3.451 \(\int \frac{(a+b \tan (c+d x))^{5/2} (\frac{3 b B}{2 a}+B \tan (c+d x))}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=253 \[ -\frac{2 B \left (a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{2 b^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B (2 a-3 i b) (-b+i a)^{5/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{2 a d}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{B (2 a+3 i b) (b+i a)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{2 a d} \]

[Out]

((I*a - b)^(5/2)*(2*a - (3*I)*b)*B*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*a*d
) + (2*b^(5/2)*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((2*a + (3*I)*b)*(I*a + b
)^(5/2)*B*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*a*d) - (2*(a^2 + 3*b^2)*B*S
qrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (b*B*(a + b*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(3/2))

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Rubi [A]  time = 2.45648, antiderivative size = 253, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.233, Rules used = {3605, 3645, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ -\frac{2 B \left (a^2+3 b^2\right ) \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}+\frac{2 b^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{B (2 a-3 i b) (-b+i a)^{5/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{2 a d}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}-\frac{B (2 a+3 i b) (b+i a)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Tan[c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

((I*a - b)^(5/2)*(2*a - (3*I)*b)*B*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*a*d
) + (2*b^(5/2)*B*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((2*a + (3*I)*b)*(I*a + b
)^(5/2)*B*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(2*a*d) - (2*(a^2 + 3*b^2)*B*S
qrt[a + b*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (b*B*(a + b*Tan[c + d*x])^(3/2))/(d*Tan[c + d*x]^(3/2))

Rule 3605

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 3645

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*T
an[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 3655

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 6725

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]

Rule 63

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 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 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 93

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 205

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(a+b \tan (c+d x))^{5/2} \left (\frac{3 b B}{2 a}+B \tan (c+d x)\right )}{\tan ^{\frac{5}{2}}(c+d x)} \, dx &=-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{2}{3} \int \frac{\sqrt{a+b \tan (c+d x)} \left (\frac{3}{2} \left (a^2+3 b^2\right ) B+\frac{3}{4} b \left (a+\frac{3 b^2}{a}\right ) B \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 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4}{3} \int \frac{\frac{9}{8} b \left (a^2+3 b^2\right ) B-\frac{3 \left (2 a^4+3 a^2 b^2-3 b^4\right ) B \tan (c+d x)}{8 a}+\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 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \operatorname{Subst}\left (\int \frac{\frac{9}{8} b \left (a^2+3 b^2\right ) B-\frac{3 \left (2 a^4+3 a^2 b^2-3 b^4\right ) B x}{8 a}+\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 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{4 \operatorname{Subst}\left (\int \left (\frac{3 b^3 B}{4 \sqrt{x} \sqrt{a+b x}}+\frac{3 \left (a b \left (3 a^2+7 b^2\right ) B-\left (2 a^4+3 a^2 b^2-3 b^4\right ) B x\right )}{8 a \sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{3 d}\\ &=-\frac{2 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\operatorname{Subst}\left (\int \frac{a b \left (3 a^2+7 b^2\right ) B-\left (2 a^4+3 a^2 b^2-3 b^4\right ) B x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{2 a d}+\frac{\left (b^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{2 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\operatorname{Subst}\left (\int \left (\frac{i a b \left (3 a^2+7 b^2\right ) B+\left (2 a^4+3 a^2 b^2-3 b^4\right ) B}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{i a b \left (3 a^2+7 b^2\right ) B-\left (2 a^4+3 a^2 b^2-3 b^4\right ) B}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{2 a d}+\frac{\left (2 b^3 B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=-\frac{2 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\left ((a+i b)^3 (2 a-3 i b) B\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{4 a d}-\frac{\left ((a-i b)^3 (2 a+3 i b) B\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{4 a d}+\frac{\left (2 b^3 B\right ) \operatorname{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{2 b^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{2 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}+\frac{\left ((a+i b)^3 (2 a-3 i b) B\right ) \operatorname{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 )}{2 a d}-\frac{\left ((a-i b)^3 (2 a+3 i b) B\right ) \operatorname{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 )}{2 a d}\\ &=\frac{(i a-b)^{5/2} (2 a-3 i b) B \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{2 a 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 )}{d}-\frac{(2 a+3 i b) (i a+b)^{5/2} B \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{2 a d}-\frac{2 \left (a^2+3 b^2\right ) B \sqrt{a+b \tan (c+d x)}}{d \sqrt{\tan (c+d x)}}-\frac{b B (a+b \tan (c+d x))^{3/2}}{d \tan ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 4.19159, size = 355, normalized size = 1.4 \[ \frac{B \cos (c+d x) (2 a \tan (c+d x)+3 b) \left (4 \sqrt{a} b^{5/2} \tan ^{\frac{3}{2}}(c+d x) \sqrt{a+b \tan (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )+\sqrt{\frac{b \tan (c+d x)}{a}+1} \left (-2 a \sqrt{a+b \tan (c+d x)} \left (\left (2 a^2+7 b^2\right ) \tan (c+d x)+a b\right )+\sqrt [4]{-1} (2 a-3 i b) (-a-i b)^{5/2} \tan ^{\frac{3}{2}}(c+d x) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+\sqrt [4]{-1} (a-i b)^{5/2} (2 a+3 i b) \tan ^{\frac{3}{2}}(c+d x) \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )\right )\right )}{2 a d \tan ^{\frac{3}{2}}(c+d x) \sqrt{\frac{b \tan (c+d x)}{a}+1} (2 a \sin (c+d x)+3 b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Tan[c + d*x])^(5/2)*((3*b*B)/(2*a) + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

(B*Cos[c + d*x]*(3*b + 2*a*Tan[c + d*x])*(4*Sqrt[a]*b^(5/2)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Tan[
c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]] + Sqrt[1 + (b*Tan[c + d*x])/a]*((-1)^(1/4)*(-a - I*b)^(5/2)*(2*a - (3*
I)*b)*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Tan[c + d*x]^(3/2) + (-
1)^(1/4)*(a - I*b)^(5/2)*(2*a + (3*I)*b)*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[
c + d*x]]]*Tan[c + d*x]^(3/2) - 2*a*Sqrt[a + b*Tan[c + d*x]]*(a*b + (2*a^2 + 7*b^2)*Tan[c + d*x]))))/(2*a*d*(3
*b*Cos[c + d*x] + 2*a*Sin[c + d*x])*Tan[c + d*x]^(3/2)*Sqrt[1 + (b*Tan[c + d*x])/a])

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Maple [B]  time = 0.658, size = 1490268, normalized size = 5890.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, \int \frac{{\left (2 \, B \tan \left (d x + c\right ) + \frac{3 \, B b}{a}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\tan \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

1/2*integrate((2*B*tan(d*x + c) + 3*B*b/a)*(b*tan(d*x + c) + a)^(5/2)/tan(d*x + c)^(5/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

Timed out

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(5/2)*(3/2*b*B/a+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

Timed out