\(\int \frac {\tan ^3(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx\) [530]

Optimal result

Integrand size = 23, antiderivative size = 140 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}-\frac {4 a \sqrt {a+b \tan (c+d x)}}{3 b^2 d}+\frac {2 \tan (c+d x) \sqrt {a+b \tan (c+d x)}}{3 b d} \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.14 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {3 \sqrt {a-i b} (a+i b) b^2 \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )-(a-i b) \left (-3 \sqrt {a+i b} b^2 \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 (a+i b) (2 a-b \tan (c+d x)) \sqrt {a+b \tan (c+d x)}\right )}{3 b^2 \left (a^2+b^2\right ) d} \] Input:

Integrate[Tan[c + d*x]^3/Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

(3*Sqrt[a - I*b]*(a + I*b)*b^2*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I 
*b]] - (a - I*b)*(-3*Sqrt[a + I*b]*b^2*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sq 
rt[a + I*b]] + 2*(a + I*b)*(2*a - b*Tan[c + d*x])*Sqrt[a + b*Tan[c + d*x]] 
))/(3*b^2*(a^2 + b^2)*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.73 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 4049, 27, 3042, 4113, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4113

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

Int[Tan[c + d*x]^3/Sqrt[a + b*Tan[c + d*x]],x]
 

Output:

(2*Tan[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(3*b*d) - (3*b*(((-I)*ArcTan[Tan 
[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) + (I*ArcTan[Tan[c + d*x]/Sqrt[ 
a + I*b]])/(Sqrt[a + I*b]*d)) + (4*a*Sqrt[a + b*Tan[c + d*x]])/(b*d))/(3*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 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. \(374\) vs. \(2(116)=232\).

Time = 0.23 (sec) , antiderivative size = 375, normalized size of antiderivative = 2.68

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}+\frac {-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{2}}\)	\(375\)
default	\(\frac {\frac {2 \left (a +b \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 a \sqrt {a +b \tan \left (d x +c \right )}+2 b^{2} \left (\frac {\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a +\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}+\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}+\frac {-\frac {\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}\, \ln \left (b \tan \left (d x +c \right )+a -\sqrt {a +b \tan \left (d x +c \right )}\, \sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}+\sqrt {a^{2}+b^{2}}\right )}{2}+\frac {2 \left (a -\sqrt {a^{2}+b^{2}}\right ) \arctan \left (\frac {2 \sqrt {a +b \tan \left (d x +c \right )}-\sqrt {2 \sqrt {a^{2}+b^{2}}+2 a}}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}\right )}{\sqrt {2 \sqrt {a^{2}+b^{2}}-2 a}}}{4 \sqrt {a^{2}+b^{2}}}\right )}{d \,b^{2}}\)	\(375\)

Input:

int(tan(d*x+c)^3/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 743 vs. \(2 (112) = 224\).

Time = 0.10 (sec) , antiderivative size = 743, normalized size of antiderivative = 5.31 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/6*(3*b^2*d*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4 
)) + a)/((a^2 + b^2)*d^2))*log(((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^ 
2 + b^4)*d^4)) - a*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + 
b^4)*d^4)) + a)/((a^2 + b^2)*d^2)) + sqrt(b*tan(d*x + c) + a)) - 3*b^2*d*s 
qrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a)/((a^2 + 
 b^2)*d^2))*log(-((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) 
 - a*d)*sqrt(((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a 
)/((a^2 + b^2)*d^2)) + sqrt(b*tan(d*x + c) + a)) - 3*b^2*d*sqrt(-((a^2 + b 
^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))*l 
og(((a^2 + b^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a*d)*sqrt(- 
((a^2 + b^2)*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2 
)*d^2)) + sqrt(b*tan(d*x + c) + a)) + 3*b^2*d*sqrt(-((a^2 + b^2)*d^2*sqrt( 
-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2))*log(-((a^2 + b 
^2)*d^3*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) + a*d)*sqrt(-((a^2 + b^2) 
*d^2*sqrt(-b^2/((a^4 + 2*a^2*b^2 + b^4)*d^4)) - a)/((a^2 + b^2)*d^2)) + sq 
rt(b*tan(d*x + c) + a)) - 4*sqrt(b*tan(d*x + c) + a)*(b*tan(d*x + c) - 2*a 
))/(b^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate(tan(d*x+c)^3/(a+b*tan(d*x+c))^(1/2),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,[0,14,5]%%%}+%%%{6,[0,12,5]%%%}+%%%{15,[0,10,5]%%%}+ 
%%%{20,[0
 

