\(\int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) [529]

Optimal result

Integrand size = 23, antiderivative size = 500 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {b \text {arctanh}\left (\frac {\sqrt {a+\sqrt {a^2+b^2}}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}} d}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+\sqrt {a^2+b^2}}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}} d}-\frac {b \log \left (a+\sqrt {a^2+b^2}+b \tan (c+d x)-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} \sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {b \log \left (a+\sqrt {a^2+b^2}+b \tan (c+d x)+\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} \sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \]

[Out]

1/2*b*arctanh(((a+(a^2+b^2)^(1/2))^(1/2)-2^(1/2)*(a+b*tan(d*x+c))^(1/2))/(a-(a^2+b^2)^(1/2))^(1/2))/d*2^(1/2)/
(a^2+b^2)^(1/2)/(a-(a^2+b^2)^(1/2))^(1/2)-1/2*b*arctanh(((a+(a^2+b^2)^(1/2))^(1/2)+2^(1/2)*(a+b*tan(d*x+c))^(1
/2))/(a-(a^2+b^2)^(1/2))^(1/2))/d*2^(1/2)/(a^2+b^2)^(1/2)/(a-(a^2+b^2)^(1/2))^(1/2)-1/4*b*ln(a+(a^2+b^2)^(1/2)
-2^(1/2)*(a+(a^2+b^2)^(1/2))^(1/2)*(a+b*tan(d*x+c))^(1/2)+b*tan(d*x+c))/d*2^(1/2)/(a^2+b^2)^(1/2)/(a+(a^2+b^2)
^(1/2))^(1/2)+1/4*b*ln(a+(a^2+b^2)^(1/2)+2^(1/2)*(a+(a^2+b^2)^(1/2))^(1/2)*(a+b*tan(d*x+c))^(1/2)+b*tan(d*x+c)
)/d*2^(1/2)/(a^2+b^2)^(1/2)/(a+(a^2+b^2)^(1/2))^(1/2)+2/15*(8*a^2-15*b^2)*(a+b*tan(d*x+c))^(1/2)/b^3/d-8/15*a*
(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)/b^2/d+2/5*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^2/b/d

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 500, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3647, 3728, 3712, 3566, 722, 1108, 648, 632, 212, 642} \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {b \text {arctanh}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}}}-\frac {b \text {arctanh}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}}}-\frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {a^2+b^2} \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {a^2+b^2} \sqrt {\sqrt {a^2+b^2}+a}}+\frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \]

[In]

Int[Tan[c + d*x]^4/Sqrt[a + b*Tan[c + d*x]],x]

[Out]

(b*ArcTanh[(Sqrt[a + Sqrt[a^2 + b^2]] - Sqrt[2]*Sqrt[a + b*Tan[c + d*x]])/Sqrt[a - Sqrt[a^2 + b^2]]])/(Sqrt[2]
*Sqrt[a^2 + b^2]*Sqrt[a - Sqrt[a^2 + b^2]]*d) - (b*ArcTanh[(Sqrt[a + Sqrt[a^2 + b^2]] + Sqrt[2]*Sqrt[a + b*Tan
[c + d*x]])/Sqrt[a - Sqrt[a^2 + b^2]]])/(Sqrt[2]*Sqrt[a^2 + b^2]*Sqrt[a - Sqrt[a^2 + b^2]]*d) - (b*Log[a + Sqr
t[a^2 + b^2] + b*Tan[c + d*x] - Sqrt[2]*Sqrt[a + Sqrt[a^2 + b^2]]*Sqrt[a + b*Tan[c + d*x]]])/(2*Sqrt[2]*Sqrt[a
^2 + b^2]*Sqrt[a + Sqrt[a^2 + b^2]]*d) + (b*Log[a + Sqrt[a^2 + b^2] + b*Tan[c + d*x] + Sqrt[2]*Sqrt[a + Sqrt[a
^2 + b^2]]*Sqrt[a + b*Tan[c + d*x]]])/(2*Sqrt[2]*Sqrt[a^2 + b^2]*Sqrt[a + Sqrt[a^2 + b^2]]*d) + (2*(8*a^2 - 15
*b^2)*Sqrt[a + b*Tan[c + d*x]])/(15*b^3*d) - (8*a*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(15*b^2*d) + (2*Tan[c
+ d*x]^2*Sqrt[a + b*Tan[c + d*x]])/(5*b*d)

Rule 212

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

Rule 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 722

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2*e, Subst[Int[1/(c*d^2 + a*e^2 - 2*c
*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1108

