\(\int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx\) [410]

Optimal result

Integrand size = 29, antiderivative size = 460 \[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\frac {a \left (a^2-b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac {a \left (a^2+b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac {b^4 c^2 \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}-\frac {b \left (a^2-b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)}-\frac {b \left (a^2+b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)} \] Output:

1/2*a*(a^2-b^2*(-c^2)^(1/2))*hypergeom([1, 1+m],[2+m],-c*tan(f*x+e)/(-c^2) 
^(1/2))*tan(f*x+e)*(d*tan(f*x+e))^m/(b^4*c^2+a^4)/f/(1+m)+1/2*a*(a^2+b^2*( 
-c^2)^(1/2))*hypergeom([1, 1+m],[2+m],c*tan(f*x+e)/(-c^2)^(1/2))*tan(f*x+e 
)*(d*tan(f*x+e))^m/(b^4*c^2+a^4)/f/(1+m)+b^4*c^2*hypergeom([1, 2+2*m],[3+2 
*m],-b*(c*tan(f*x+e))^(1/2)/a)*tan(f*x+e)*(d*tan(f*x+e))^m/a/(b^4*c^2+a^4) 
/f/(1+m)-b*(a^2-b^2*(-c^2)^(1/2))*hypergeom([1, 3/2+m],[5/2+m],-c*tan(f*x+ 
e)/(-c^2)^(1/2))*(c*tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/(b^4*c^2+a^4)/f/( 
3+2*m)-b*(a^2+b^2*(-c^2)^(1/2))*hypergeom([1, 3/2+m],[5/2+m],c*tan(f*x+e)/ 
(-c^2)^(1/2))*(c*tan(f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/(b^4*c^2+a^4)/f/(3+2 
*m)
 

Mathematica [A] (verified)

Time = 3.05 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.57 \[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^m \left (\frac {a^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(e+f x)\right )}{1+m}+b \left (\frac {b^3 c^2 \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right )}{a+a m}+\frac {a b c \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{2+m}+2 \sqrt {c \tan (e+f x)} \left (-\frac {a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right )}{3+2 m}-\frac {b^2 c \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (5+2 m),\frac {1}{4} (9+2 m),-\tan ^2(e+f x)\right ) \tan (e+f x)}{5+2 m}\right )\right )\right )}{\left (a^4+b^4 c^2\right ) f} \] Input:

Integrate[(d*Tan[e + f*x])^m/(a + b*Sqrt[c*Tan[e + f*x]]),x]
 

Output:

(Tan[e + f*x]*(d*Tan[e + f*x])^m*((a^3*Hypergeometric2F1[1, (1 + m)/2, (3 
+ m)/2, -Tan[e + f*x]^2])/(1 + m) + b*((b^3*c^2*Hypergeometric2F1[1, 2*(1 
+ m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)])/(a + a*m) + (a*b*c*Hypergeo 
metric2F1[1, (2 + m)/2, (4 + m)/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(2 + m) 
+ 2*Sqrt[c*Tan[e + f*x]]*(-((a^2*Hypergeometric2F1[1, (3 + 2*m)/4, (7 + 2* 
m)/4, -Tan[e + f*x]^2])/(3 + 2*m)) - (b^2*c*Hypergeometric2F1[1, (5 + 2*m) 
/4, (9 + 2*m)/4, -Tan[e + f*x]^2]*Tan[e + f*x])/(5 + 2*m)))))/((a^4 + b^4* 
c^2)*f)
 

Rubi [A] (warning: unable to verify)

Time = 1.40 (sec) , antiderivative size = 482, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 4153, 7267, 30, 7276, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

\(\displaystyle \frac {c \int \frac {(d \tan (e+f x))^m}{\left (\tan ^2(e+f x) c^2+c^2\right ) \left (a+b \sqrt {c \tan (e+f x)}\right )}d(c \tan (e+f x))}{f}\)

\(\Big \downarrow \) 7267

\(\displaystyle \frac {2 c \int \frac {\sqrt {c \tan (e+f x)} \left (c d \tan ^2(e+f x)\right )^m}{(a+b c \tan (e+f x)) \left (c^4 \tan ^4(e+f x)+c^2\right )}d\sqrt {c \tan (e+f x)}}{f}\)

\(\Big \downarrow \) 30

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

\(\Big \downarrow \) 7276

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \left (\frac {b^4 (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right )}{2 a (m+1) \left (a^4+b^4 c^2\right )}+\frac {a \left (a^2-b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1) \left (a^4+b^4 c^2\right )}+\frac {a \left (a^2+b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1) \left (a^4+b^4 c^2\right )}-\frac {b \left (a^2-b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{2 c^2 (2 m+3) \left (a^4+b^4 c^2\right )}-\frac {b \left (a^2+b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{2 c^2 (2 m+3) \left (a^4+b^4 c^2\right )}\right )}{f}\)

Input:

Int[(d*Tan[e + f*x])^m/(a + b*Sqrt[c*Tan[e + f*x]]),x]
 

Output:

