3.399 \(\int \cot ^2(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=111 \[ -\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac {(a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f} \]

[Out]

-a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)^(1/2))/f+b^(3/2)*arctanh(b^(1/2)*tan(f*x+e)/(a+b+b*tan
(f*x+e)^2)^(1/2))/f-(a+b)*cot(f*x+e)*(a+b+b*tan(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.23, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {4141, 1975, 474, 523, 217, 206, 377, 203} \[ -\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac {(a+b) \cot (e+f x) \sqrt {a+b \tan ^2(e+f x)+b}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

-((a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f) + (b^(3/2)*ArcTanh[(Sqrt[b]*Tan[e
+ f*x])/Sqrt[a + b + b*Tan[e + f*x]^2]])/f - ((a + b)*Cot[e + f*x]*Sqrt[a + b + b*Tan[e + f*x]^2])/f

Rule 203

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

Rule 206

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 474

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(c*(e*x)^
(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)
*(a + b*x^n)^p*(c + d*x^n)^(q - 2)*Simp[c*(c*b - a*d)*(m + 1) + c*n*(b*c*(p + 1) + a*d*(q - 1)) + d*((c*b - a*
d)*(m + 1) + c*b*n*(p + q))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0]
&& GtQ[q, 1] && LtQ[m, -1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

\begin {align*} \int \cot ^2(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \left (1+x^2\right )\right )^{3/2}}{x^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b+b x^2\right )^{3/2}}{x^2 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a+b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}+\frac {\operatorname {Subst}\left (\int \frac {-a^2+b^2+b^2 x^2}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a+b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {(a+b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}-\frac {a^2 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}\\ &=-\frac {a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}+\frac {b^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac {(a+b) \cot (e+f x) \sqrt {a+b+b \tan ^2(e+f x)}}{f}\\ \end {align*}

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Mathematica [C]  time = 6.12, size = 410, normalized size = 3.69 \[ \frac {\sqrt {2} e^{i (e+f x)} \cos ^3(e+f x) \sqrt {4 b+a e^{-2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^2} \left (\frac {-2 \left (a^{3/2} f x+b^{3/2} \log \left (\frac {f \left (\sqrt {b} \left (-1+e^{2 i (e+f x)}\right )-i \sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}\right )}{b^2 \left (1+e^{2 i (e+f x)}\right )}\right )\right )+i a^{3/2} \log \left (\sqrt {a} \sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}+a e^{2 i (e+f x)}+a+2 b\right )-i a^{3/2} \log \left (\sqrt {a} \sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}+a e^{2 i (e+f x)}+a+2 b e^{2 i (e+f x)}\right )}{\sqrt {a \left (1+e^{2 i (e+f x)}\right )^2+4 b e^{2 i (e+f x)}}}-\frac {2 i (a+b)}{-1+e^{2 i (e+f x)}}\right ) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{f (a \cos (2 e+2 f x)+a+2 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2*(a + b*Sec[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[2]*E^(I*(e + f*x))*Sqrt[4*b + (a*(1 + E^((2*I)*(e + f*x)))^2)/E^((2*I)*(e + f*x))]*Cos[e + f*x]^3*(((-2*
I)*(a + b))/(-1 + E^((2*I)*(e + f*x))) + (I*a^(3/2)*Log[a + 2*b + a*E^((2*I)*(e + f*x)) + Sqrt[a]*Sqrt[4*b*E^(
(2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - I*a^(3/2)*Log[a + a*E^((2*I)*(e + f*x)) + 2*b*E^((2*I)*(e
+ f*x)) + Sqrt[a]*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2]] - 2*(a^(3/2)*f*x + b^(3/2)*L
og[((Sqrt[b]*(-1 + E^((2*I)*(e + f*x))) - I*Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*f)/
(b^2*(1 + E^((2*I)*(e + f*x))))]))/Sqrt[4*b*E^((2*I)*(e + f*x)) + a*(1 + E^((2*I)*(e + f*x)))^2])*(a + b*Sec[e
+ f*x]^2)^(3/2))/(f*(a + 2*b + a*Cos[2*e + 2*f*x])^(3/2))

