3.5.27 \(\int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [427]

3.5.27.1 Optimal result

Integrand size = 35, antiderivative size = 264 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {\sqrt {i a-b} (i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (4 a A b-a^2 B-8 b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{4 b^{3/2} d}+\frac {\sqrt {i a+b} (i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {(4 A b-a B) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{4 b d}+\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 b d} \]

output

1/4*(4*A*a*b-B*a^2-8*B*b^2)*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)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*ta 
n(d*x+c))^(1/2))*(I*a-b)^(1/2)/d+(I*A+B)*arctanh((I*a+b)^(1/2)*tan(d*x+c)^ 
(1/2)/(a+b*tan(d*x+c))^(1/2))*(I*a+b)^(1/2)/d+1/4*(4*A*b-B*a)*tan(d*x+c)^( 
1/2)*(a+b*tan(d*x+c))^(1/2)/b/d+1/2*B*tan(d*x+c)^(1/2)*(a+b*tan(d*x+c))^(3 
/2)/b/d
 

3.5.27.2 Mathematica [A] (verified)

Time = 3.73 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.15 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {-\sqrt {a} \left (-4 a A b+a^2 B+8 b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}+\sqrt {b} \left (-4 \sqrt [4]{-1} \sqrt {-a+i b} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+4 (-1)^{3/4} \sqrt {a+i b} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {a+b \tan (c+d x)}+\sqrt {\tan (c+d x)} (a+b \tan (c+d x)) (4 A b+a B+2 b B \tan (c+d x))\right )}{4 b^{3/2} d \sqrt {a+b \tan (c+d x)}} \]

input

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

output

(-(Sqrt[a]*(-4*a*A*b + a^2*B + 8*b^2*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x] 
])/Sqrt[a]]*Sqrt[1 + (b*Tan[c + d*x])/a]) + Sqrt[b]*(-4*(-1)^(1/4)*Sqrt[-a 
 + I*b]*b*(I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/ 
Sqrt[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + 4*(-1)^(3/4)*Sqrt[a + 
 I*b]*b*(A + I*B)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqr 
t[a + b*Tan[c + d*x]]]*Sqrt[a + b*Tan[c + d*x]] + Sqrt[Tan[c + d*x]]*(a + 
b*Tan[c + d*x])*(4*A*b + a*B + 2*b*B*Tan[c + d*x])))/(4*b^(3/2)*d*Sqrt[a + 
 b*Tan[c + d*x]])
 

3.5.27.3 Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.314, Rules used = {3042, 4090, 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.

\(\displaystyle \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4138

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

\(\Big \downarrow \) 2035

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

\(\Big \downarrow \) 2257

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

\(\Big \downarrow \) 2009

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

input

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

output

(B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))/(2*b*d) - ((-4*Sqrt[I*a 
- b]*b*(I*A - B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[ 
c + d*x]]] - ((4*a*A*b - a^2*B - 8*b^2*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d* 
x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[b] - 4*b*Sqrt[I*a + b]*(I*A + B)*ArcT 
anh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((4* 
A*b - a*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d)/(4*b)
 

3.5.27.3.1 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 4090

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

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

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

Time = 1.76 (sec) , antiderivative size = 2183234, normalized size of antiderivative = 8269.83

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

input

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

output

result too large to display
 

3.5.27.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 8056 vs. \(2 (212) = 424\).

Time = 3.48 (sec) , antiderivative size = 16118, normalized size of antiderivative = 61.05 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

integrate((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)*(A+B*tan(d*x+c)),x, algo 
rithm="fricas")
 

output

Too large to include
 

3.5.27.6 Sympy [F]

\[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {a + b \tan {\left (c + d x \right )}} \tan ^{\frac {3}{2}}{\left (c + d x \right )}\, dx \]

input

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

output

Integral((A + B*tan(c + d*x))*sqrt(a + b*tan(c + d*x))*tan(c + d*x)**(3/2) 
, x)
 

3.5.27.7 Maxima [F]

\[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {b \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {3}{2}} \,d x } \]

input

integrate((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)*(A+B*tan(d*x+c)),x, algo 
rithm="maxima")
 

output

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

3.5.27.8 Giac [F(-1)]

Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

input

integrate((a+b*tan(d*x+c))^(1/2)*tan(d*x+c)^(3/2)*(A+B*tan(d*x+c)),x, algo 
rithm="giac")
 

output

Timed out
 

3.5.27.9 Mupad [F(-1)]

Timed out. \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]

input

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

output

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