Mupad [B] (verification not implemented)

Time = 2.42 (sec) , antiderivative size = 827, normalized size of antiderivative = 5.91 \[ \int \frac {\tan ^3(c+d x)}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {2\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{3\,b^2\,d}-\frac {2\,a\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{b^2\,d}+\mathrm {atan}\left (-\frac {b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,32{}\mathrm {i}}{\frac {16\,b^2}{d}-\frac {64\,a^2\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^3\,d^2\,64{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {a^2\,b^2\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,128{}\mathrm {i}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}+\frac {128\,a\,b^3\,\sqrt {\frac {a}{4\,a^2\,d^2+4\,b^2\,d^2}-\frac {b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}}{\frac {64\,b^4}{d}+\frac {64\,a^2\,b^2}{d}-\frac {256\,a^2\,b^4\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a^3\,b^3\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}-\frac {256\,a^4\,b^2\,d^2}{4\,a^2\,d^3+4\,b^2\,d^3}+\frac {a\,b^5\,d^2\,256{}\mathrm {i}}{4\,a^2\,d^3+4\,b^2\,d^3}}\right )\,\sqrt {\frac {a-b\,1{}\mathrm {i}}{4\,a^2\,d^2+4\,b^2\,d^2}}\,2{}\mathrm {i}-\mathrm {atan}\left (\frac {b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,16{}\mathrm {i}}{\frac {16\,b^2}{d}-\frac {16\,a\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}+\frac {a\,b^2\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,\sqrt {a+b\,\mathrm {tan}\left (c+d\,x\right )}\,16{}\mathrm {i}}{\frac {b^3\,16{}\mathrm {i}}{d}-\frac {16\,a\,b^2}{d}-\frac {a\,b^3\,d^2\,16{}\mathrm {i}}{a\,d^3-b\,d^3\,1{}\mathrm {i}}+\frac {16\,a^2\,b^2\,d^2}{a\,d^3-b\,d^3\,1{}\mathrm {i}}}\right )\,\sqrt {\frac {1}{a\,d^2-b\,d^2\,1{}\mathrm {i}}}\,1{}\mathrm {i} \] Input:

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

Output:

atan((a^2*b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2*d^2)) 
^(1/2)*(a + b*tan(c + d*x))^(1/2)*128i)/((64*b^4)/d + (64*a^2*b^2)/d - (25 
6*a^2*b^4*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4 
*b^2*d^3) - (256*a^4*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/( 
4*a^2*d^3 + 4*b^2*d^3)) - (b^2*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2* 
d^2 + 4*b^2*d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2)*32i)/((16*b^2)/d - (64* 
a^2*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^3*d^2*64i)/(4*a^2*d^3 + 4*b^2* 
d^3)) + (128*a*b^3*(a/(4*a^2*d^2 + 4*b^2*d^2) - (b*1i)/(4*a^2*d^2 + 4*b^2* 
d^2))^(1/2)*(a + b*tan(c + d*x))^(1/2))/((64*b^4)/d + (64*a^2*b^2)/d - (25 
6*a^2*b^4*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a^3*b^3*d^2*256i)/(4*a^2*d^3 + 4 
*b^2*d^3) - (256*a^4*b^2*d^2)/(4*a^2*d^3 + 4*b^2*d^3) + (a*b^5*d^2*256i)/( 
4*a^2*d^3 + 4*b^2*d^3)))*((a - b*1i)/(4*a^2*d^2 + 4*b^2*d^2))^(1/2)*2i - a 
tan((b^2*(1/(a*d^2 - b*d^2*1i))^(1/2)*(a + b*tan(c + d*x))^(1/2)*16i)/((16 
*b^2)/d - (16*a*b^2*d^2)/(a*d^3 - b*d^3*1i)) + (a*b^2*(1/(a*d^2 - b*d^2*1i 
))^(1/2)*(a + b*tan(c + d*x))^(1/2)*16i)/((b^3*16i)/d - (16*a*b^2)/d - (a* 
b^3*d^2*16i)/(a*d^3 - b*d^3*1i) + (16*a^2*b^2*d^2)/(a*d^3 - b*d^3*1i)))*(1 
/(a*d^2 - b*d^2*1i))^(1/2)*1i + (2*(a + b*tan(c + d*x))^(3/2))/(3*b^2*d) - 
 (2*a*(a + b*tan(c + d*x))^(1/2))/(b^2*d)
 

Reduce [F]

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

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

Output:

int((sqrt(tan(c + d*x)*b + a)*tan(c + d*x)**3)/(tan(c + d*x)*b + a),x)