3.562 \(\int \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx\)

Optimal. Leaf size=268 \[ \frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d} \]

[Out]

((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a^2 - 2*a*b - b^2)*ArcTan[1 + Sqr
t[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]
])/(2*Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (4
*a*b*Sqrt[Tan[c + d*x]])/d + (2*(a^2 - b^2)*Tan[c + d*x]^(3/2))/(3*d) + (4*a*b*Tan[c + d*x]^(5/2))/(5*d) + (2*
b^2*Tan[c + d*x]^(7/2))/(7*d)

________________________________________________________________________________________

Rubi [A]  time = 0.25578, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {3543, 3528, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{\sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2,x]

[Out]

((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a^2 - 2*a*b - b^2)*ArcTan[1 + Sqr
t[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a^2 + 2*a*b - b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]
])/(2*Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) - (4
*a*b*Sqrt[Tan[c + d*x]])/d + (2*(a^2 - b^2)*Tan[c + d*x]^(3/2))/(3*d) + (4*a*b*Tan[c + d*x]^(5/2))/(5*d) + (2*
b^2*Tan[c + d*x]^(7/2))/(7*d)

Rule 3543

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 1168

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

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

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 ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^2 \, dx &=\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac{5}{2}}(c+d x) \left (a^2-b^2+2 a b \tan (c+d x)\right ) \, dx\\ &=\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \tan ^{\frac{3}{2}}(c+d x) \left (-2 a b+\left (a^2-b^2\right ) \tan (c+d x)\right ) \, dx\\ &=\frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \sqrt{\tan (c+d x)} \left (-a^2+b^2-2 a b \tan (c+d x)\right ) \, dx\\ &=-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\int \frac{2 a b-\left (a^2-b^2\right ) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}+\frac{2 \operatorname{Subst}\left (\int \frac{2 a b+\left (-a^2+b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}-\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left (a^2+2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}-\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}-\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}-\frac{\left (a^2+2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}\\ &=-\frac{\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}-\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}+\frac{\left (a^2-2 a b-b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}\\ &=\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}+\frac{\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt{2} d}-\frac{4 a b \sqrt{\tan (c+d x)}}{d}+\frac{2 \left (a^2-b^2\right ) \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a b \tan ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{2 b^2 \tan ^{\frac{7}{2}}(c+d x)}{7 d}\\ \end{align*}

Mathematica [C]  time = 1.10473, size = 133, normalized size = 0.5 \[ \frac{2 \sqrt{\tan (c+d x)} \left (35 \left (a^2-b^2\right ) \tan (c+d x)+42 a b \tan ^2(c+d x)-210 a b+15 b^2 \tan ^3(c+d x)\right )-105 (-1)^{3/4} (a-i b)^2 \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )+105 (-1)^{3/4} (a+i b)^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{105 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^(5/2)*(a + b*Tan[c + d*x])^2,x]

[Out]

(-105*(-1)^(3/4)*(a - I*b)^2*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 105*(-1)^(3/4)*(a + I*b)^2*ArcTanh[(-1)^(
3/4)*Sqrt[Tan[c + d*x]]] + 2*Sqrt[Tan[c + d*x]]*(-210*a*b + 35*(a^2 - b^2)*Tan[c + d*x] + 42*a*b*Tan[c + d*x]^
2 + 15*b^2*Tan[c + d*x]^3))/(105*d)

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 399, normalized size = 1.5 \begin{align*}{\frac{2\,{b}^{2}}{7\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{7}{2}}}}+{\frac{4\,ab}{5\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}}}+{\frac{2\,{a}^{2}}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-{\frac{2\,{b}^{2}}{3\,d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}-4\,{\frac{ab\sqrt{\tan \left ( dx+c \right ) }}{d}}+{\frac{ab\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{ab\sqrt{2}}{2\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{ab\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{a}^{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{\sqrt{2}{b}^{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{\sqrt{2}{a}^{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{\sqrt{2}{b}^{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{\sqrt{2}{a}^{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{\sqrt{2}{b}^{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.7469, size = 293, normalized size = 1.09 \begin{align*} \frac{120 \, b^{2} \tan \left (d x + c\right )^{\frac{7}{2}} + 336 \, a b \tan \left (d x + c\right )^{\frac{5}{2}} - 210 \, \sqrt{2}{\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) - 210 \, \sqrt{2}{\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - 2 \, \sqrt{\tan \left (d x + c\right )}\right )}\right ) + 105 \, \sqrt{2}{\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 105 \, \sqrt{2}{\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\sqrt{2} \sqrt{\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 1680 \, a b \sqrt{\tan \left (d x + c\right )} + 280 \,{\left (a^{2} - b^{2}\right )} \tan \left (d x + c\right )^{\frac{3}{2}}}{420 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

