3.9.52 \(\int \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) [852]

Optimal. Leaf size=286 \[ \frac {(i a-b)^{3/2} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\left (3 a^2-8 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{4 \sqrt {b} d}+\frac {(i a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}} \]

[Out]

(I*a-b)^(3/2)*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+(I*a+b)^(3/2)*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)*tan(d*x+c)^(1/
2)/d+1/4*(3*a^2-8*b^2)*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/b^(1/2)+3/4*a*(a+b*tan(d*x+c))^(1/2)/d/cot(d*x+c)^(1/2)+1/2*(a+b*tan(d*x+c))^(3/2)/d/cot(d*x+c)^(1/2)

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Rubi [A]
time = 1.02, antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {4326, 3651, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} \frac {\left (3 a^2-8 b^2\right ) \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 )}{4 \sqrt {b} d}+\frac {(-b+i a)^{3/2} \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 {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}+\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(b+i a)^{3/2} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[T
an[c + d*x]])/d + ((3*a^2 - 8*b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c +
d*x]]*Sqrt[Tan[c + d*x]])/(4*Sqrt[b]*d) + ((I*a + b)^(3/2)*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 + (3*a*Sqrt[a + b*Tan[c + d*x]])/(4*d*Sqrt[Cot[c +
d*x]]) + (a + b*Tan[c + d*x])^(3/2)/(2*d*Sqrt[Cot[c + d*x]])

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 3651

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

Rule 3728

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

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 \frac {(a+b \tan (c+d x))^{3/2}}{\cot ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2} \, dx\\ &=\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}+\frac {1}{2} \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+b \tan (c+d x)} \left (-\frac {a}{2}-2 b \tan (c+d x)+\frac {3}{2} a \tan ^2(c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx\\ &=\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}+\frac {1}{2} \left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {5 a^2}{4}-4 a b \tan (c+d x)+\frac {1}{4} \left (3 a^2-8 b^2\right ) \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {-\frac {5 a^2}{4}-4 a b x+\frac {1}{4} \left (3 a^2-8 b^2\right ) x^2}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {3 a^2-8 b^2}{4 \sqrt {x} \sqrt {a+b x}}-\frac {2 \left (a^2-b^2+2 a b x\right )}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \frac {a^2-b^2+2 a b x}{\sqrt {x} \sqrt {a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (\left (3 a^2-8 b^2\right ) \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 )}{8 d}\\ &=\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {Subst}\left (\int \left (\frac {-2 a b+i \left (a^2-b^2\right )}{2 (i-x) \sqrt {x} \sqrt {a+b x}}+\frac {2 a b+i \left (a^2-b^2\right )}{2 \sqrt {x} (i+x) \sqrt {a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (\left (3 a^2-8 b^2\right ) \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 )}{4 d}\\ &=\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}-\frac {\left (i (a-i b)^2 \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 (i (a+i b)^2 \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 {\left (\left (3 a^2-8 b^2\right ) \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 )}{4 d}\\ &=\frac {\left (3 a^2-8 b^2\right ) \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)}}{4 \sqrt {b} d}+\frac {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}-\frac {\left (i (a-i b)^2 \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 (i (a+i b)^2 \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)^{3/2} \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 {\left (3 a^2-8 b^2\right ) \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)}}{4 \sqrt {b} d}+\frac {(i a+b)^{3/2} \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 {3 a \sqrt {a+b \tan (c+d x)}}{4 d \sqrt {\cot (c+d x)}}+\frac {(a+b \tan (c+d x))^{3/2}}{2 d \sqrt {\cot (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 6.00, size = 312, normalized size = 1.09 \begin {gather*} \frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (-4 \sqrt [4]{-1} \sqrt {-a+i b} (i a+b) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+4 (-1)^{3/4} (a+i b)^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\frac {\left (3 a^2-8 b^2\right ) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) (a+b \tan (c+d x))}{\sqrt {a} \sqrt {b} \sqrt {1+\frac {b \tan (c+d x)}{a}}}+\sqrt {\tan (c+d x)} \left (5 a^2+7 a b \tan (c+d x)+2 b^2 \tan ^2(c+d x)\right )\right )}{4 d \sqrt {a+b \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-4*(-1)^(1/4)*Sqrt[-a + I*b]*(I*a + b)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*
b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + 4*(-1)^(3/4)*(a + I*b)^(3/2)*ArcTa
n[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + ((3*a^2 -
8*b^2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*(a + b*Tan[c + d*x]))/(Sqrt[a]*Sqrt[b]*Sqrt[1 + (b*Tan[c
+ d*x])/a]) + Sqrt[Tan[c + d*x]]*(5*a^2 + 7*a*b*Tan[c + d*x] + 2*b^2*Tan[c + d*x]^2)))/(4*d*Sqrt[a + b*Tan[c
+ d*x]])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 35.70, size = 16676, normalized size = 58.31

method	result	size
default	\(\text {Expression too large to display}\)	\(16676\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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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((a+b*tan(d*x+c))^(3/2)/cot(d*x+c)^(3/2),x, algorithm="giac")

[Out]

Timed out

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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 )}^{3/2}}{{\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))^(3/2)/cot(c + d*x)^(3/2),x)

[Out]

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

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