3.5.60 \(\int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx\) [460]

3.5.60.1 Optimal result

Integrand size = 35, antiderivative size = 175 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\frac {(i A-B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}+\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]

output

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

3.5.60.2 Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.15 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\frac {\sqrt [4]{-1} (-i a+b) (A-i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} (a-i b) (-i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 b (A b-a B) \sqrt {\tan (c+d x)}}{a \sqrt {a+b \tan (c+d x)}}}{\left (a^2+b^2\right ) d} \]

input

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

output

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

3.5.60.3 Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4092, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4092

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

input

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

output

((a*(a - I*b)*(A + I*B)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + 
 b*Tan[c + d*x]]])/(Sqrt[I*a - b]*d) + (a*(a + I*b)*(A - I*B)*ArcTanh[(Sqr 
t[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/(Sqrt[I*a + b]*d 
))/(a*(a^2 + b^2)) + (2*b*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(a*(a^2 + b^2)*d 
*Sqrt[a + b*Tan[c + d*x]])
 

3.5.60.3.1 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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[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] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

rule 4092

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*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1) 
/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^ 
2 + b^2))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b* 
B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 
)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n 
+ 2)*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] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] 
 || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4098

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[A^2/f   Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e + f* 
x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && 
 NeQ[a^2 + b^2, 0] && EqQ[A^2 + B^2, 0]
 

rule 4099

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[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

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

Leaf count of result is larger than twice the leaf count of optimal. \(4926\) vs. \(2(149)=298\).

Time = 29.70 (sec) , antiderivative size = 4927, normalized size of antiderivative = 28.15

method	result	size
default	\(\text {Expression too large to display}\)	\(4927\)
parts	\(\text {Expression too large to display}\)	\(1559557\)

input

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

output

1/4/d*2^(1/2)/a/(b+(a^2+b^2)^(1/2))^(1/2)/(a^2+b^2)^(3/2)*(A*ln(1/(1-cos(d 
*x+c))*(a*(1-cos(d*x+c))^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))-2*( 
(1-cos(d*x+c))*(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-2*b*(csc(d*x+c)-cot(d*x+c) 
)-a)*csc(d*x+c))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*b*(1-cos(d* 
x+c))-a*sin(d*x+c)))*a^2*(-b+(a^2+b^2)^(1/2))^(1/2)*((1-cos(d*x+c))*(a*(1- 
cos(d*x+c))^2*csc(d*x+c)^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*csc(d*x+c))^(1/2 
)*(b+(a^2+b^2)^(1/2))^(1/2)-A*ln(1/(1-cos(d*x+c))*(a*(1-cos(d*x+c))^2*csc( 
d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))-2*((1-cos(d*x+c))*(a*(1-cos(d*x+c) 
)^2*csc(d*x+c)^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*csc(d*x+c))^(1/2)*(-b+(a^2 
+b^2)^(1/2))^(1/2)*sin(d*x+c)-2*b*(1-cos(d*x+c))-a*sin(d*x+c)))*b^2*(-b+(a 
^2+b^2)^(1/2))^(1/2)*((1-cos(d*x+c))*(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-2*b* 
(csc(d*x+c)-cot(d*x+c))-a)*csc(d*x+c))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)-A*( 
a^2+b^2)^(1/2)*ln(1/(1-cos(d*x+c))*(a*(1-cos(d*x+c))^2*csc(d*x+c)+2*(a^2+b 
^2)^(1/2)*(1-cos(d*x+c))-2*((1-cos(d*x+c))*(a*(1-cos(d*x+c))^2*csc(d*x+c)^ 
2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*csc(d*x+c))^(1/2)*(-b+(a^2+b^2)^(1/2))^(1 
/2)*sin(d*x+c)-2*b*(1-cos(d*x+c))-a*sin(d*x+c)))*((1-cos(d*x+c))*(a*(1-cos 
(d*x+c))^2*csc(d*x+c)^2-2*b*(csc(d*x+c)-cot(d*x+c))-a)*csc(d*x+c))^(1/2)*( 
-b+(a^2+b^2)^(1/2))^(1/2)*(b+(a^2+b^2)^(1/2))^(1/2)*b-A*ln(1/(1-cos(d*x+c) 
)*(a*(1-cos(d*x+c))^2*csc(d*x+c)+2*(a^2+b^2)^(1/2)*(1-cos(d*x+c))+2*((1-co 
s(d*x+c))*(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-2*b*(csc(d*x+c)-cot(d*x+c))-...
 

3.5.60.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 18563 vs. \(2 (143) = 286\).

Time = 6.95 (sec) , antiderivative size = 18563, normalized size of antiderivative = 106.07 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

input

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

output

Too large to include
 

3.5.60.6 Sympy [F]

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

input

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

output

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

3.5.60.7 Maxima [F]

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

input

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

output

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

3.5.60.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.5.60.9 Mupad [F(-1)]

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

input

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

output

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