Integrand size = 29, antiderivative size = 460 \[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\frac {a \left (a^2-b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac {a \left (a^2+b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 \left (a^4+b^4 c^2\right ) f (1+m)}+\frac {b^4 c^2 \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right ) \tan (e+f x) (d \tan (e+f x))^m}{a \left (a^4+b^4 c^2\right ) f (1+m)}-\frac {b \left (a^2-b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)}-\frac {b \left (a^2+b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (3+2 m),\frac {1}{2} (5+2 m),\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c \left (a^4+b^4 c^2\right ) f (3+2 m)} \]
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)+1/2*a*hypergeom([1, 1+m],[2+m],- c*tan(f*x+e)/(-c^2)^(1/2))*(a^2-b^2*(-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*hypergeom([1, 1+m],[2+m],c*tan(f*x+e)/(-c^ 2)^(1/2))*(a^2+b^2*(-c^2)^(1/2))*tan(f*x+e)*(d*tan(f*x+e))^m/(b^4*c^2+a^4) /f/(1+m)-b*hypergeom([1, 3/2+m],[5/2+m],-c*tan(f*x+e)/(-c^2)^(1/2))*(a^2-b ^2*(-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*hypergeom([1, 3/2+m],[5/2+m],c*tan(f*x+e)/(-c^2)^(1/2))*(a^2+b^2* (-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)
Time = 5.05 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.57 \[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^m \left (\frac {a^3 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(e+f x)\right )}{1+m}+b \left (\frac {b^3 c^2 \operatorname {Hypergeometric2F1}\left (1,2 (1+m),3+2 m,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right )}{a+a m}+\frac {a b c \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{2+m}+2 \sqrt {c \tan (e+f x)} \left (-\frac {a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right )}{3+2 m}-\frac {b^2 c \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (5+2 m),\frac {1}{4} (9+2 m),-\tan ^2(e+f x)\right ) \tan (e+f x)}{5+2 m}\right )\right )\right )}{\left (a^4+b^4 c^2\right ) f} \]
(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)
Time = 1.41 (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 |
| \(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \left (\frac {b^4 (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,2 (m+1),2 m+3,-\frac {b \sqrt {c \tan (e+f x)}}{a}\right )}{2 a (m+1) \left (a^4+b^4 c^2\right )}+\frac {a \left (a^2-b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1) \left (a^4+b^4 c^2\right )}+\frac {a \left (a^2+b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1) \left (a^4+b^4 c^2\right )}-\frac {b \left (a^2-b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{2 c^2 (2 m+3) \left (a^4+b^4 c^2\right )}-\frac {b \left (a^2+b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{2} (2 m+3),\frac {1}{2} (2 m+5),\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{2 c^2 (2 m+3) \left (a^4+b^4 c^2\right )}\right )}{f}\) |
(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)
3.5.8.3.1 Defintions of rubi rules used
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]
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]))
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]]
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]
\[\int \frac {\left (d \tan \left (f x +e \right )\right )^{m}}{a +b \sqrt {c \tan \left (f x +e \right )}}d x\]
\[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{m}}{\sqrt {c \tan \left (f x + e\right )} b + a} \,d x } \]
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)
\[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\int \frac {\left (d \tan {\left (e + f x \right )}\right )^{m}}{a + b \sqrt {c \tan {\left (e + f x \right )}}}\, dx \]
\[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\int { \frac {\left (d \tan \left (f x + e\right )\right )^{m}}{\sqrt {c \tan \left (f x + e\right )} b + a} \,d x } \]
Exception generated. \[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Not invertible Error: Bad Argument Value
Timed out. \[ \int \frac {(d \tan (e+f x))^m}{a+b \sqrt {c \tan (e+f x)}} \, dx=\int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m}{a+b\,\sqrt {c\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \]