3.9.12 \(\int \frac {(a+b \tan (c+d x))^2}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\) [812]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.21, antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3754, 3623, 3610, 3615, 1182, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \text {ArcTan}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\cot (c+d x)}}+\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - 2*a*b - b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d)) + ((a^2 - 2*a*b - b^2)*ArcTan[1 +
Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) + (2*b^2)/(7*d*Cot[c + d*x]^(7/2)) + (4*a*b)/(5*d*Cot[c + d*x]^(5/2))
+ (2*(a^2 - b^2))/(3*d*Cot[c + d*x]^(3/2)) - (4*a*b)/(d*Sqrt[Cot[c + d*x]]) - ((a^2 + 2*a*b - b^2)*Log[1 - Sq
rt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d) + ((a^2 + 2*a*b - b^2)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*
x]] + Cot[c + d*x]])/(2*Sqrt[2]*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 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 3623

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

Rule 3754

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {(a+b \tan (c+d x))^2}{\cot ^{\frac {5}{2}}(c+d x)} \, dx &=\int \frac {(b+a \cot (c+d x))^2}{\cot ^{\frac {9}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\int \frac {2 a b+\left (a^2-b^2\right ) \cot (c+d x)}{\cot ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\int \frac {a^2-b^2-2 a b \cot (c+d x)}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\int \frac {-2 a b-\left (a^2-b^2\right ) \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\cot (c+d x)}}+\int \frac {-a^2+b^2+2 a b \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\cot (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {a^2-b^2-2 a b x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\cot (c+d x)}}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}+\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\cot (c+d x)}}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}-\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}-\frac {\left (a^2+2 a b-b^2\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}\\ &=\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\cot (c+d x)}}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}\\ &=-\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2-2 a b-b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b^2}{7 d \cot ^{\frac {7}{2}}(c+d x)}+\frac {4 a b}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{3 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {4 a b}{d \sqrt {\cot (c+d x)}}-\frac {\left (a^2+2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.40, size = 80, normalized size = 0.30 \begin {gather*} \frac {2 \left (-5 \left (a^2-b^2\right ) \, _2F_1\left (-\frac {7}{4},1;-\frac {3}{4};-\cot ^2(c+d x)\right )+a \left (5 a+14 b \cot (c+d x) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(c+d x)\right )\right )\right )}{35 d \cot ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-5*(a^2 - b^2)*Hypergeometric2F1[-7/4, 1, -3/4, -Cot[c + d*x]^2] + a*(5*a + 14*b*Cot[c + d*x]*Hypergeometr
ic2F1[-5/4, 1, -1/4, -Cot[c + d*x]^2])))/(35*d*Cot[c + d*x]^(7/2))

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 55.30, size = 1903, normalized size = 7.10

method	result	size
default	\(\text {Expression too large to display}\)	\(1903\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/210/d*(-1+cos(d*x+c))*(210*I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2)
)*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-s
in(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*a*b+105*I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1
/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)*a^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))
^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+105*I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+c)*b^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin
(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-210*I*EllipticPi((-(cos(
d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((
cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*a*b-105*
I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)*b^2*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/
2)*cos(d*x+c)^3-105*I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+
c)*a^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x
+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+105*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2
*2^(1/2))*sin(d*x+c)*a^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(co
s(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+210*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(
1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c
))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*a*b-105*EllipticPi((-(cos(d*x+c)-1-sin(d*x
+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)*b^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1
+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-420*EllipticF((-(cos
(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c
)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*a*b+105*EllipticP
i((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+c)*a^2*((-1+cos(d*x+c))/sin(d*x
+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+
c)^3+210*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin(d*x+c)*((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^
(1/2)*cos(d*x+c)^3*a*b-105*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2-1/2*I,1/2*2^(1/2))*sin
(d*x+c)*b^2*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-si
n(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+504*2^(1/2)*cos(d*x+c)^4*a*b-70*sin(d*x+c)*2^(1/2)*a^2*cos(d*x+c)^3+1
00*sin(d*x+c)*2^(1/2)*b^2*cos(d*x+c)^3-504*2^(1/2)*cos(d*x+c)^3*a*b+70*sin(d*x+c)*2^(1/2)*a^2*cos(d*x+c)^2-100
*sin(d*x+c)*2^(1/2)*b^2*cos(d*x+c)^2-84*2^(1/2)*cos(d*x+c)^2*a*b-30*sin(d*x+c)*2^(1/2)*b^2*cos(d*x+c)+84*2^(1/
2)*cos(d*x+c)*a*b+30*sin(d*x+c)*2^(1/2)*b^2)*(cos(d*x+c)+1)^2/(cos(d*x+c)/sin(d*x+c))^(5/2)/cos(d*x+c)/sin(d*x
+c)^6*2^(1/2)

________________________________________________________________________________________

Maxima [A]
time = 0.54, size = 224, normalized size = 0.84 \begin {gather*} \frac {8 \, {\left (15 \, b^{2} + \frac {42 \, a b}{\tan \left (d x + c\right )} - \frac {210 \, a b}{\tan \left (d x + c\right )^{3}} + \frac {35 \, {\left (a^{2} - b^{2}\right )}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {7}{2}} + 210 \, \sqrt {2} {\left (a^{2} - 2 \, a b - b^{2}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {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} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 105 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 105 \, \sqrt {2} {\left (a^{2} + 2 \, a b - b^{2}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{420 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/420*(8*(15*b^2 + 42*a*b/tan(d*x + c) - 210*a*b/tan(d*x + c)^3 + 35*(a^2 - b^2)/tan(d*x + c)^2)*tan(d*x + c)^
(7/2) + 210*sqrt(2)*(a^2 - 2*a*b - b^2)*arctan(1/2*sqrt(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/sqrt(tan(d*x + c)))) + 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(s
qrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 105*sqrt(2)*(a^2 + 2*a*b - b^2)*log(-sqrt(2)/sqrt(tan(d*x +
c)) + 1/tan(d*x + c) + 1))/d

________________________________________________________________________________________

Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{2}}{\cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^2}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________