\(\int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [573]

Optimal result

Integrand size = 23, antiderivative size = 245 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}+\frac {2 b \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 (a+b \tan (c+d x))}{d \sqrt {\tan (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3646, 3711, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b \left (a^2+b^2\right ) \sqrt {\tan (c+d x)}}{d}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a^2 (a+b \tan (c+d x))}{d \sqrt {\tan (c+d x)}} \]

[In]

Int[(a + b*Tan[c + d*x])^3/Tan[c + d*x]^(3/2),x]

[Out]

((a + b)*(a^2 - 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a + b)*(a^2 - 4*a*b + b^2
)*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Tan[
c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d) + ((a - b)*(a^2 + 4*a*b + b^2)*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + T
an[c + d*x]])/(2*Sqrt[2]*d) + (2*b*(a^2 + b^2)*Sqrt[Tan[c + d*x]])/d - (2*a^2*(a + b*Tan[c + d*x]))/(d*Sqrt[Ta
n[c + d*x]])

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 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]

Rule 3646

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

Rule 3711

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

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.03 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {-32 a b^2-8 a \left (a^2-3 b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\tan ^2(c+d x)\right )+\sqrt {2} b \left (-3 a^2+b^2\right ) \left (2 \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-2 \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right ) \sqrt {\tan (c+d x)}+8 b^2 (a+b \tan (c+d x))}{4 d \sqrt {\tan (c+d x)}} \]

[In]

Integrate[(a + b*Tan[c + d*x])^3/Tan[c + d*x]^(3/2),x]

[Out]

(-32*a*b^2 - 8*a*(a^2 - 3*b^2)*Hypergeometric2F1[-1/4, 1, 3/4, -Tan[c + d*x]^2] + Sqrt[2]*b*(-3*a^2 + b^2)*(2*
ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - 2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Tan[c
+ d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])*Sqrt[Tan[c + d*x]] + 8*b^2*(a +
b*Tan[c + d*x]))/(4*d*Sqrt[Tan[c + d*x]])

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 226, normalized size of antiderivative = 0.92

method	result	size
derivativedivides	\(\frac {2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2 a^{3}}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(226\)
default	\(\frac {2 b^{3} \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (3 a^{2} b -b^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-a^{3}+3 a \,b^{2}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2 a^{3}}{\sqrt {\tan \left (d x +c \right )}}}{d}\)	\(226\)
parts	\(\frac {a^{3} \left (-\frac {2}{\sqrt {\tan \left (d x +c \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {b^{3} \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {3 a \,b^{2} \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {3 a^{2} b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}\)	\(392\)

[In]

int((a+b*tan(d*x+c))^3/tan(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1398 vs. \(2 (213) = 426\).

Time = 0.29 (sec) , antiderivative size = 1398, normalized size of antiderivative = 5.71 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

1/2*(d*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*
a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2)*log(((a^3 - 3*a*b^2)*d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 45
2*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) - (3*a^8*b - 46*a^6*b^3 + 60*a^4*b^5 - 18*a^2*b^7 + b^9)*d)
*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^
8 - 30*a^2*b^10 + b^12)/d^4))/d^2) - (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27*a^4*b^8 + 12*a^2*b^10 - b^12)*sqrt(
tan(d*x + c)))*tan(d*x + c) - d*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8
*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2)*log(-((a^3 - 3*a*b^2)*d^3*sqrt(-(a^12 - 30*a
^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) - (3*a^8*b - 46*a^6*b^3 + 60*a^4*
b^5 - 18*a^2*b^7 + b^9)*d)*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 + d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4
- 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2) - (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27*a^4*b^8 +
12*a^2*b^10 - b^12)*sqrt(tan(d*x + c)))*tan(d*x + c) - d*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 - d^2*sqrt(-(a^
12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2)*log(((a^3 - 3*a*b^
2)*d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4) + (3*a^8
*b - 46*a^6*b^3 + 60*a^4*b^5 - 18*a^2*b^7 + b^9)*d)*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a*b^5 - d^2*sqrt(-(a^12 - 3
0*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2) - (a^12 - 12*a^10*b^2 -
27*a^8*b^4 + 27*a^4*b^8 + 12*a^2*b^10 - b^12)*sqrt(tan(d*x + c)))*tan(d*x + c) + d*sqrt((6*a^5*b - 20*a^3*b^3
+ 6*a*b^5 - d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4)
)/d^2)*log(-((a^3 - 3*a*b^2)*d^3*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*
b^10 + b^12)/d^4) + (3*a^8*b - 46*a^6*b^3 + 60*a^4*b^5 - 18*a^2*b^7 + b^9)*d)*sqrt((6*a^5*b - 20*a^3*b^3 + 6*a
*b^5 - d^2*sqrt(-(a^12 - 30*a^10*b^2 + 255*a^8*b^4 - 452*a^6*b^6 + 255*a^4*b^8 - 30*a^2*b^10 + b^12)/d^4))/d^2
) - (a^12 - 12*a^10*b^2 - 27*a^8*b^4 + 27*a^4*b^8 + 12*a^2*b^10 - b^12)*sqrt(tan(d*x + c)))*tan(d*x + c) + 4*(
b^3*tan(d*x + c) - a^3)*sqrt(tan(d*x + c)))/(d*tan(d*x + c))