1/420*(120*b^2*tan(d*x + c)^(7/2) + 336*a*b*tan(d*x + c)^(5/2) - 210*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sq
rt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) - 210*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqr
t(tan(d*x + c)))) + 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 105*s
qrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 1680*a*b*sqrt(tan(d*x + c)) +
280*(a^2 - b^2)*tan(d*x + c)^(3/2))/d

________________________________________________________________________________________

Fricas [B]  time = 6.41934, size = 11221, normalized size = 41.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/420*(420*sqrt(2)*d^5*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 +
4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*
a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4)
*arctan(-((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*a^6*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 + 4
*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) - sq
rt(2)*(2*a*b*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4
- 12*a^2*b^6 + b^8)/d^4) + (a^6 + a^4*b^2 - a^2*b^4 - b^6)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b
^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*
b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(((a^12 - 10*
a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 +
4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + sqrt(2)*((a^10 - 13*a^8*b^2 + 50*a^6*b^4 - 50*a^4*b^6 + 13*a^2*b^8 - b^10
)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + 2*(a^13*b - 10*a^11*b^3 + 15*a^
9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*
a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6
*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b
^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*b^6 + 134*a^8*b^8 + 72*a^6*b^10 - 4*a^4*b^12
- 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4) +
sqrt(2)*(2*(a^9*b - 4*a^7*b^3 - 10*a^5*b^5 - 4*a^3*b^7 + a*b^9)*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2
*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^14 - 3*a^12*b^2 - 15*a^10*b
^4 - 11*a^8*b^6 + 11*a^6*b^8 + 15*a^4*b^10 + 3*a^2*b^12 - b^14)*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a
^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*
a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x
+ c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4))/(a^24 - 4*a^22*b^2 - 30*a^20*b
^4 + 12*a^18*b^6 + 367*a^16*b^8 + 1016*a^14*b^10 + 1372*a^12*b^12 + 1016*a^10*b^14 + 367*a^8*b^16 + 12*a^6*b^1
8 - 30*a^4*b^20 - 4*a^2*b^22 + b^24))*cos(d*x + c)^3 + 420*sqrt(2)*d^5*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a
^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*
b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4)*sqrt((a^8 -
12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4)*arctan(((a^16 - 20*a^12*b^4 - 64*a^10*b^6 - 90*a^8*b^8 - 64*a
^6*b^10 - 20*a^4*b^12 + b^16)*d^4*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6
*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + sqrt(2)*(2*a*b*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6
+ b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4) + (a^6 + a^4*b^2 - a^2*b^4 - b^6)*d^5
*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 +
b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38
*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 +
b^12)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) - sqrt(2)*((a^10 - 13*a^8*b^
2 + 50*a^6*b^4 - 50*a^4*b^6 + 13*a^2*b^8 - b^10)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)
*cos(d*x + c) + 2*(a^13*b - 10*a^11*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d
*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6
*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*
x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^1