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(-1), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}
, Dist[1/(2*c*q*r), Int[(r - x)/(q - r*x + x^2), x], x] + Dist[1/(2*c*q*r), Int[(r + x)/(q + r*x + x^2), x], x
]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && NegQ[b^2 - 4*a*c]

Rule 3566

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[(a + x)^n/(b^2 + x^2), x], x
, b*Tan[c + d*x]], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[a^2 + b^2, 0]

Rule 3647

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3712

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp
[C*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Dist[A - C, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[
{a, b, e, f, A, C, m}, x] && NeQ[A*b^2 + a^2*C, 0] &&  !LeQ[m, -1]

Rule 3728

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps \begin{align*} \text {integral}& = \frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {2 \int \frac {\tan (c+d x) \left (-2 a-\frac {5}{2} b \tan (c+d x)-2 a \tan ^2(c+d x)\right )}{\sqrt {a+b \tan (c+d x)}} \, dx}{5 b} \\ & = -\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {4 \int \frac {2 a^2+\frac {1}{4} \left (8 a^2-15 b^2\right ) \tan ^2(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx}{15 b^2} \\ & = \frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\int \frac {1}{\sqrt {a+b \tan (c+d x)}} \, dx \\ & = \frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a+x} \left (b^2+x^2\right )} \, dx,x,b \tan (c+d x)\right )}{d} \\ & = \frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {(2 b) \text {Subst}\left (\int \frac {1}{a^2+b^2-2 a x^2+x^4} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{d} \\ & = \frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {b \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}}-x}{\sqrt {a^2+b^2}-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} x+x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{\sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {b \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}}+x}{\sqrt {a^2+b^2}+\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} x+x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{\sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d} \\ & = \frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} x+x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {a^2+b^2} d}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a^2+b^2}+\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} x+x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {a^2+b^2} d}-\frac {b \text {Subst}\left (\int \frac {-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}}+2 x}{\sqrt {a^2+b^2}-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} x+x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {b \text {Subst}\left (\int \frac {\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}}+2 x}{\sqrt {a^2+b^2}+\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} x+x^2} \, dx,x,\sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d} \\ & = -\frac {b \log \left (a+\sqrt {a^2+b^2}+b \tan (c+d x)-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} \sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {b \log \left (a+\sqrt {a^2+b^2}+b \tan (c+d x)+\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} \sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d}-\frac {b \text {Subst}\left (\int \frac {1}{2 \left (a-\sqrt {a^2+b^2}\right )-x^2} \, dx,x,-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}}+2 \sqrt {a+b \tan (c+d x)}\right )}{\sqrt {a^2+b^2} d}-\frac {b \text {Subst}\left (\int \frac {1}{2 \left (a-\sqrt {a^2+b^2}\right )-x^2} \, dx,x,\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}}+2 \sqrt {a+b \tan (c+d x)}\right )}{\sqrt {a^2+b^2} d} \\ & = \frac {b \text {arctanh}\left (\frac {\sqrt {a+\sqrt {a^2+b^2}}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}} d}-\frac {b \text {arctanh}\left (\frac {\sqrt {a+\sqrt {a^2+b^2}}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}} d}-\frac {b \log \left (a+\sqrt {a^2+b^2}+b \tan (c+d x)-\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} \sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {b \log \left (a+\sqrt {a^2+b^2}+b \tan (c+d x)+\sqrt {2} \sqrt {a+\sqrt {a^2+b^2}} \sqrt {a+b \tan (c+d x)}\right )}{2 \sqrt {2} \sqrt {a^2+b^2} \sqrt {a+\sqrt {a^2+b^2}} d}+\frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.37 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {-\frac {15 \sqrt {-b^2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {a-\sqrt {-b^2}}}+\frac {-\frac {15 \left (-b^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {a+\sqrt {-b^2}}}+2 \sqrt {a+b \tan (c+d x)} \left (8 a^2-15 b^2-4 a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)\right )}{b^2}}{15 b d} \]

[In]

Integrate[Tan[c + d*x]^4/Sqrt[a + b*Tan[c + d*x]],x]

[Out]

((-15*Sqrt[-b^2]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/Sqrt[a - Sqrt[-b^2]] + ((-15*(-b^2)^(
3/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + Sqrt[-b^2]]])/Sqrt[a + Sqrt[-b^2]] + 2*Sqrt[a + b*Tan[c + d*x]]
*(8*a^2 - 15*b^2 - 4*a*b*Tan[c + d*x] + 3*b^2*Tan[c + d*x]^2))/b^2)/(15*b*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1640\) vs. \(2(409)=818\).

