3.7.31 \(\int \cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [631]

Optimal. Leaf size=300 \[ -\frac {(i a-b)^{5/2} (i A-B) \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {2 b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {(i a+b)^{5/2} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a (2 A b+a B) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d} \]

[Out]

-(I*a-b)^(5/2)*(I*A-B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+
c)^(1/2)/d+2*b^(5/2)*B*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1
/2)/d-(I*a+b)^(5/2)*(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*ta
n(d*x+c)^(1/2)/d-2*a*(2*A*b+B*a)*cot(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d-2/3*a*A*cot(d*x+c)^(3/2)*(a+b*tan(d
*x+c))^(3/2)/d

________________________________________________________________________________________

Rubi [A]
time = 1.61, antiderivative size = 300, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {4326, 3686, 3726, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} -\frac {(-b+i a)^{5/2} (-B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a (a B+2 A b) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {(b+i a)^{5/2} (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}}{3 d}+\frac {2 b^{5/2} B \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*a - b)^(5/2)*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c +
d*x]]*Sqrt[Tan[c + d*x]])/d) + (2*b^(5/2)*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqr
t[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - ((I*a + b)^(5/2)*(I*A + B)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/
Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/d - (2*a*(2*A*b + a*B)*Sqrt[Cot[c + d*x]]*Sqr
t[a + b*Tan[c + d*x]])/d - (2*a*A*Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2))/(3*d)

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 211

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

Rule 212

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

Rule 214

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

Rule 223

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 3686

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

Rule 3726

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

Rule 3736

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x
]}, Dist[ff/f, Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^2)/(1 + ff^2*x^2)), x], x, Tan[
e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]

Rule 4326

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 38.94, size = 130606, normalized size = 435.35 \begin {gather*} \text {Result too large to show} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 27.88, size = 47378, normalized size = 157.93

method	result	size
default	\(\text {Expression too large to display}\)	\(47378\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

[Out]

Timed out

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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