Sympy [F]

\[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{3}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+b*tan(d*x+c))**3/tan(d*x+c)**(3/2),x)

[Out]

Integral((a + b*tan(c + d*x))**3/tan(c + d*x)**(3/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.87 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {8 \, b^{3} \sqrt {\tan \left (d x + c\right )} - 2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - 2 \, \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \frac {8 \, a^{3}}{\sqrt {\tan \left (d x + c\right )}}}{4 \, d} \]

[In]

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

1/4*(8*b^3*sqrt(tan(d*x + c)) - 2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt
(tan(d*x + c)))) - 2*sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c
)))) + sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - sqrt(2)*(a
^3 + 3*a^2*b - 3*a*b^2 - b^3)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 8*a^3/sqrt(tan(d*x + c)))/
d

Giac [F(-1)]

Timed out. \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+b*tan(d*x+c))^3/tan(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 5.83 (sec) , antiderivative size = 1767, normalized size of antiderivative = 7.21 \[ \int \frac {(a+b \tan (c+d x))^3}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

int((a + b*tan(c + d*x))^3/tan(c + d*x)^(3/2),x)

[Out]

(2*b^3*tan(c + d*x)^(1/2))/d - atan((a^6*d^3*tan(c + d*x)^(1/2)*((a^6*1i)/(4*d^2) - (b^6*1i)/(4*d^2) + (3*a*b^
5)/(2*d^2) + (3*a^5*b)/(2*d^2) + (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 - (a^4*b^2*15i)/(4*d^2))^(1/2)*32i)/(
16*a^9*d^2 - b^9*d^2*16i + 48*a*b^8*d^2 - a^8*b*d^2*48i + a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 - a^4*b^5*d^2*960
i + 960*a^5*b^4*d^2 + a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) - (b^6*d^3*tan(c + d*x)^(1/2)*((a^6*1i)/(4*d^2) - (b
^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) + (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 - (a^4*b^2*15
i)/(4*d^2))^(1/2)*32i)/(16*a^9*d^2 - b^9*d^2*16i + 48*a*b^8*d^2 - a^8*b*d^2*48i + a^2*b^7*d^2*288i - 736*a^3*b
^6*d^2 - a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 + a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) + (a^2*b^4*d^3*tan(c + d*x)^
(1/2)*((a^6*1i)/(4*d^2) - (b^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) + (a^2*b^4*15i)/(4*d^2) - (
5*a^3*b^3)/d^2 - (a^4*b^2*15i)/(4*d^2))^(1/2)*480i)/(16*a^9*d^2 - b^9*d^2*16i + 48*a*b^8*d^2 - a^8*b*d^2*48i +
a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 - a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 + a^6*b^3*d^2*736i - 288*a^7*b^2*d^2)
- (a^4*b^2*d^3*tan(c + d*x)^(1/2)*((a^6*1i)/(4*d^2) - (b^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2
) + (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 - (a^4*b^2*15i)/(4*d^2))^(1/2)*480i)/(16*a^9*d^2 - b^9*d^2*16i + 4
8*a*b^8*d^2 - a^8*b*d^2*48i + a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 - a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 + a^6*b^
3*d^2*736i - 288*a^7*b^2*d^2))*((6*a*b^5 + 6*a^5*b + a^6*1i - b^6*1i + a^2*b^4*15i - 20*a^3*b^3 - a^4*b^2*15i)
/(4*d^2))^(1/2)*2i - (2*a^3)/(d*tan(c + d*x)^(1/2)) - atan((a^6*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^
6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^2*15i
)/(4*d^2))^(1/2)*32i)/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^
6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) - (b^6*d^3*tan(c + d*x)^(1/2)
*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3
*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2)*32i)/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b
^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2) + (a^
2*b^4*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*d^2) + (3*a^5*b)/(2*d^2) - (a
^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2)*480i)/(16*a^9*d^2 + b^9*d^2*16i + 48*a*b^
8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 960*a^5*b^4*d^2 - a^6*b^3*d^2*
736i - 288*a^7*b^2*d^2) - (a^4*b^2*d^3*tan(c + d*x)^(1/2)*((b^6*1i)/(4*d^2) - (a^6*1i)/(4*d^2) + (3*a*b^5)/(2*
d^2) + (3*a^5*b)/(2*d^2) - (a^2*b^4*15i)/(4*d^2) - (5*a^3*b^3)/d^2 + (a^4*b^2*15i)/(4*d^2))^(1/2)*480i)/(16*a^
9*d^2 + b^9*d^2*16i + 48*a*b^8*d^2 + a^8*b*d^2*48i - a^2*b^7*d^2*288i - 736*a^3*b^6*d^2 + a^4*b^5*d^2*960i + 9
60*a^5*b^4*d^2 - a^6*b^3*d^2*736i - 288*a^7*b^2*d^2))*((6*a*b^5 + 6*a^5*b - a^6*1i + b^6*1i - a^2*b^4*15i - 20
*a^3*b^3 + a^4*b^2*15i)/(4*d^2))^(1/2)*2i