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

Optimal. Leaf size=500 \[ -\frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {a^2+b^2} \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {a^2+b^2} \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}}}+\frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \]

[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

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Rubi [A]  time = 0.66, antiderivative size = 500, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3566, 3647, 3631, 3485, 708, 1094, 634, 618, 206, 628} \[ \frac {2 \left (8 a^2-15 b^2\right ) \sqrt {a+b \tan (c+d x)}}{15 b^3 d}-\frac {b \log \left (-\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {a^2+b^2} \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \log \left (\sqrt {2} \sqrt {\sqrt {a^2+b^2}+a} \sqrt {a+b \tan (c+d x)}+\sqrt {a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt {2} d \sqrt {a^2+b^2} \sqrt {\sqrt {a^2+b^2}+a}}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}-\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {\sqrt {a^2+b^2}+a}+\sqrt {2} \sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {a^2+b^2}}}\right )}{\sqrt {2} d \sqrt {a^2+b^2} \sqrt {a-\sqrt {a^2+b^2}}}-\frac {8 a \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{15 b^2 d}+\frac {2 \tan ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{5 b d} \]

Antiderivative was successfully verified.

[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 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 618

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 628

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 634

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 708

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 1094

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 3485

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 3566

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 3631

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 3647

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*} \int \frac {\tan ^4(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx &=\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 \operatorname {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) \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \operatorname {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 \tanh ^{-1}\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 \tanh ^{-1}\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*}

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Mathematica [A]  time = 3.38, size = 184, normalized size = 0.37 \[ \frac {\frac {2 \sqrt {a+b \tan (c+d x)} \left (8 a^2-4 a b \tan (c+d x)+3 b^2 \tan ^2(c+d x)-15 b^2\right )-\frac {15 \left (-b^2\right )^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{\sqrt {a+\sqrt {-b^2}}}}{b^2}-\frac {15 \sqrt {-b^2} \tanh ^{-1}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{\sqrt {a-\sqrt {-b^2}}}}{15 b d} \]

Antiderivative was successfully verified.

[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)

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fricas [B]  time = 1.04, size = 1886, normalized size = 3.77 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(60*sqrt(2)*(a^2*b^3 + b^5)*d^5*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*sqrt
(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(3/4)*arctan((sqrt(2)*(a^4 + 2*a^2*b^2 + b^4)*d^7*sq
rt((sqrt(2)*b^3*d*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 +
b^2)*d^4)) + a^2 + b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4)*cos(d*x + c) + (a^2*b^2 + b^4)*d^2*sqrt(1/((a^2 + b^2
)*d^4))*cos(d*x + c) + a*b^2*cos(d*x + c) + b^3*sin(d*x + c))/cos(d*x + c))*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a
^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(5/4) - sqrt(2
)*(a^4*b + 2*a^2*b^3 + b^5)*d^7*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(((a^3 + a*b^2)*d^2*s
qrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(5/4)
- (a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) - (a^3 + a*b^
2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)))/b^2)*cos(d*x + c)^2 + 60*sqrt(2)*(a^2*b^3 + b^5)*d^5*sqrt(((a^
3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 +
b^2)*d^4))^(3/4)*arctan((sqrt(2)*(a^4 + 2*a^2*b^2 + b^4)*d^7*sqrt(-(sqrt(2)*b^3*d*sqrt((a*cos(d*x + c) + b*sin
(d*x + c))/cos(d*x + c))*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*(1/((a^2 + b^2)*d
^4))^(1/4)*cos(d*x + c) - (a^2*b^2 + b^4)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) - a*b^2*cos(d*x + c) - b^
3*sin(d*x + c))/cos(d*x + c))*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*sqrt(b^2/((a
^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(5/4) - sqrt(2)*(a^4*b + 2*a^2*b^3 + b^5)*d^7*sqrt((a*cos(d*
x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*sqr
t(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))*(1/((a^2 + b^2)*d^4))^(5/4) + (a^4 + 2*a^2*b^2 + b^4)*d^4*sqrt(b^2/((a^4
+ 2*a^2*b^2 + b^4)*d^4))*sqrt(1/((a^2 + b^2)*d^4)) + (a^3 + a*b^2)*d^2*sqrt(b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4))
)/b^2)*cos(d*x + c)^2 + 15*sqrt(2)*(a*b^3*d^3*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c)^2 - b^3*d*cos(d*x + c)^2)
*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4)*log((sqrt(2)*
b^3*d*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) +
a^2 + b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4)*cos(d*x + c) + (a^2*b^2 + b^4)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(
d*x + c) + a*b^2*cos(d*x + c) + b^3*sin(d*x + c))/cos(d*x + c)) - 15*sqrt(2)*(a*b^3*d^3*sqrt(1/((a^2 + b^2)*d^
4))*cos(d*x + c)^2 - b^3*d*cos(d*x + c)^2)*sqrt(((a^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)
*(1/((a^2 + b^2)*d^4))^(1/4)*log(-(sqrt(2)*b^3*d*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt(((a
^3 + a*b^2)*d^2*sqrt(1/((a^2 + b^2)*d^4)) + a^2 + b^2)/b^2)*(1/((a^2 + b^2)*d^4))^(1/4)*cos(d*x + c) - (a^2*b^
2 + b^4)*d^2*sqrt(1/((a^2 + b^2)*d^4))*cos(d*x + c) - a*b^2*cos(d*x + c) - b^3*sin(d*x + c))/cos(d*x + c)) + 8
*(4*a*b*cos(d*x + c)*sin(d*x + c) - 2*(4*a^2 - 9*b^2)*cos(d*x + c)^2 - 3*b^2)*sqrt((a*cos(d*x + c) + b*sin(d*x
+ c))/cos(d*x + c)))/(b^3*d*cos(d*x + c)^2)

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.29, size = 1641, normalized size = 3.28 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 7.94, size = 791, normalized size = 1.58 \[ \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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