3.9.11 \(\int \frac {(a+b \tan (c+d x))^2}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) [811]

Optimal. Leaf size=249 \[ -\frac {\left (a^2+2 a b-b^2\right ) \text {ArcTan}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {\left (a^2+2 a b-b^2\right ) \text {ArcTan}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\cot (c+d x)}}+\frac {\left (a^2-2 a b-b^2\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {\left (a^2-2 a b-b^2\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d} \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 14, 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 )}{d \sqrt {\cot (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}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^2/Cot[c + d*x]^(3/2),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)/(5*d*Cot[c + d*x]^(5/2)) + (4*a*b)/(3*d*Cot[c + d*x]^(3/2))
+ (2*(a^2 - b^2))/(d*Sqrt[Cot[c + d*x]]) + ((a^2 - 2*a*b - b^2)*Log[1 - Sqrt[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 {3}{2}}(c+d x)} \, dx &=\int \frac {(b+a \cot (c+d x))^2}{\cot ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\int \frac {2 a b+\left (a^2-b^2\right ) \cot (c+d x)}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\int \frac {a^2-b^2-2 a b \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\cot (c+d x)}}+\int \frac {-2 a b-\left (a^2-b^2\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\cot (c+d x)}}+\frac {2 \text {Subst}\left (\int \frac {2 a b+\left (a^2-b^2\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{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}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{d \sqrt {\cot (c+d x)}}+\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 {\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 {2 b^2}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{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}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {4 a b}{3 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2-b^2\right )}{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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.29, size = 81, normalized size = 0.33 \begin {gather*} \frac {-6 \left (a^2-b^2\right ) \, _2F_1\left (-\frac {5}{4},1;-\frac {1}{4};-\cot ^2(c+d x)\right )+2 a \left (3 a+10 b \cot (c+d x) \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )\right )}{15 d \cot ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(a^2 - b^2)*Hypergeometric2F1[-5/4, 1, -1/4, -Cot[c + d*x]^2] + 2*a*(3*a + 10*b*Cot[c + d*x]*Hypergeometri
c2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2]))/(15*d*Cot[c + d*x]^(5/2))

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 59.00, size = 1975, normalized size = 7.93

method	result	size
default	\(\text {Expression too large to display}\)	\(1975\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30/d*(-1+cos(d*x+c))*(-15*I*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)^2*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))-30*I*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)^2*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))*a*b+15*I*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)^
2*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))+30*I*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)^2*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))*a*b+15*I*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)^2*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))-15*I*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)^2*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))-30*cos(d*x+c)^2*sin(d*x+c)*(-(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2),1/2*2^(1/2))*a^2+30*cos(d*x+c)^2*sin(d*x+c)*(-(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*
x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2+15*cos(d*x+c)^2*sin(d*x+c)*(-(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))*a^2-30*cos(d*x+c)^2*sin(d*x+c)*(-(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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))*a*b-15*cos(d*x+c)^2*sin(d*x+c)*(-(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)*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)*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))*b^2+15*cos(d*x+c)^2*si
n(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*a^2-3
0*cos(d*x+c)^2*sin(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*
x+c))^(1/2)*a*b-15*cos(d*x+c)^2*sin(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*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))*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)*b^2+30*cos(d*x+c)^3*2^(1/2)*a^2-36*cos(d*x+c)^3*2^(1/2)*b^2+20*cos(d*x+c)^2*sin(d
*x+c)*2^(1/2)*a*b-30*cos(d*x+c)^2*2^(1/2)*a^2+36*cos(d*x+c)^2*2^(1/2)*b^2-20*cos(d*x+c)*sin(d*x+c)*2^(1/2)*a*b
+6*cos(d*x+c)*2^(1/2)*b^2-6*2^(1/2)*b^2)*(cos(d*x+c)+1)^2/(cos(d*x+c)/sin(d*x+c))^(3/2)/cos(d*x+c)/sin(d*x+c)^
5*2^(1/2)

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Maxima [A]
time = 0.54, size = 212, normalized size = 0.85 \begin {gather*} \frac {8 \, {\left (3 \, b^{2} + \frac {10 \, a b}{\tan \left (d x + c\right )} + \frac {15 \, {\left (a^{2} - b^{2}\right )}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}} + 30 \, \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 ) + 30 \, \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 ) - 15 \, \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 ) + 15 \, \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 )}{60 \, 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)^(3/2),x, algorithm="maxima")

[Out]

1/60*(8*(3*b^2 + 10*a*b/tan(d*x + c) + 15*(a^2 - b^2)/tan(d*x + c)^2)*tan(d*x + c)^(5/2) + 30*sqrt(2)*(a^2 + 2
*a*b - b^2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 30*sqrt(2)*(a^2 + 2*a*b - b^2)*arctan(-1/2*
sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - 15*sqrt(2)*(a^2 - 2*a*b - b^2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/
tan(d*x + c) + 1) + 15*sqrt(2)*(a^2 - 2*a*b - b^2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/d

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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)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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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 {3}{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)**(3/2),x)

[Out]

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

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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)^(3/2),x, algorithm="giac")

[Out]

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

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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 )}^{3/2}} \,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)^(3/2),x)

[Out]

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

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