\(\int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [383]

Optimal result

Integrand size = 31, antiderivative size = 229 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(b (A-B)+a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(b (A-B)+a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}+\frac {(a (A-B)-b (A+B)) \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d}-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (A b+a B)}{d \sqrt {\tan (c+d x)}} \]

[Out]

-1/2*(b*(A-B)+a*(A+B))*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)-1/2*(b*(A-B)+a*(A+B))*arctan(1+2^(1/2)*ta
n(d*x+c)^(1/2))/d*2^(1/2)+1/4*(a*(A-B)-b*(A+B))*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-1/4*(a*(A-
B)-b*(A+B))*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)-2*(A*b+B*a)/d/tan(d*x+c)^(1/2)-2/3*a*A/d/tan(d
*x+c)^(3/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3672, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {(a (A+B)+b (A-B)) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d}-\frac {(a (A+B)+b (A-B)) \arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d}-\frac {2 (a B+A b)}{d \sqrt {\tan (c+d x)}}+\frac {(a (A-B)-b (A+B)) \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a (A-B)-b (A+B)) \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {2 a A}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

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

[Out]

((b*(A - B) + a*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) - ((b*(A - B) + a*(A + B))*ArcTan
[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d) + ((a*(A - B) - b*(A + B))*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] +
Tan[c + d*x]])/(2*Sqrt[2]*d) - ((a*(A - B) - b*(A + B))*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2
*Sqrt[2]*d) - (2*a*A)/(3*d*Tan[c + d*x]^(3/2)) - (2*(A*b + a*B))/(d*Sqrt[Tan[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 3610

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

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 3672

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

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.78 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {6 \sqrt {2} (b (A-B)+a (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+3 \sqrt {2} (a (A-B)-b (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )-\frac {8 a A}{\tan ^{\frac {3}{2}}(c+d x)}-\frac {24 (A b+a B)}{\sqrt {\tan (c+d x)}}}{12 d} \]

[In]

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

[Out]

(6*Sqrt[2]*(b*(A - B) + a*(A + B))*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d
*x]]]) + 3*Sqrt[2]*(a*(A - B) - b*(A + B))*(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]]) - (8*a*A)/Tan[c + d*x]^(3/2) - (24*(A*b + a*B))/Sqrt[Tan[c + d*x]])/(12
*d)

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 222, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (A b +B a \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(222\)
default	\(\frac {-\frac {2 \left (A b +B a \right )}{\sqrt {\tan \left (d x +c \right )}}-\frac {2 a A}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {\left (-a A +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (-A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}}{d}\)	\(222\)
parts	\(\frac {\left (A b +B a \right ) \left (-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}-\frac {2}{\sqrt {\tan \left (d x +c \right )}}\right )}{d}+\frac {B b \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {a A \left (-\frac {2}{3 \tan \left (d x +c \right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(301\)

[In]

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

[Out]

1/d*(-2*(A*b+B*a)/tan(d*x+c)^(1/2)-2/3*a*A/tan(d*x+c)^(3/2)+1/4*(-A*a+B*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*b-B*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. 2275 vs. \(2 (195) = 390\).