0*b^6 + 134*a^8*b^8 + 72*a^6*b^10 - 4*a^4*b^12 - 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c))*((a^8 + 4*a^6*
b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(3/4) - sqrt(2)*(2*(a^9*b - 4*a^7*b^3 - 10*a^5*b^5 - 4*a^3*b^7 + a*b^9
)*d^7*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b
^6 + b^8)/d^4) + (a^14 - 3*a^12*b^2 - 15*a^10*b^4 - 11*a^8*b^6 + 11*a^6*b^8 + 15*a^4*b^10 + 3*a^2*b^12 - b^14)
*d^5*sqrt((a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8)/d^4))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b
^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2
+ 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 +
b^8)/d^4)^(3/4))/(a^24 - 4*a^22*b^2 - 30*a^20*b^4 + 12*a^18*b^6 + 367*a^16*b^8 + 1016*a^14*b^10 + 1372*a^12*b^
12 + 1016*a^10*b^14 + 367*a^8*b^16 + 12*a^6*b^18 - 30*a^4*b^20 - 4*a^2*b^22 + b^24))*cos(d*x + c)^3 + 105*sqrt
(2)*(4*(a^3*b - a*b^3)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c)^3 + (a^8 + 4
*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d*cos(d*x + c)^3)*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8
- 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*
b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4)*log(((a^12 - 10*a^10*b^2
+ 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6
+ b^8)/d^4)*cos(d*x + c) + sqrt(2)*((a^10 - 13*a^8*b^2 + 50*a^6*b^4 - 50*a^4*b^6 + 13*a^2*b^8 - b^10)*d^3*sqr
t((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) + 2*(a^13*b - 10*a^11*b^3 + 15*a^9*b^5 + 5
2*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x + c))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 +
b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38
*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)
/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*b^6 + 134*a^8*b^8 + 72*a^6*b^10 - 4*a^4*b^12 - 8*a^2*b
^14 + b^16)*sin(d*x + c))/cos(d*x + c)) - 105*sqrt(2)*(4*(a^3*b - a*b^3)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4
+ 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c)^3 + (a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d*cos(d*x + c)^3)*sq
rt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 +
4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a
^2*b^6 + b^8)/d^4)^(1/4)*log(((a^12 - 10*a^10*b^2 + 15*a^8*b^4 + 52*a^6*b^6 + 15*a^4*b^8 - 10*a^2*b^10 + b^12)
*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d*x + c) - sqrt(2)*((a^10 - 13*a^8*b^2 + 50
*a^6*b^4 - 50*a^4*b^6 + 13*a^2*b^8 - b^10)*d^3*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)*cos(d
*x + c) + 2*(a^13*b - 10*a^11*b^3 + 15*a^9*b^5 + 52*a^7*b^7 + 15*a^5*b^9 - 10*a^3*b^11 + a*b^13)*d*cos(d*x + c
))*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8 - 4*(a^3*b - a*b^3)*d^2*sqrt((a^8 + 4*a^6*b^2 + 6*a^4*b
^4 + 4*a^2*b^6 + b^8)/d^4))/(a^8 - 12*a^6*b^2 + 38*a^4*b^4 - 12*a^2*b^6 + b^8))*sqrt(sin(d*x + c)/cos(d*x + c)
)*((a^8 + 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)/d^4)^(1/4) + (a^16 - 8*a^14*b^2 - 4*a^12*b^4 + 72*a^10*b^6
+ 134*a^8*b^8 + 72*a^6*b^10 - 4*a^4*b^12 - 8*a^2*b^14 + b^16)*sin(d*x + c))/cos(d*x + c)) - 8*(252*(a^9*b + 4*
a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^9)*cos(d*x + c)^3 - 42*(a^9*b + 4*a^7*b^3 + 6*a^5*b^5 + 4*a^3*b^7 + a*b^
9)*cos(d*x + c) - 5*(3*a^8*b^2 + 12*a^6*b^4 + 18*a^4*b^6 + 12*a^2*b^8 + 3*b^10 + (7*a^10 + 18*a^8*b^2 + 2*a^6*
b^4 - 32*a^4*b^6 - 33*a^2*b^8 - 10*b^10)*cos(d*x + c)^2)*sin(d*x + c))*sqrt(sin(d*x + c)/cos(d*x + c)))/((a^8
+ 4*a^6*b^2 + 6*a^4*b^4 + 4*a^2*b^6 + b^8)*d*cos(d*x + c)^3)

________________________________________________________________________________________

Sympy [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(tan(d*x+c)**(5/2)*(a+b*tan(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

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(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

Timed out