3.7.25 \(\int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx\) [625]

3.7.25.1 Optimal result

Integrand size = 25, antiderivative size = 231 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=-\frac {(i a-b)^{5/2} \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\sqrt {b} \left (15 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 d}+\frac {(i a+b)^{5/2} \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 d}+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d} \]

output

-(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+(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+1/4*(15*a^2-8*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x 
+c))^(1/2))*b^(1/2)/d+9/4*a*b*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(1/2)/d+1/ 
2*b^2*(a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)/d
 

3.7.25.2 Mathematica [A] (verified)

Time = 2.17 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.14 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\frac {4 \sqrt [4]{-1} (-a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 \sqrt [4]{-1} (a+i b)^{5/2} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}+\frac {\sqrt {a} \sqrt {b} \left (15 a^2-8 b^2\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {a+b \tan (c+d x)}}}{4 d} \]

input

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

output

(4*(-1)^(1/4)*(-a + I*b)^(5/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[ 
c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 4*(-1)^(1/4)*(a + I*b)^(5/2)*ArcTan 
[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] + 
 9*a*b*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 2*b^2*Tan[c + d*x]^(3 
/2)*Sqrt[a + b*Tan[c + d*x]] + (Sqrt[a]*Sqrt[b]*(15*a^2 - 8*b^2)*ArcSinh[( 
Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a])/Sqrt[a 
+ b*Tan[c + d*x]])/(4*d)
 

3.7.25.3 Rubi [A] (verified)

Time = 1.38 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 4049, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \frac {1}{4} \left (\frac {\int -\frac {9 a^2 b^2-\left (15 a^2-8 b^2\right ) \tan ^2(c+d x) b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{b}+\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{4} \left (\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {9 a^2 b^2-\left (15 a^2-8 b^2\right ) \tan ^2(c+d x) b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}\right )+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{4} \left (\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {9 a^2 b^2-\left (15 a^2-8 b^2\right ) \tan (c+d x)^2 b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx}{2 b}\right )+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 4138

\(\displaystyle \frac {1}{4} \left (\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {9 a^2 b^2-\left (15 a^2-8 b^2\right ) \tan ^2(c+d x) b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 b d}\right )+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 2035

\(\displaystyle \frac {1}{4} \left (\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \frac {9 a^2 b^2-\left (15 a^2-8 b^2\right ) \tan ^2(c+d x) b^2-8 a \left (a^2-3 b^2\right ) \tan (c+d x) b}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{b d}\right )+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 2257

\(\displaystyle \frac {1}{4} \left (\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {\int \left (\frac {8 \left (-b^4+3 a^2 b^2-a \left (a^2-3 b^2\right ) \tan (c+d x) b\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}-\frac {b^2 \left (15 a^2-8 b^2\right )}{\sqrt {a+b \tan (c+d x)}}\right )d\sqrt {\tan (c+d x)}}{b d}\right )+\frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {b^2 \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}}{2 d}+\frac {1}{4} \left (\frac {9 a b \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}-\frac {-b^{3/2} \left (15 a^2-8 b^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+4 b (-b+i a)^{5/2} \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-4 b (b+i a)^{5/2} \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{b d}\right )\)

input

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

output

(b^2*Tan[c + d*x]^(3/2)*Sqrt[a + b*Tan[c + d*x]])/(2*d) + (-((4*(I*a - b)^ 
(5/2)*b*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]] 
] - b^(3/2)*(15*a^2 - 8*b^2)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
 b*Tan[c + d*x]]] - 4*b*(I*a + b)^(5/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c 
+ d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(b*d)) + (9*a*b*Sqrt[Tan[c + d*x]]*Sqr 
t[a + b*Tan[c + d*x]])/d)/4
 

3.7.25.3.1 Defintions of rubi rules used

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

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 4130

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] + Simp[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] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4138

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[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]
 

3.7.25.4 Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 0.65 (sec) , antiderivative size = 1344637, normalized size of antiderivative = 5820.94

\[\text {output too large to display}\]

input

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

output

result too large to display
 

3.7.25.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4656 vs. \(2 (185) = 370\).

Time = 1.50 (sec) , antiderivative size = 9318, normalized size of antiderivative = 40.34 \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \]

input

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
 

output

Too large to include
 

3.7.25.6 Sympy [F]

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

input

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

output

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

3.7.25.7 Maxima [F]

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

input

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
 

output

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

3.7.25.8 Giac [F(-1)]

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

input

integrate(tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
 

output

Timed out
 

3.7.25.9 Mupad [F(-1)]

Timed out. \[ \int \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2} \, dx=\int \sqrt {\mathrm {tan}\left (c+d\,x\right )}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2} \,d x \]

input

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

output

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