Time = 0.41 (sec) , antiderivative size = 2275, normalized size of antiderivative = 9.93 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/6*(3*d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B
- A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/
d^4))/d^2)*log(((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2
*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) + ((A^3 - A*B^2)*a^3 - (5*A^
2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*
a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 +
8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b
- 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 - 3*d*sqrt(-(2*A*B*a^2 - 2*A*
B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2
*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(-((B*a + A*b)*d^3*
sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B -
A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) + ((A^3 - A*B^2)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^
2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b + d^2*sqrt(-((A^4 - 2*A^2*B^
2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 -
2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^
4 - B^4)*b^4)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 - 3*d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*
sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B -
A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)*log(((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^
4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2
+ B^4)*b^4)/d^4) - ((A^3 - A*B^2)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*
sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*
a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2)
+ ((A^4 - B^4)*a^4 - 4*(A^3*B + A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(tan(d*x + c)))
*tan(d*x + c)^2 + 3*d*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^
4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2
+ B^4)*b^4)/d^4))/d^2)*log(-((B*a + A*b)*d^3*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*
(A^4 - 10*A^2*B^2 + B^4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4) - ((A^3 - A*B^2
)*a^3 - (5*A^2*B - B^3)*a^2*b - (A^3 - 5*A*B^2)*a*b^2 + (A^2*B - B^3)*b^3)*d)*sqrt(-(2*A*B*a^2 - 2*A*B*b^2 + 2
*(A^2 - B^2)*a*b - d^2*sqrt(-((A^4 - 2*A^2*B^2 + B^4)*a^4 - 8*(A^3*B - A*B^3)*a^3*b - 2*(A^4 - 10*A^2*B^2 + B^
4)*a^2*b^2 + 8*(A^3*B - A*B^3)*a*b^3 + (A^4 - 2*A^2*B^2 + B^4)*b^4)/d^4))/d^2) + ((A^4 - B^4)*a^4 - 4*(A^3*B +
A*B^3)*a^3*b - 4*(A^3*B + A*B^3)*a*b^3 - (A^4 - B^4)*b^4)*sqrt(tan(d*x + c)))*tan(d*x + c)^2 + 4*(A*a + 3*(B*
a + A*b)*tan(d*x + c))*sqrt(tan(d*x + c)))/(d*tan(d*x + c)^2)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.84 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {6 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 6 \, \sqrt {2} {\left ({\left (A + B\right )} a + {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 3 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - 3 \, \sqrt {2} {\left ({\left (A - B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \frac {8 \, {\left (A a + 3 \, {\left (B a + A b\right )} \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \]

[In]

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

[Out]

-1/12*(6*sqrt(2)*((A + B)*a + (A - B)*b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 6*sqrt(2)*((A
+ B)*a + (A - B)*b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) + 3*sqrt(2)*((A - B)*a - (A + B)*b)*
log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) - 3*sqrt(2)*((A - B)*a - (A + B)*b)*log(-sqrt(2)*sqrt(tan(d
*x + c)) + tan(d*x + c) + 1) + 8*(A*a + 3*(B*a + A*b)*tan(d*x + c))/tan(d*x + c)^(3/2))/d

Giac [F(-1)]

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

[In]

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

[Out]

Timed out

Mupad [B] (verification not implemented)

Time = 11.75 (sec) , antiderivative size = 1448, normalized size of antiderivative = 6.32 \[ \int \frac {(a+b \tan (c+d x)) (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}-\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}-16\,A^3\,b^3\,d^2+16\,A^3\,a^2\,b\,d^2}-\frac {32\,A^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}-\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}-16\,A^3\,b^3\,d^2+16\,A^3\,a^2\,b\,d^2}\right )\,\sqrt {-\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}-\frac {A^2\,a\,b}{2\,d^2}}-2\,\mathrm {atanh}\left (\frac {32\,A^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}-\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}+16\,A^3\,b^3\,d^2-16\,A^3\,a^2\,b\,d^2}-\frac {32\,A^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}-\frac {A^2\,a\,b}{2\,d^2}}}{16\,A\,a\,\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}+16\,A^3\,b^3\,d^2-16\,A^3\,a^2\,b\,d^2}\right )\,\sqrt {\frac {\sqrt {-A^4\,a^4\,d^4+2\,A^4\,a^2\,b^2\,d^4-A^4\,b^4\,d^4}}{4\,d^4}-\frac {A^2\,a\,b}{2\,d^2}}+2\,\mathrm {atanh}\left (\frac {32\,B^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,a\,b}{2\,d^2}-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}}}{16\,B\,b\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}+16\,B^3\,a^3\,d^2-16\,B^3\,a\,b^2\,d^2}-\frac {32\,B^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {B^2\,a\,b}{2\,d^2}-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}}}{16\,B\,b\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}+16\,B^3\,a^3\,d^2-16\,B^3\,a\,b^2\,d^2}\right )\,\sqrt {\frac {B^2\,a\,b}{2\,d^2}-\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}}-2\,\mathrm {atanh}\left (\frac {32\,B^2\,a^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}+\frac {B^2\,a\,b}{2\,d^2}}}{16\,B\,b\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}-16\,B^3\,a^3\,d^2+16\,B^3\,a\,b^2\,d^2}-\frac {32\,B^2\,b^2\,d^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}+\frac {B^2\,a\,b}{2\,d^2}}}{16\,B\,b\,\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}-16\,B^3\,a^3\,d^2+16\,B^3\,a\,b^2\,d^2}\right )\,\sqrt {\frac {\sqrt {-B^4\,a^4\,d^4+2\,B^4\,a^2\,b^2\,d^4-B^4\,b^4\,d^4}}{4\,d^4}+\frac {B^2\,a\,b}{2\,d^2}}-\frac {\frac {2\,A\,a}{3}+2\,A\,b\,\mathrm {tan}\left (c+d\,x\right )}{d\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}-\frac {2\,B\,a}{d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}} \]

