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

Optimal. Leaf size=332 \[ -\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (240 a^2 b^2+5 a^4-128 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}+\frac{(-b+i a)^{5/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}-\frac{(b+i a)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

[Out]

((I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((5*a^4 + 240*a^2*b^
2 - 128*b^4)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(64*b^(3/2)*d) - ((I*a + b)^(5/2)
*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (a*(5*a^2 + 112*b^2)*Sqrt[Tan[c + d
*x]]*Sqrt[a + b*Tan[c + d*x]])/(64*b*d) - ((5*a^2 + 48*b^2)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(96
*b*d) - (a*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2))/(24*b*d) + (Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^
(7/2))/(4*b*d)

________________________________________________________________________________________

Rubi [A]  time = 2.57827, antiderivative size = 332, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {3566, 3647, 3655, 6725, 63, 217, 206, 93, 205, 208} \[ -\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (240 a^2 b^2+5 a^4-128 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}+\frac{(-b+i a)^{5/2} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}-\frac{(b+i a)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I*a - b)^(5/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((5*a^4 + 240*a^2*b^
2 - 128*b^4)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(64*b^(3/2)*d) - ((I*a + b)^(5/2)
*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (a*(5*a^2 + 112*b^2)*Sqrt[Tan[c + d
*x]]*Sqrt[a + b*Tan[c + d*x]])/(64*b*d) - ((5*a^2 + 48*b^2)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(96
*b*d) - (a*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2))/(24*b*d) + (Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^
(7/2))/(4*b*d)

Rule 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3647

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 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 \tan ^{\frac{5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx &=\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\int \frac{(a+b \tan (c+d x))^{5/2} \left (-\frac{a}{2}-4 b \tan (c+d x)-\frac{1}{2} a \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{4 b}\\ &=-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\int \frac{(a+b \tan (c+d x))^{3/2} \left (-\frac{5 a^2}{4}-12 a b \tan (c+d x)-\frac{1}{4} \left (5 a^2+48 b^2\right ) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{12 b}\\ &=-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\int \frac{\sqrt{a+b \tan (c+d x)} \left (-\frac{3}{8} a \left (5 a^2-16 b^2\right )-24 b \left (a^2-b^2\right ) \tan (c+d x)-\frac{3}{8} a \left (5 a^2+112 b^2\right ) \tan ^2(c+d x)\right )}{\sqrt{\tan (c+d x)}} \, dx}{24 b}\\ &=-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\int \frac{-\frac{3}{16} a^2 \left (5 a^2-144 b^2\right )-24 a b \left (a^2-3 b^2\right ) \tan (c+d x)-\frac{3}{16} \left (5 a^4+240 a^2 b^2-128 b^4\right ) \tan ^2(c+d x)}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx}{24 b}\\ &=-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\operatorname{Subst}\left (\int \frac{-\frac{3}{16} a^2 \left (5 a^2-144 b^2\right )-24 a b \left (a^2-3 b^2\right ) x-\frac{3}{16} \left (5 a^4+240 a^2 b^2-128 b^4\right ) x^2}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{24 b d}\\ &=-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\operatorname{Subst}\left (\int \left (-\frac{3 \left (5 a^4+240 a^2 b^2-128 b^4\right )}{16 \sqrt{x} \sqrt{a+b x}}+\frac{24 \left (3 a^2 b^2-b^4-a b \left (a^2-3 b^2\right ) x\right )}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )}\right ) \, dx,x,\tan (c+d x)\right )}{24 b d}\\ &=-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\operatorname{Subst}\left (\int \frac{3 a^2 b^2-b^4-a b \left (a^2-3 b^2\right ) x}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{b d}-\frac{\left (5 a^4+240 a^2 b^2-128 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{128 b d}\\ &=-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}+\frac{\operatorname{Subst}\left (\int \left (\frac{a b \left (a^2-3 b^2\right )+i \left (3 a^2 b^2-b^4\right )}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{-a b \left (a^2-3 b^2\right )+i \left (3 a^2 b^2-b^4\right )}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{b d}-\frac{\left (5 a^4+240 a^2 b^2-128 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sqrt{\tan (c+d x)}\right )}{64 b d}\\ &=-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac{(a-i b)^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{(a+i b)^3 \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}-\frac{\left (5 a^4+240 a^2 b^2-128 b^4\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 )}{64 b d}\\ &=-\frac{\left (5 a^4+240 a^2 b^2-128 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}-\frac{(a-i b)^3 \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 )}{d}+\frac{(a+i b)^3 \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 )}{d}\\ &=\frac{(i a-b)^{5/2} \tan ^{-1}\left (\frac{\sqrt{i a-b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\left (5 a^4+240 a^2 b^2-128 b^4\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{64 b^{3/2} d}-\frac{(i a+b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{i a+b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}}{64 b d}-\frac{\left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{96 b d}-\frac{a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{24 b d}+\frac{\sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}}{4 b d}\\ \end{align*}

Mathematica [A]  time = 2.642, size = 349, normalized size = 1.05 \[ \frac{-2 \left (5 a^2+48 b^2\right ) \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{3/2}-3 a \left (5 a^2+112 b^2\right ) \sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}-\frac{3 \sqrt{a} \left (240 a^2 b^2+5 a^4-128 b^4\right ) \sqrt{\frac{b \tan (c+d x)}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} \sqrt{\tan (c+d x)}}{\sqrt{a}}\right )}{\sqrt{b} \sqrt{a+b \tan (c+d x)}}+48 \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{7/2}-8 a \sqrt{\tan (c+d x)} (a+b \tan (c+d x))^{5/2}+192 \sqrt [4]{-1} b (-a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )+192 \sqrt [4]{-1} b (a-i b)^{5/2} \tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{192 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(192*(-1)^(1/4)*(-a - I*b)^(5/2)*b*ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d
*x]]] + 192*(-1)^(1/4)*(a - I*b)^(5/2)*b*ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[
c + d*x]]] - 3*a*(5*a^2 + 112*b^2)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] - 2*(5*a^2 + 48*b^2)*Sqrt[Tan[c
+ d*x]]*(a + b*Tan[c + d*x])^(3/2) - 8*a*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(5/2) + 48*Sqrt[Tan[c + d*x]
]*(a + b*Tan[c + d*x])^(7/2) - (3*Sqrt[a]*(5*a^4 + 240*a^2*b^2 - 128*b^4)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])
/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]]))/(192*b*d)

________________________________________________________________________________________

Maple [B]  time = 0.378, size = 1347722, normalized size = 4059.4 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\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(tan(d*x+c)^(5/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError