\(\int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [519]

Optimal result

Integrand size = 23, antiderivative size = 211 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}-\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}-\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}-\frac {4 a (a+b \tan (c+d x))^{7/2}}{63 b^2 d}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d} \] Output:

(a-I*b)^(5/2)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/d+(a+I*b)^(5/2 
)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/d-2*(a^2-b^2)*(a+b*tan(d*x 
+c))^(1/2)/d-2/3*a*(a+b*tan(d*x+c))^(3/2)/d-2/5*(a+b*tan(d*x+c))^(5/2)/d-4 
/63*a*(a+b*tan(d*x+c))^(7/2)/b^2/d+2/9*tan(d*x+c)*(a+b*tan(d*x+c))^(7/2)/b 
/d
 

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.94 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-10 a^4-483 a^2 b^2+315 b^4+a b \left (5 a^2-231 b^2\right ) \tan (c+d x)+\left (75 a^2 b^2-63 b^4\right ) \tan ^2(c+d x)+95 a b^3 \tan ^3(c+d x)+35 b^4 \tan ^4(c+d x)\right )}{315 b^2 d} \] Input:

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

Output:

((a - I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/d + ((a 
+ I*b)^(5/2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/d + (2*Sqrt[ 
a + b*Tan[c + d*x]]*(-10*a^4 - 483*a^2*b^2 + 315*b^4 + a*b*(5*a^2 - 231*b^ 
2)*Tan[c + d*x] + (75*a^2*b^2 - 63*b^4)*Tan[c + d*x]^2 + 95*a*b^3*Tan[c + 
d*x]^3 + 35*b^4*Tan[c + d*x]^4))/(315*b^2*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.41 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.08, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.826, Rules used = {3042, 4049, 27, 3042, 4113, 27, 3042, 4011, 3042, 4011, 3042, 4011, 3042, 4022, 3042, 4020, 25, 73, 221}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x)^3 (a+b \tan (c+d x))^{5/2}dx\)

\(\Big \downarrow \) 4049

\(\displaystyle \frac {2 \int -\frac {1}{2} (a+b \tan (c+d x))^{5/2} \left (2 a \tan ^2(c+d x)+9 b \tan (c+d x)+2 a\right )dx}{9 b}+\frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{5/2} \left (2 a \tan ^2(c+d x)+9 b \tan (c+d x)+2 a\right )dx}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int (a+b \tan (c+d x))^{5/2} \left (2 a \tan (c+d x)^2+9 b \tan (c+d x)+2 a\right )dx}{9 b}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\int 9 b \tan (c+d x) (a+b \tan (c+d x))^{5/2}dx+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \int \tan (c+d x) (a+b \tan (c+d x))^{5/2}dx+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \int \tan (c+d x) (a+b \tan (c+d x))^{5/2}dx+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \left (\int (a \tan (c+d x)-b) (a+b \tan (c+d x))^{3/2}dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}\right )+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \left (\int (a \tan (c+d x)-b) (a+b \tan (c+d x))^{3/2}dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}\right )+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \left (\int \sqrt {a+b \tan (c+d x)} \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \left (\int \sqrt {a+b \tan (c+d x)} \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 4011

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \left (\int \frac {a \left (a^2-3 b^2\right ) \tan (c+d x)-b \left (3 a^2-b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {9 b \left (\int \frac {a \left (a^2-3 b^2\right ) \tan (c+d x)-b \left (3 a^2-b^2\right )}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )+\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}}{9 b}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}+9 b \left (-\frac {1}{2} (-b+i a)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )}{9 b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}+9 b \left (-\frac {1}{2} (-b+i a)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} (b+i a)^3 \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )}{9 b}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}+9 b \left (\frac {i (-b+i a)^3 \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}+\frac {i (b+i a)^3 \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )}{9 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}+9 b \left (-\frac {i (-b+i a)^3 \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}-\frac {i (b+i a)^3 \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )}{9 b}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}+9 b \left (-\frac {(-b+i a)^3 \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {(b+i a)^3 \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )}{9 b}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 \tan (c+d x) (a+b \tan (c+d x))^{7/2}}{9 b d}-\frac {\frac {4 a (a+b \tan (c+d x))^{7/2}}{7 b d}+9 b \left (\frac {2 \left (a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{d}-\frac {(-b+i a)^3 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}+\frac {(b+i a)^3 \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}+\frac {2 (a+b \tan (c+d x))^{5/2}}{5 d}+\frac {2 a (a+b \tan (c+d x))^{3/2}}{3 d}\right )}{9 b}\)

Input:

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

Output:

(2*Tan[c + d*x]*(a + b*Tan[c + d*x])^(7/2))/(9*b*d) - ((4*a*(a + b*Tan[c + 
 d*x])^(7/2))/(7*b*d) + 9*b*(((I*a + b)^3*ArcTan[Tan[c + d*x]/Sqrt[a - I*b 
]])/(Sqrt[a - I*b]*d) - ((I*a - b)^3*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/( 
Sqrt[a + I*b]*d) + (2*(a^2 - b^2)*Sqrt[a + b*Tan[c + d*x]])/d + (2*a*(a + 
b*Tan[c + d*x])^(3/2))/(3*d) + (2*(a + b*Tan[c + d*x])^(5/2))/(5*d)))/(9*b 
)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[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] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 221

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4011

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

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[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] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1239\) vs. \(2(177)=354\).

Time = 0.24 (sec) , antiderivative size = 1240, normalized size of antiderivative = 5.88

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1240\)
default	\(\text {Expression too large to display}\)	\(1240\)

Input:

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

Output:

2/9/d/b^2*(a+b*tan(d*x+c))^(9/2)-2/7*a*(a+b*tan(d*x+c))^(7/2)/b^2/d-2/5*(a 
+b*tan(d*x+c))^(5/2)/d-2/3*a*(a+b*tan(d*x+c))^(3/2)/d-2/d*a^2*(a+b*tan(d*x 
+c))^(1/2)+2*b^2*(a+b*tan(d*x+c))^(1/2)/d-1/2/d*ln(b*tan(d*x+c)+a+(a+b*tan 
(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(a^2+b^2)^(1 
/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+3/4/d*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c 
))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2) 
+2*a)^(1/2)*a^2-1/4/d*b^2*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2 
+b^2)^(1/2)+2*a)^(1/2)+(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/d/ 
(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2 
)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)*a^2-1/d 
*b^2/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^ 
2+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*(a^2+b^2)^(1/2)-1/ 
d/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b 
^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a^3+3/d*b^2/(2*(a^2+b 
^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2+b^2)^(1/2)+2 
*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*a+1/2/d*ln((a+b*tan(d*x+c))^(1/2 
)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(a^2+b^2)^ 
(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-3/4/d*ln((a+b*tan(d*x+c))^(1/2)*(2*( 
a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan(d*x+c)-a-(a^2+b^2)^(1/2))*(2*(a^2+b^2)^(1/ 
2)+2*a)^(1/2)*a^2+1/4/d*b^2*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1193 vs. \(2 (173) = 346\).

Time = 0.10 (sec) , antiderivative size = 1193, normalized size of antiderivative = 5.65 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/630*(315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 
 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 
14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-(2 
5*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 
 15*a^4*b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2* 
sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d 
^2)) - 315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 
- 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 1 
4*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) - (2*a*d^3*sqrt(-(25 
*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 
15*a^4*b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*s 
qrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^ 
2)) - 315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 
 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14 
*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-(25* 
a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4) + (5*a^6 - 1 
5*a^4*b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sq 
rt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2 
)) + 315*b^2*d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 
100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - ...
 

Sympy [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\int \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \tan ^{3}{\left (c + d x \right )}\, dx \] Input:

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

Output:

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

Maxima [F(-1)]

Timed out. \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[0,29,11]%%%}+%%%{12,[0,27,11]%%%}+%%%{66,[0,25,11]% 
%%}+%%%{2
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 47.72 (sec) , antiderivative size = 2384, normalized size of antiderivative = 11.30 \[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

(a + b*tan(c + d*x))^(3/2)*((2*a*((2*a^2)/(b^2*d) - (2*(a^2 + b^2))/(b^2*d 
)))/3 - (2*a^3)/(3*b^2*d) + (2*a*(a^2 + b^2))/(3*b^2*d)) - atan(((((8*(4*b 
^6*d^2 - 4*a^4*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b 
^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*( 
(5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1 
/2) - (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6* 
b^2))/d^2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2) 
/(4*d^2))^(1/2)*1i - (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + 
 b*tan(c + d*x))^(1/2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 
 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3 
*10i - 10*a^3*b^2)/(4*d^2))^(1/2) + (16*(a + b*tan(c + d*x))^(1/2)*(b^8 - 
15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5* 
1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2)*1i)/((((8*(4*b^6*d^2 - 4*a^4 
*b^2*d^2))/d^3 - 64*a*b^2*(a + b*tan(c + d*x))^(1/2)*((5*a*b^4 + a^4*b*5i 
+ a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2))*((5*a*b^4 + a^4 
*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2) - (16*(a + 
 b*tan(c + d*x))^(1/2)*(b^8 - 15*a^2*b^6 + 15*a^4*b^4 - a^6*b^2))/d^2)*((5 
*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(4*d^2))^(1/2 
) + (((8*(4*b^6*d^2 - 4*a^4*b^2*d^2))/d^3 + 64*a*b^2*(a + b*tan(c + d*x))^ 
(1/2)*((5*a*b^4 + a^4*b*5i + a^5 + b^5*1i - a^2*b^3*10i - 10*a^3*b^2)/(...
 

Reduce [F]

\[ \int \tan ^3(c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{5}d x \right ) b^{2}+2 \left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{4}d x \right ) a b +\left (\int \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right )^{3}d x \right ) a^{2} \] Input:

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

Output:

int(sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**5,x)*b**2 + 2*int(sqrt(tan(c + 
d*x)*b + a)*tan(c + d*x)**4,x)*a*b + int(sqrt(tan(c + d*x)*b + a)*tan(c + 
d*x)**3,x)*a**2