[In]

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

[Out]

2*atanh((32*A^2*a^2*d^3*tan(c + d*x)^(1/2)*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) -
(A^2*a*b)/(2*d^2))^(1/2))/(16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) - 16*A^3*b^3*d^2 + 16*
A^3*a^2*b*d^2) - (32*A^2*b^2*d^3*tan(c + d*x)^(1/2)*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(
4*d^4) - (A^2*a*b)/(2*d^2))^(1/2))/(16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) - 16*A^3*b^3*
d^2 + 16*A^3*a^2*b*d^2))*(- (2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a*b)/(2*d^2))
^(1/2) - 2*atanh((32*A^2*a^2*d^3*tan(c + d*x)^(1/2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*
d^4) - (A^2*a*b)/(2*d^2))^(1/2))/(16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) + 16*A^3*b^3*d^
2 - 16*A^3*a^2*b*d^2) - (32*A^2*b^2*d^3*tan(c + d*x)^(1/2)*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1
/2)/(4*d^4) - (A^2*a*b)/(2*d^2))^(1/2))/(16*A*a*(2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2) + 16*A^3
*b^3*d^2 - 16*A^3*a^2*b*d^2))*((2*A^4*a^2*b^2*d^4 - A^4*b^4*d^4 - A^4*a^4*d^4)^(1/2)/(4*d^4) - (A^2*a*b)/(2*d^
2))^(1/2) + 2*atanh((32*B^2*a^2*d^3*tan(c + d*x)^(1/2)*((B^2*a*b)/(2*d^2) - (2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 -
B^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2))/(16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2) + 16*B^3*a^3
*d^2 - 16*B^3*a*b^2*d^2) - (32*B^2*b^2*d^3*tan(c + d*x)^(1/2)*((B^2*a*b)/(2*d^2) - (2*B^4*a^2*b^2*d^4 - B^4*b^
4*d^4 - B^4*a^4*d^4)^(1/2)/(4*d^4))^(1/2))/(16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2) + 16*
B^3*a^3*d^2 - 16*B^3*a*b^2*d^2))*((B^2*a*b)/(2*d^2) - (2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2)/(4
*d^4))^(1/2) - 2*atanh((32*B^2*a^2*d^3*tan(c + d*x)^(1/2)*((2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/
2)/(4*d^4) + (B^2*a*b)/(2*d^2))^(1/2))/(16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2) - 16*B^3*
a^3*d^2 + 16*B^3*a*b^2*d^2) - (32*B^2*b^2*d^3*tan(c + d*x)^(1/2)*((2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d
^4)^(1/2)/(4*d^4) + (B^2*a*b)/(2*d^2))^(1/2))/(16*B*b*(2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2) -
16*B^3*a^3*d^2 + 16*B^3*a*b^2*d^2))*((2*B^4*a^2*b^2*d^4 - B^4*b^4*d^4 - B^4*a^4*d^4)^(1/2)/(4*d^4) + (B^2*a*b)
/(2*d^2))^(1/2) - ((2*A*a)/3 + 2*A*b*tan(c + d*x))/(d*tan(c + d*x)^(3/2)) - (2*B*a)/(d*tan(c + d*x)^(1/2))