) - (a^2 + 2*a*b + b^2)*arctan(-(sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a) + sqrt(b))/sqrt(-a - b))/s
qrt(-a - b) - 1/16*(3*a^2*sqrt(b) + 12*a*b^(3/2) + 8*b^(5/2))*log(abs(-sqrt(b)*tan(x)^2 + sqrt(b*tan(x)^4 + a)
))/b

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maple [B]  time = 0.21, size = 374, normalized size = 2.53 \[ \frac {b \left (\tan ^{6}\relax (x )\right ) \sqrt {a +b \left (\tan ^{4}\relax (x )\right )}}{8}+\frac {5 a \left (\tan ^{2}\relax (x )\right ) \sqrt {a +b \left (\tan ^{4}\relax (x )\right )}}{16}+\frac {3 a^{2} \ln \left (\sqrt {b}\, \left (\tan ^{2}\relax (x )\right )+\sqrt {a +b \left (\tan ^{4}\relax (x )\right )}\right )}{16 \sqrt {b}}-\frac {b \left (\tan ^{4}\relax (x )\right ) \sqrt {a +b \left (\tan ^{4}\relax (x )\right )}}{6}-\frac {2 a \sqrt {a +b \left (\tan ^{4}\relax (x )\right )}}{3}+\frac {b \left (\tan ^{2}\relax (x )\right ) \sqrt {a +b \left (\tan ^{4}\relax (x )\right )}}{4}+\frac {3 a \sqrt {b}\, \ln \left (\sqrt {b}\, \left (\tan ^{2}\relax (x )\right )+\sqrt {a +b \left (\tan ^{4}\relax (x )\right )}\right )}{4}-\frac {b \sqrt {a +b \left (\tan ^{4}\relax (x )\right )}}{2}+\frac {b^{\frac {3}{2}} \ln \left (\sqrt {b}\, \left (\tan ^{2}\relax (x )\right )+\sqrt {a +b \left (\tan ^{4}\relax (x )\right )}\right )}{2}+\frac {\ln \left (\frac {2 a +2 b -2 \left (1+\tan ^{2}\relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\tan ^{2}\relax (x )\right )^{2} b -2 \left (1+\tan ^{2}\relax (x )\right ) b +a +b}}{1+\tan ^{2}\relax (x )}\right ) a^{2}}{2 \sqrt {a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 \left (1+\tan ^{2}\relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\tan ^{2}\relax (x )\right )^{2} b -2 \left (1+\tan ^{2}\relax (x )\right ) b +a +b}}{1+\tan ^{2}\relax (x )}\right ) a b}{\sqrt {a +b}}+\frac {\ln \left (\frac {2 a +2 b -2 \left (1+\tan ^{2}\relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\tan ^{2}\relax (x )\right )^{2} b -2 \left (1+\tan ^{2}\relax (x )\right ) b +a +b}}{1+\tan ^{2}\relax (x )}\right ) b^{2}}{2 \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3*(a+b*tan(x)^4)^(3/2),x)

[Out]

1/8*b*tan(x)^6*(a+b*tan(x)^4)^(1/2)+5/16*a*tan(x)^2*(a+b*tan(x)^4)^(1/2)+3/16*a^2*ln(b^(1/2)*tan(x)^2+(a+b*tan
(x)^4)^(1/2))/b^(1/2)-1/6*b*tan(x)^4*(a+b*tan(x)^4)^(1/2)-2/3*a*(a+b*tan(x)^4)^(1/2)+1/4*b*tan(x)^2*(a+b*tan(x
)^4)^(1/2)+3/4*a*b^(1/2)*ln(b^(1/2)*tan(x)^2+(a+b*tan(x)^4)^(1/2))-1/2*b*(a+b*tan(x)^4)^(1/2)+1/2*b^(3/2)*ln(b
^(1/2)*tan(x)^2+(a+b*tan(x)^4)^(1/2))+1/2/(a+b)^(1/2)*ln((2*a+2*b-2*(1+tan(x)^2)*b+2*(a+b)^(1/2)*((1+tan(x)^2)
^2*b-2*(1+tan(x)^2)*b+a+b)^(1/2))/(1+tan(x)^2))*a^2+1/(a+b)^(1/2)*ln((2*a+2*b-2*(1+tan(x)^2)*b+2*(a+b)^(1/2)*(
(1+tan(x)^2)^2*b-2*(1+tan(x)^2)*b+a+b)^(1/2))/(1+tan(x)^2))*a*b+1/2/(a+b)^(1/2)*ln((2*a+2*b-2*(1+tan(x)^2)*b+2
*(a+b)^(1/2)*((1+tan(x)^2)^2*b-2*(1+tan(x)^2)*b+a+b)^(1/2))/(1+tan(x)^2))*b^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \relax (x)^{4} + a\right )}^{\frac {3}{2}} \tan \relax (x)^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)^3*(a+b*tan(x)^4)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(x)^4 + a)^(3/2)*tan(x)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\relax (x)}^3\,{\left (b\,{\mathrm {tan}\relax (x)}^4+a\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)^3*(a + b*tan(x)^4)^(3/2),x)

[Out]

int(tan(x)^3*(a + b*tan(x)^4)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan ^{4}{\relax (x )}\right )^{\frac {3}{2}} \tan ^{3}{\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)**3*(a+b*tan(x)**4)**(3/2),x)

[Out]

Integral((a + b*tan(x)**4)**(3/2)*tan(x)**3, x)

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