Time = 0.11 (sec) , antiderivative size = 1641, normalized size of antiderivative = 3.28

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1641\)
default	\(\text {Expression too large to display}\)	\(1641\)

[In]

int(tan(d*x+c)^4/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/5/d/b^3*(a+b*tan(d*x+c))^(5/2)-4/3/d/b^3*a*(a+b*tan(d*x+c))^(3/2)+2/d/b^3*a^2*(a+b*tan(d*x+c))^(1/2)-2*(a+b*
tan(d*x+c))^(1/2)/b/d-1/4/d/b/(a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a
-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/d*b/(a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(
1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/4/d/b/(a^2+b^2)^(3/2)*ln((a+b*
tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*
a^3+1/4/d*b/(a^2+b^2)^(3/2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(
1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-3/d*b/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)
^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2-2/d*b^3/(a^2+b^2)^(3/2)/(2*(a^2
+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)
^(1/2))+1/d/b/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d
*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2+1/d*b/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((
2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))-1/d/b/(a^2+b^2)^(3/2)/(2
*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)
-2*a)^(1/2))*a^4+1/4/d/b/(a^2+b^2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2
+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2+1/4/d*b/(a^2+b^2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(
a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-1/4/d/b/(a^2+b^2)^(3/2)*ln(b*tan(d*x+
c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3-1
/4/d*b/(a^2+b^2)^(3/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))
*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+3/d*b/(a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c)
)^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2+2/d*b^3/(a^2+b^2)^(3/2)/(2*(a^2+b^2)
^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2
))-1/d/b/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a
)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^2-1/d*b/(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b
*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))+1/d/b/(a^2+b^2)^(3/2)/(2*(a^2
+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)
^(1/2))*a^4

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 789, normalized size of antiderivative = 1.58 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {15 \, b^{3} d \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left (\sqrt {b \tan \left (d x + c\right ) + a} b + {\left ({\left (a^{3} + a b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + b^{2} d\right )} \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}}\right ) - 15 \, b^{3} d \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left (\sqrt {b \tan \left (d x + c\right ) + a} b - {\left ({\left (a^{3} + a b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + b^{2} d\right )} \sqrt {-\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} + a}{{\left (a^{2} + b^{2}\right )} d^{2}}}\right ) - 15 \, b^{3} d \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left (\sqrt {b \tan \left (d x + c\right ) + a} b + {\left ({\left (a^{3} + a b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - b^{2} d\right )} \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}}\right ) + 15 \, b^{3} d \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}} \log \left (\sqrt {b \tan \left (d x + c\right ) + a} b - {\left ({\left (a^{3} + a b^{2}\right )} d^{3} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - b^{2} d\right )} \sqrt {\frac {{\left (a^{2} + b^{2}\right )} d^{2} \sqrt {-\frac {b^{2}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{4}}} - a}{{\left (a^{2} + b^{2}\right )} d^{2}}}\right ) + 4 \, {\left (3 \, b^{2} \tan \left (d x + c\right )^{2} - 4 \, a b \tan \left (d x + c\right ) + 8 \, a^{2} - 15 \, b^{2}\right )} \sqrt {b \tan \left (d x + c\right ) + a}}{30 \, b^{3} d} \]

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/30*(15*b^3*d*sqrt(-((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/((a^2 + b^2)*d^2))*log(sqr
t(b*tan(d*x + c) + a)*b + ((a^3 + a*b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + b^2*d)*sqrt(-((a^2 + b
^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/((a^2 + b^2)*d^2))) - 15*b^3*d*sqrt(-((a^2 + b^2)*d^2*sq
rt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/((a^2 + b^2)*d^2))*log(sqrt(b*tan(d*x + c) + a)*b - ((a^3 + a*b^2)
*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + b^2*d)*sqrt(-((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^
4)*d^4)) + a)/((a^2 + b^2)*d^2))) - 15*b^3*d*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) -
a)/((a^2 + b^2)*d^2))*log(sqrt(b*tan(d*x + c) + a)*b + ((a^3 + a*b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d
^4)) - b^2*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))) + 15*b^3
*d*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))*log(sqrt(b*tan(d*x +
c) + a)*b - ((a^3 + a*b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - b^2*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-
b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))) + 4*(3*b^2*tan(d*x + c)^2 - 4*a*b*tan(d*x + c) + 8
*a^2 - 15*b^2)*sqrt(b*tan(d*x + c) + a))/(b^3*d)

Sympy [F]

\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int \frac {\tan ^{4}{\left (c + d x \right )}}{\sqrt {a + b \tan {\left (c + d x \right )}}}\, dx \]

