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

Optimal. Leaf size=456 \[ \frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac{b \log \left (-\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}-\frac{b \log \left (\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}-\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}+\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d} \]

[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 - 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 - 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 + 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 + Sqrt[a^2 + b^2]]*d) + (2*(8*a^2 - 35*b^2)*(a + b*Tan[c + d*x])^(3/2))/(105*b^3*d) - (8*a*Tan[c + d*x
]*(a + b*Tan[c + d*x])^(3/2))/(35*b^2*d) + (2*Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(3/2))/(7*b*d)

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Rubi [A]  time = 0.703612, antiderivative size = 456, normalized size of antiderivative = 1., 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, 700, 1129, 634, 618, 206, 628} \[ \frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}+\frac{b \log \left (-\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}-\frac{b \log \left (\sqrt{2} \sqrt{\sqrt{a^2+b^2}+a} \sqrt{a+b \tan (c+d x)}+\sqrt{a^2+b^2}+a+b \tan (c+d x)\right )}{2 \sqrt{2} d \sqrt{\sqrt{a^2+b^2}+a}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}-\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}}-\frac{b \tanh ^{-1}\left (\frac{\sqrt{\sqrt{a^2+b^2}+a}+\sqrt{2} \sqrt{a+b \tan (c+d x)}}{\sqrt{a-\sqrt{a^2+b^2}}}\right )}{\sqrt{2} d \sqrt{a-\sqrt{a^2+b^2}}}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d} \]

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 - 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 - 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 + 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 + Sqrt[a^2 + b^2]]*d) + (2*(8*a^2 - 35*b^2)*(a + b*Tan[c + d*x])^(3/2))/(105*b^3*d) - (8*a*Tan[c + d*x
]*(a + b*Tan[c + d*x])^(3/2))/(35*b^2*d) + (2*Tan[c + d*x]^2*(a + b*Tan[c + d*x])^(3/2))/(7*b*d)

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 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])))

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 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 700

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(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 1129

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/
c, 2]}, Dist[1/(2*c*r), Int[x^(m - 1)/(q - r*x + x^2), x], x] - Dist[1/(2*c*r), Int[x^(m - 1)/(q + r*x + x^2),
x], x]]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 1] && LtQ[m, 3] && NegQ[b^2 - 4*a*c]

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 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 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 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]

Rubi steps

\begin{align*} \int \tan ^4(c+d x) \sqrt{a+b \tan (c+d x)} \, dx &=\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac{2 \int \tan (c+d x) \sqrt{a+b \tan (c+d x)} \left (-2 a-\frac{7}{2} b \tan (c+d x)-2 a \tan ^2(c+d x)\right ) \, dx}{7 b}\\ &=-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac{4 \int \sqrt{a+b \tan (c+d x)} \left (2 a^2+\frac{1}{4} \left (8 a^2-35 b^2\right ) \tan ^2(c+d x)\right ) \, dx}{35 b^2}\\ &=\frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\int \sqrt{a+b \tan (c+d x)} \, dx\\ &=\frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac{b \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{b^2+x^2} \, dx,x,b \tan (c+d x)\right )}{d}\\ &=\frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{x^2}{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^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}+\frac{b \operatorname{Subst}\left (\int \frac{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+\sqrt{a^2+b^2}} d}-\frac{b \operatorname{Subst}\left (\int \frac{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+\sqrt{a^2+b^2}} d}\\ &=\frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 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 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 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+\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+\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+\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+\sqrt{a^2+b^2}} d}+\frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 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 )}{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 )}{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-\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-\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+\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+\sqrt{a^2+b^2}} d}+\frac{2 \left (8 a^2-35 b^2\right ) (a+b \tan (c+d x))^{3/2}}{105 b^3 d}-\frac{8 a \tan (c+d x) (a+b \tan (c+d x))^{3/2}}{35 b^2 d}+\frac{2 \tan ^2(c+d x) (a+b \tan (c+d x))^{3/2}}{7 b d}\\ \end{align*}

Mathematica [C]  time = 19.2857, size = 167, normalized size = 0.37 \[ \frac{\frac{2 \sqrt{a+b \tan (c+d x)} \left (-2 b \left (2 a^2+25 b^2\right ) \tan (c+d x)+8 a^3+3 b^2 \sec ^2(c+d x) (a+5 b \tan (c+d x))-38 a b^2\right )}{b^3}-105 i \sqrt{a-i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a-i b}}\right )+105 i \sqrt{a+i b} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan (c+d x)}}{\sqrt{a+i b}}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-105*I)*Sqrt[a - I*b]*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]] + (105*I)*Sqrt[a + I*b]*ArcTanh[Sqrt[a
+ b*Tan[c + d*x]]/Sqrt[a + I*b]] + (2*Sqrt[a + b*Tan[c + d*x]]*(8*a^3 - 38*a*b^2 - 2*b*(2*a^2 + 25*b^2)*Tan[c
+ d*x] + 3*b^2*Sec[c + d*x]^2*(a + 5*b*Tan[c + d*x])))/b^3)/(105*d)

