\(\int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx\) [555]

Optimal result

Integrand size = 21, antiderivative size = 202 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\frac {(a-b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(a+b) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 b \sqrt {\tan (c+d x)}}{d}+\frac {2 a \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b \tan ^{\frac {5}{2}}(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3609, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\frac {(a-b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {(a+b) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a+b) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {2 a \tan ^{\frac {3}{2}}(c+d x)}{3 d}+\frac {2 b \tan ^{\frac {5}{2}}(c+d x)}{5 d}-\frac {2 b \sqrt {\tan (c+d x)}}{d} \]

[In]

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

[Out]

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

Rule 210

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

Rule 631

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 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 1176

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 1179

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 1182

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 3609

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 3615

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]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.56 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.52 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\frac {-15 \sqrt [4]{-1} (i a+b) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+15 (-1)^{3/4} (a+i b) \text {arctanh}\left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+2 \sqrt {\tan (c+d x)} \left (-15 b+5 a \tan (c+d x)+3 b \tan ^2(c+d x)\right )}{15 d} \]

[In]

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

[Out]

(-15*(-1)^(1/4)*(I*a + b)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + 15*(-1)^(3/4)*(a + I*b)*ArcTanh[(-1)^(3/4)*S
qrt[Tan[c + d*x]]] + 2*Sqrt[Tan[c + d*x]]*(-15*b + 5*a*Tan[c + d*x] + 3*b*Tan[c + d*x]^2))/(15*d)

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\frac {2 b \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 a \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 b \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(211\)
default	\(\frac {\frac {2 b \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}+\frac {2 a \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 b \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {a \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(211\)
parts	\(\frac {a \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {b \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(214\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 608 vs. \(2 (164) = 328\).

Time = 0.24 (sec) , antiderivative size = 608, normalized size of antiderivative = 3.01 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\frac {15 \, d \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} \log \left ({\left (a d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - {\left (a^{2} b - b^{3}\right )} d\right )} \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) - 15 \, d \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} \log \left (-{\left (a d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - {\left (a^{2} b - b^{3}\right )} d\right )} \sqrt {\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) - 15 \, d \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} \log \left ({\left (a d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + {\left (a^{2} b - b^{3}\right )} d\right )} \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) + 15 \, d \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} \log \left (-{\left (a d^{3} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} + {\left (a^{2} b - b^{3}\right )} d\right )} \sqrt {-\frac {d^{2} \sqrt {-\frac {a^{4} - 2 \, a^{2} b^{2} + b^{4}}{d^{4}}} - 2 \, a b}{d^{2}}} - {\left (a^{4} - b^{4}\right )} \sqrt {\tan \left (d x + c\right )}\right ) + 4 \, {\left (3 \, b \tan \left (d x + c\right )^{2} + 5 \, a \tan \left (d x + c\right ) - 15 \, b\right )} \sqrt {\tan \left (d x + c\right )}}{30 \, d} \]

[In]

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

[Out]

1/30*(15*d*sqrt((d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 2*a*b)/d^2)*log((a*d^3*sqrt(-(a^4 - 2*a^2*b^2 + b^4)
/d^4) - (a^2*b - b^3)*d)*sqrt((d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 2*a*b)/d^2) - (a^4 - b^4)*sqrt(tan(d*x
+ c))) - 15*d*sqrt((d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 2*a*b)/d^2)*log(-(a*d^3*sqrt(-(a^4 - 2*a^2*b^2 +
b^4)/d^4) - (a^2*b - b^3)*d)*sqrt((d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) + 2*a*b)/d^2) - (a^4 - b^4)*sqrt(ta
n(d*x + c))) - 15*d*sqrt(-(d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) - 2*a*b)/d^2)*log((a*d^3*sqrt(-(a^4 - 2*a^2*
b^2 + b^4)/d^4) + (a^2*b - b^3)*d)*sqrt(-(d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) - 2*a*b)/d^2) - (a^4 - b^4)*s
qrt(tan(d*x + c))) + 15*d*sqrt(-(d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) - 2*a*b)/d^2)*log(-(a*d^3*sqrt(-(a^4 -
2*a^2*b^2 + b^4)/d^4) + (a^2*b - b^3)*d)*sqrt(-(d^2*sqrt(-(a^4 - 2*a^2*b^2 + b^4)/d^4) - 2*a*b)/d^2) - (a^4 -
b^4)*sqrt(tan(d*x + c))) + 4*(3*b*tan(d*x + c)^2 + 5*a*tan(d*x + c) - 15*b)*sqrt(tan(d*x + c)))/d

Sympy [F]

\[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right ) \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.78 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\frac {24 \, b \tan \left (d x + c\right )^{\frac {5}{2}} - 30 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 30 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 15 \, \sqrt {2} {\left (a + b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 15 \, \sqrt {2} {\left (a + b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + 40 \, a \tan \left (d x + c\right )^{\frac {3}{2}} - 120 \, b \sqrt {\tan \left (d x + c\right )}}{60 \, d} \]

[In]

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

[Out]

1/60*(24*b*tan(d*x + c)^(5/2) - 30*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) - 30*s
qrt(2)*(a - b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 15*sqrt(2)*(a + b)*log(sqrt(2)*sqrt(tan
(d*x + c)) + tan(d*x + c) + 1) - 15*sqrt(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + 40*a
*tan(d*x + c)^(3/2) - 120*b*sqrt(tan(d*x + c)))/d

Giac [F(-1)]

Timed out. \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 6.95 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.64 \[ \int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x)) \, dx=\frac {2\,a\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{3\,d}-\frac {2\,b\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}+\frac {2\,b\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}}{5\,d}-\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}-\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,1{}\mathrm {i}\right )}{d} \]

[In]

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

[Out]

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