(2*c*(c*d*Tan[e + f*x]^2)^m*((a*(a^2 - b^2*Sqrt[-c^2])*Hypergeometric2F1[1 
, 1 + m, 2 + m, -((c^2*Tan[e + f*x]^2)/Sqrt[-c^2])]*(c*Tan[e + f*x])^(1 + 
m))/(4*c^2*(a^4 + b^4*c^2)*(1 + m)) + (a*(a^2 + b^2*Sqrt[-c^2])*Hypergeome 
tric2F1[1, 1 + m, 2 + m, (c^2*Tan[e + f*x]^2)/Sqrt[-c^2]]*(c*Tan[e + f*x]) 
^(1 + m))/(4*c^2*(a^4 + b^4*c^2)*(1 + m)) + (b^4*Hypergeometric2F1[1, 2*(1 
 + m), 3 + 2*m, -((b*Sqrt[c*Tan[e + f*x]])/a)]*(c*Tan[e + f*x])^(1 + m))/( 
2*a*(a^4 + b^4*c^2)*(1 + m)) - (b*(a^2 - b^2*Sqrt[-c^2])*Hypergeometric2F1 
[1, (3 + 2*m)/2, (5 + 2*m)/2, -((c^2*Tan[e + f*x]^2)/Sqrt[-c^2])]*(c*Tan[e 
 + f*x])^((3 + 2*m)/2))/(2*c^2*(a^4 + b^4*c^2)*(3 + 2*m)) - (b*(a^2 + b^2* 
Sqrt[-c^2])*Hypergeometric2F1[1, (3 + 2*m)/2, (5 + 2*m)/2, (c^2*Tan[e + f* 
x]^2)/Sqrt[-c^2]]*(c*Tan[e + f*x])^((3 + 2*m)/2))/(2*c^2*(a^4 + b^4*c^2)*( 
3 + 2*m))))/(f*(c*Tan[e + f*x])^m)
 

Defintions of rubi rules used

rule 30

Int[(u_.)*((a_.)*(x_))^(m_.)*((b_.)*(x_)^(i_.))^(p_), x_Symbol] :> Simp[b^I 
ntPart[p]*((b*x^i)^FracPart[p]/(a^(i*IntPart[p])*(a*x)^(i*FracPart[p]))) 
Int[u*(a*x)^(m + i*p), x], x] /; FreeQ[{a, b, i, m, p}, x] && IntegerQ[i] & 
&  !IntegerQ[p]
 

rule 2009

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

rule 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [F]

\[\int \frac {\left (d \tan \left (f x +e \right )\right )^{m}}{a +b \sqrt {c \tan \left (f x +e \right )}}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((sqrt(c*tan(f*x + e))*(d*tan(f*x + e))^m*b - (d*tan(f*x + e))^m*a 
)/(b^2*c*tan(f*x + e) - a^2), x)
 

Sympy [F]

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

integrate((d*tan(f*x+e))**m/(a+b*(c*tan(f*x+e))**(1/2)),x)
 

Output:

Integral((d*tan(e + f*x))**m/(a + b*sqrt(c*tan(e + f*x))), x)
 

Maxima [F]

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

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

Output:

integrate((d*tan(f*x + e))^m/(sqrt(c*tan(f*x + e))*b + a), x)
 

Giac [F]

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

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

Output:

integrate((d*tan(f*x + e))^m/(sqrt(c*tan(f*x + e))*b + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{a+b\,\sqrt {c\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(d**m*(2*sqrt(c)*tan(e + f*x)**((2*m + 1)/2)*b*m - 2*tan(e + f*x)**m*a*m - 
 tan(e + f*x)**m*a + 2*sqrt(c)*int(tan(e + f*x)**((2*m + 1)/2)/(tan(e + f* 
x)**2*b**2*c - tan(e + f*x)*a**2),x)*a**2*b*f*m**2 + sqrt(c)*int(tan(e + f 
*x)**((2*m + 1)/2)/(tan(e + f*x)**2*b**2*c - tan(e + f*x)*a**2),x)*a**2*b* 
f*m - 2*sqrt(c)*int((tan(e + f*x)**((2*m + 1)/2)*tan(e + f*x)**2)/(tan(e + 
 f*x)*b**2*c - a**2),x)*b**3*c*f*m**2 - sqrt(c)*int((tan(e + f*x)**((2*m + 
 1)/2)*tan(e + f*x)**2)/(tan(e + f*x)*b**2*c - a**2),x)*b**3*c*f*m + 2*sqr 
t(c)*int((tan(e + f*x)**((2*m + 1)/2)*tan(e + f*x))/(tan(e + f*x)*b**2*c - 
 a**2),x)*a**2*b*f*m**2 + sqrt(c)*int((tan(e + f*x)**((2*m + 1)/2)*tan(e + 
 f*x))/(tan(e + f*x)*b**2*c - a**2),x)*a**2*b*f*m - 2*int(tan(e + f*x)**m/ 
(tan(e + f*x)**2*b**2*c - tan(e + f*x)*a**2),x)*a**3*f*m**2 - int(tan(e + 
f*x)**m/(tan(e + f*x)**2*b**2*c - tan(e + f*x)*a**2),x)*a**3*f*m + 2*int(( 
tan(e + f*x)**m*tan(e + f*x)**2)/(tan(e + f*x)*b**2*c - a**2),x)*a*b**2*c* 
f*m**2 + int((tan(e + f*x)**m*tan(e + f*x)**2)/(tan(e + f*x)*b**2*c - a**2 
),x)*a*b**2*c*f*m - 2*int((tan(e + f*x)**m*tan(e + f*x))/(tan(e + f*x)*b** 
2*c - a**2),x)*a**3*f*m**2 - int((tan(e + f*x)**m*tan(e + f*x))/(tan(e + f 
*x)*b**2*c - a**2),x)*a**3*f*m))/(b**2*c*f*m*(2*m + 1))