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fricas [B]  time = 1.21, size = 1446, normalized size = 13.03 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(sqrt(-a)*a*log(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*
b^2)*cos(f*x + e)^4 + a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*co
s(f*x + e)^2 + 8*(16*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos
(f*x + e)^3 - (a^3 - 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^
2)*sin(f*x + e))*sin(f*x + e) + 2*b^(3/2)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x + e)
^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x
+ e) + 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 8*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x +
e))/(f*sin(f*x + e)), 1/8*(4*sqrt(-b)*b*arctan(-1/2*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt(
(a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^2)*sin(f*x + e)))*sin(f*x + e) + sqrt(-a)*a*lo
g(128*a^4*cos(f*x + e)^8 - 256*(a^4 - a^3*b)*cos(f*x + e)^6 + 32*(5*a^4 - 14*a^3*b + 5*a^2*b^2)*cos(f*x + e)^4
+ a^4 - 28*a^3*b + 70*a^2*b^2 - 28*a*b^3 + b^4 - 32*(a^4 - 7*a^3*b + 7*a^2*b^2 - a*b^3)*cos(f*x + e)^2 + 8*(1
6*a^3*cos(f*x + e)^7 - 24*(a^3 - a^2*b)*cos(f*x + e)^5 + 2*(5*a^3 - 14*a^2*b + 5*a*b^2)*cos(f*x + e)^3 - (a^3
- 7*a^2*b + 7*a*b^2 - b^3)*cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e))*si
n(f*x + e) - 8*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/(f*sin(f*x + e)), 1/4*(a^(3/2
)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 - 6*a*b + b^2)*cos(f*x + e))*sqrt(a)*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b + a*b^2 - (a^3 - 3*a^2*b)*cos(f*x +
e)^2)*sin(f*x + e)))*sin(f*x + e) + b^(3/2)*log(((a^2 - 6*a*b + b^2)*cos(f*x + e)^4 + 8*(a*b - b^2)*cos(f*x +
e)^2 + 4*((a - b)*cos(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(
f*x + e) + 8*b^2)/cos(f*x + e)^4)*sin(f*x + e) - 4*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x
+ e))/(f*sin(f*x + e)), 1/4*(a^(3/2)*arctan(1/4*(8*a^2*cos(f*x + e)^5 - 8*(a^2 - a*b)*cos(f*x + e)^3 + (a^2 -
6*a*b + b^2)*cos(f*x + e))*sqrt(a)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*a^3*cos(f*x + e)^4 - a^2*b
+ a*b^2 - (a^3 - 3*a^2*b)*cos(f*x + e)^2)*sin(f*x + e)))*sin(f*x + e) + 2*sqrt(-b)*b*arctan(-1/2*((a - b)*cos
(f*x + e)^3 + 2*b*cos(f*x + e))*sqrt(-b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((a*b*cos(f*x + e)^2 + b^
2)*sin(f*x + e)))*sin(f*x + e) - 4*(a + b)*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e))/(f*sin(f*
x + e))]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^2, x)

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maple [C]  time = 2.03, size = 1952, normalized size = 17.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x)

[Out]

-1/f*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(3/2)*cos(f*x+e)^3*(2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^
(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*
x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+
e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*cos(f*x+e)*sin(f*x+e)*a^2-2^(1/2)
*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b
^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*
I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)
^2)^(1/2))*b^2*sin(f*x+e)*cos(f*x+e)+2*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b
)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e
))/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),1/(2*I*a^(1/2)*b
^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*b^2*sin(f*
x+e)*cos(f*x+e)-2*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b
))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*Ellip
ticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-
(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*cos(f*x+e)*sin(f*x+e)*a^2+a^2*
2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^
(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e
))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2
)/(a+b)^2)^(1/2))*sin(f*x+e)-2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f
*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^
(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*
a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*b^2*sin(f*x+e)+2*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/
2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*c
os(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/si
n(f*x+e),1/(2*I*a^(1/2)*b^(1/2)+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)
/(a+b))^(1/2))*b^2*sin(f*x+e)-2*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+co
s(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b
))^(1/2)*EllipticPi((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),-1/(2*I*a^(1/2)*b^(1/2)
+a-b)*(a+b),(-(2*I*a^(1/2)*b^(1/2)-a+b)/(a+b))^(1/2)/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2))*sin(f*x+e)*a^2+(
(2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a^2+((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a
*b+((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b+((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*b^2)/sin(f*x+e)/(b+a*co
s(f*x+e)^2)^2/((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \cot \left (f x + e\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2*(a+b*sec(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(f*x + e)^2 + a)^(3/2)*cot(f*x + e)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (e+f\,x\right )}^2\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^2*(a + b/cos(e + f*x)^2)^(3/2),x)

[Out]

int(cot(e + f*x)^2*(a + b/cos(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}} \cot ^{2}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**2*(a+b*sec(f*x+e)**2)**(3/2),x)

[Out]

Integral((a + b*sec(e + f*x)**2)**(3/2)*cot(e + f*x)**2, x)

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