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Maple [A]  time = 0.129, size = 556, normalized size = 1.2 \begin{align*}{\frac{2}{7\,d{b}^{3}} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}}-{\frac{4\,a}{5\,d{b}^{3}} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{a}^{2}}{3\,d{b}^{3}} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{2}{3\,bd} \left ( a+b\tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{a}{4\,bd}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\ln \left ( b\tan \left ( dx+c \right ) +a+\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ) }-{\frac{1}{4\,bd}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\sqrt{{a}^{2}+{b}^{2}}\ln \left ( b\tan \left ( dx+c \right ) +a+\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ) }+{\frac{b}{d}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }+\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}}-{\frac{a}{4\,bd}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\ln \left ( b\tan \left ( dx+c \right ) +a-\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ) }+{\frac{1}{4\,bd}\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}\sqrt{{a}^{2}+{b}^{2}}\ln \left ( b\tan \left ( dx+c \right ) +a-\sqrt{a+b\tan \left ( dx+c \right ) }\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a}+\sqrt{{a}^{2}+{b}^{2}} \right ) }+{\frac{b}{d}\arctan \left ({ \left ( 2\,\sqrt{a+b\tan \left ( dx+c \right ) }-\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}+2\,a} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \right ){\frac{1}{\sqrt{2\,\sqrt{{a}^{2}+{b}^{2}}-2\,a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7/d/b^3*(a+b*tan(d*x+c))^(7/2)-4/5/d/b^3*a*(a+b*tan(d*x+c))^(5/2)+2/3/d/b^3*(a+b*tan(d*x+c))^(3/2)*a^2-2/3*(
a+b*tan(d*x+c))^(3/2)/b/d+1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a*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))-1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*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))+1/d*b/(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*(2*(a
^2+b^2)^(1/2)+2*a)^(1/2)*a*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))+1/4/d/b*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*(a^2+b^2)^(1/2)*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))+1/d*b/(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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.633, size = 3893, normalized size = 8.54 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*(420*sqrt(2)*b^3*d^5*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) + a^2 + b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4
)^(3/4)*arctan(-(sqrt(2)*b*d^5*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt((a*d^2*sqrt((a^2 + b^
2)/d^4) + a^2 + b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)^(3/4) - sqrt(2)*d^5*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4
) + a^2 + b^2)/b^2)*sqrt((sqrt(2)*b^3*d^3*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt((a*d^2*sqr
t((a^2 + b^2)/d^4) + a^2 + b^2)/b^2)*((a^2 + b^2)/d^4)^(3/4)*cos(d*x + c) + (a^2*b^2 + b^4)*d^2*sqrt((a^2 + b^
2)/d^4)*cos(d*x + c) + (a^3*b^2 + a*b^4)*cos(d*x + c) + (a^2*b^3 + b^5)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c
)))*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)^(3/4) + (a^2 + b^2)*d^4*sqrt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) + (a^3 + a*b^2
)*d^2*sqrt(b^2/d^4))/(a^2*b^2 + b^4))*cos(d*x + c)^3 + 420*sqrt(2)*b^3*d^5*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) +
a^2 + b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)^(3/4)*arctan(-(sqrt(2)*b*d^5*sqrt((a*cos(d*x + c) + b*sin(d*x
+ c))/cos(d*x + c))*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) + a^2 + b^2)/b^2)*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)^(3/4)
- sqrt(2)*d^5*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) + a^2 + b^2)/b^2)*sqrt(-(sqrt(2)*b^3*d^3*sqrt((a*cos(d*x + c)
+ b*sin(d*x + c))/cos(d*x + c))*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) + a^2 + b^2)/b^2)*((a^2 + b^2)/d^4)^(3/4)*c
os(d*x + c) - (a^2*b^2 + b^4)*d^2*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) - (a^3*b^2 + a*b^4)*cos(d*x + c) - (a^2*b
^3 + b^5)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c)))*sqrt(b^2/d^4)*((a^2 + b^2)/d^4)^(3/4) - (a^2 + b^2)*d^4*sq
rt(b^2/d^4)*sqrt((a^2 + b^2)/d^4) - (a^3 + a*b^2)*d^2*sqrt(b^2/d^4))/(a^2*b^2 + b^4))*cos(d*x + c)^3 - 105*sqr
t(2)*(a*b^3*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c)^3 - (a^2*b^3 + b^5)*d*cos(d*x + c)^3)*sqrt((a*d^2*sqrt((a^2
+ b^2)/d^4) + a^2 + b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4)*log((sqrt(2)*b^3*d^3*sqrt((a*cos(d*x + c) + b*sin(d*x +
c))/cos(d*x + c))*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) + a^2 + b^2)/b^2)*((a^2 + b^2)/d^4)^(3/4)*cos(d*x + c) +
(a^2*b^2 + b^4)*d^2*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) + (a^3*b^2 + a*b^4)*cos(d*x + c) + (a^2*b^3 + b^5)*sin(
d*x + c))/((a^2 + b^2)*cos(d*x + c))) + 105*sqrt(2)*(a*b^3*d^3*sqrt((a^2 + b^2)/d^4)*cos(d*x + c)^3 - (a^2*b^3
+ b^5)*d*cos(d*x + c)^3)*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) + a^2 + b^2)/b^2)*((a^2 + b^2)/d^4)^(1/4)*log(-(sq
rt(2)*b^3*d^3*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c))*sqrt((a*d^2*sqrt((a^2 + b^2)/d^4) + a^2 + b
^2)/b^2)*((a^2 + b^2)/d^4)^(3/4)*cos(d*x + c) - (a^2*b^2 + b^4)*d^2*sqrt((a^2 + b^2)/d^4)*cos(d*x + c) - (a^3*
b^2 + a*b^4)*cos(d*x + c) - (a^2*b^3 + b^5)*sin(d*x + c))/((a^2 + b^2)*cos(d*x + c))) - 8*(2*(4*a^5 - 15*a^3*b
^2 - 19*a*b^4)*cos(d*x + c)^3 + 3*(a^3*b^2 + a*b^4)*cos(d*x + c) + (15*a^2*b^3 + 15*b^5 - 2*(2*a^4*b + 27*a^2*
b^3 + 25*b^5)*cos(d*x + c)^2)*sin(d*x + c))*sqrt((a*cos(d*x + c) + b*sin(d*x + c))/cos(d*x + c)))/((a^2*b^3 +
b^5)*d*cos(d*x + c)^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \tan{\left (c + d x \right )}} \tan ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out