3.7.23 \(\int \tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [623]

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

[Out]

(I*a-b)^(5/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-1/64*(5*a^4+240*a^2*b^2-128*b^4)
*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/b^(3/2)/d-(I*a+b)^(5/2)*arctanh((I*a+b)^(1/2)*tan(d*
x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d-1/64*a*(5*a^2+112*b^2)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/b/d-1/96*(
5*a^2+48*b^2)*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3/2)/b/d-1/24*a*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2)/b/d+1
/4*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(7/2)/b/d

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Rubi [A]
time = 1.93, antiderivative size = 332, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3647, 3728, 3736, 6857, 65, 223, 212, 95, 211, 214} \begin {gather*} -\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 (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 {(-b+i a)^{5/2} \text {ArcTan}\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} \end {gather*}

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 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 3647

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 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 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 \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 {\text {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 {\text {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 {\text {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 ) \text {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 {\text {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 ) \text {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 \text {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 \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 (5 a^4+240 a^2 b^2-128 b^4\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 )}{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 \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 {(a+i b)^3 \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} \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*}

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Mathematica [A]
time = 3.42, size = 349, normalized size = 1.05 \begin {gather*} -\frac {192 \sqrt [4]{-1} (-a+i b)^{5/2} 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 )+192 \sqrt [4]{-1} (a+i b)^{5/2} 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 )+3 a \left (5 a^2+112 b^2\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 \left (5 a^2+48 b^2\right ) \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}+\frac {3 \sqrt {a} \left (5 a^4+240 a^2 b^2-128 b^4\right ) \sinh ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}}{192 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/192*(192*(-1)^(1/4)*(-a + I*b)^(5/2)*b*ArcTan[((-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*ArcTan[((-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]]))/(b*d)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 0.50, size = 1347722, normalized size = 4059.40 \[\text {output too large to display}\]

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

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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(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)

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

[Out]

Timed out

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3877 deep

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

[Out]

Timed out

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

[Out]

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

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