[In]

integrate(tan(d*x+c)**4/(a+b*tan(d*x+c))**(1/2),x)

[Out]

Integral(tan(c + d*x)**4/sqrt(a + b*tan(c + d*x)), x)

Maxima [F]

\[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\int { \frac {\tan \left (d x + c\right )^{4}}{\sqrt {b \tan \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(tan(d*x + c)^4/sqrt(b*tan(d*x + c) + a), x)

Giac [F(-1)]

Timed out. \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Timed out} \]

[In]

integrate(tan(d*x+c)^4/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 9.21 (sec) , antiderivative size = 791, normalized size of antiderivative = 1.58 \[ \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\left (\frac {4\,a^2}{b^3\,d}-\frac {2\,\left (a^2+b^2\right )}{b^3\,d}\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+\frac {\ln \left (16\,b^2\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+16\,b^3\,d\,\sqrt {-\frac {1}{d^2\,\left (a-b\,1{}\mathrm {i}\right )}}-\frac {16\,a\,b^2\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{a-b\,1{}\mathrm {i}}\right )\,\sqrt {-\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}}{2}-\ln \left (-16\,b^2\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}+16\,b^3\,d\,\sqrt {-\frac {1}{d^2\,\left (a-b\,1{}\mathrm {i}\right )}}+\frac {16\,a\,b^2\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{a-b\,1{}\mathrm {i}}\right )\,\sqrt {-\frac {1}{4\,\left (a\,d^2-b\,d^2\,1{}\mathrm {i}\right )}}+\frac {2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{5\,b^3\,d}-\frac {4\,a\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^3\,d}+\mathrm {atan}\left (-\frac {b^2\,\sqrt {-\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{-\frac {64\,a\,b^3\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {b^4\,d^2\,64{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {128\,a\,b^3\,\sqrt {-\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{-\frac {256\,a^3\,b^3\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a\,b^5\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {b^6\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^2\,b^4\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {a^2\,b^2\,\sqrt {-\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}+\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,128{}\mathrm {i}}{-\frac {256\,a^3\,b^3\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a\,b^5\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {b^6\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^2\,b^4\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}\right )\,\sqrt {-\frac {a-b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,2{}\mathrm {i} \]

[In]

int(tan(c + d*x)^4/(a + b*tan(c + d*x))^(1/2),x)

[Out]

((4*a^2)/(b^3*d) - (2*(a^2 + b^2))/(b^3*d))*(a + b*tan(c + d*x))^(1/2) + (log(16*b^2*(a + b*tan(c + d*x))^(1/2
) + 16*b^3*d*(-1/(d^2*(a - b*1i)))^(1/2) - (16*a*b^2*(a + b*tan(c + d*x))^(1/2))/(a - b*1i))*(-1/(a*d^2 - b*d^
2*1i))^(1/2))/2 - log(16*b^3*d*(-1/(d^2*(a - b*1i)))^(1/2) - 16*b^2*(a + b*tan(c + d*x))^(1/2) + (16*a*b^2*(a
+ b*tan(c + d*x))^(1/2))/(a - b*1i))*(-1/(4*(a*d^2 - b*d^2*1i)))^(1/2) + atan((128*a*b^3*((b*1i)/(4*a^2*d^2 +
4*b^2*d^2) - a/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((b^6*d^2*256i)/(4*a^2*d^3 + 4*b^2*d
^3) + (a^2*b^4*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a^3*b^3*d^2)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a*b^5*d^2)
/(4*a^2*d^3 + 4*b^2*d^3)) - (b^2*((b*1i)/(4*a^2*d^2 + 4*b^2*d^2) - a/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan
(c + d*x))^(1/2)*32i)/((b^4*d^2*64i)/(4*a^2*d^3 + 4*b^2*d^3) - (64*a*b^3*d^2)/(4*a^2*d^3 + 4*b^2*d^3)) + (a^2*
b^2*((b*1i)/(4*a^2*d^2 + 4*b^2*d^2) - a/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*128i)/((b^6*
d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) + (a^2*b^4*d^2*256i)/(4*a^2*d^3 + 4*b^2*d^3) - (256*a^3*b^3*d^2)/(4*a^2*d^3
+ 4*b^2*d^3) - (256*a*b^5*d^2)/(4*a^2*d^3 + 4*b^2*d^3)))*(-(a - b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*2i + (2*(
a + b*tan(c + d*x))^(5/2))/(5*b^3*d) - (4*a*(a + b*tan(c + d*x))^(3/2))/(3*b^3*d)