Integrand size = 29, antiderivative size = 212 \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\frac {\left (a^2-b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 f (1+m)}+\frac {\left (a^2+b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 f (1+m)}+\frac {4 a b \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c f (3+2 m)} \]
1/2*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/f/(1+m)+1/2*hypergeom([1, 1+m],[2+m],c*t an(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 /f/(1+m)+4*a*b*hypergeom([1, 3/4+1/2*m],[7/4+1/2*m],-tan(f*x+e)^2)*(c*tan( f*x+e))^(3/2)*(d*tan(f*x+e))^m/c/f/(3+2*m)
Time = 1.58 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.71 \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^m \left (\frac {a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(e+f x)\right )}{1+m}+b \left (\frac {b c \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{2+m}+\frac {4 a \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right ) \sqrt {c \tan (e+f x)}}{3+2 m}\right )\right )}{f} \]
(Tan[e + f*x]*(d*Tan[e + f*x])^m*((a^2*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[e + f*x]^2])/(1 + m) + b*((b*c*Hypergeometric2F1[1, (2 + m)/2 , (4 + m)/2, -Tan[e + f*x]^2]*Tan[e + f*x])/(2 + m) + (4*a*Hypergeometric2 F1[1, (3 + 2*m)/4, (7 + 2*m)/4, -Tan[e + f*x]^2]*Sqrt[c*Tan[e + f*x]])/(3 + 2*m))))/f
Time = 0.87 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.13, 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, 2370, 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.
| \(\displaystyle \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2dx\) |
| \(\Big \downarrow \) 4153 |
| \(\displaystyle \frac {c \int \frac {(d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2}{\tan ^2(e+f x) c^2+c^2}d(c \tan (e+f x))}{f}\) |
| \(\Big \downarrow \) 7267 |
| \(\displaystyle \frac {2 c \int \frac {\sqrt {c \tan (e+f x)} \left (c d \tan ^2(e+f x)\right )^m (a+b c \tan (e+f x))^2}{c^4 \tan ^4(e+f x)+c^2}d\sqrt {c \tan (e+f x)}}{f}\) |
| \(\Big \downarrow \) 30 |
| \(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \int \frac {(c \tan (e+f x))^{\frac {1}{2} (2 m+1)} \left (a+b \sqrt {c \tan (e+f x)}\right )^2}{c^4 \tan ^4(e+f x)+c^2}d\sqrt {c \tan (e+f x)}}{f}\) |
| \(\Big \downarrow \) 2370 |
| \(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \int \left (\frac {\left (a^2+b^2 c^2 \tan ^2(e+f x)\right ) (c \tan (e+f x))^{\frac {1}{2} (2 m+1)}}{c^4 \tan ^4(e+f x)+c^2}+\frac {2 a b (c \tan (e+f x))^{\frac {1}{2} (2 m+2)}}{c^4 \tan ^4(e+f x)+c^2}\right )d\sqrt {c \tan (e+f x)}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \left (\frac {\left (a^2-b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1)}+\frac {\left (a^2+b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1)}+\frac {2 a b (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2 m+3),\frac {1}{4} (2 m+7),-c^2 \tan ^4(e+f x)\right )}{c^2 (2 m+3)}\right )}{f}\) |
(2*c*(c*d*Tan[e + f*x]^2)^m*(((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*(1 + m)) + ((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*(1 + m)) + (2*a*b*Hypergeometric2F1[1, (3 + 2*m)/4, (7 + 2*m)/4, -(c^2*Tan[e + f*x]^4)]*(c*Tan[e + f*x])^((3 + 2*m)/2))/(c^2*(3 + 2*m))))/(f*(c*Tan[e + f*x])^m)
3.5.6.3.1 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[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[ {v = Sum[(c*x)^(m + ii)*((Coeff[Pq, x, ii] + Coeff[Pq, x, n/2 + ii]*x^(n/2) )/(c^ii*(a + b*x^n))), {ii, 0, n/2 - 1}]}, Int[v, x] /; SumQ[v]] /; FreeQ[{ a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[n/2, 0] && Expon[Pq, x] < n
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]]
\[\int \left (a +b \sqrt {c \tan \left (f x +e \right )}\right )^{2} \left (d \tan \left (f x +e \right )\right )^{m}d x\]
\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
integral(2*sqrt(c*tan(f*x + e))*(d*tan(f*x + e))^m*a*b + (b^2*c*tan(f*x + e) + a^2)*(d*tan(f*x + e))^m, x)
\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{m} \left (a + b \sqrt {c \tan {\left (e + f x \right )}}\right )^{2}\, dx \]
\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \]
Exception generated. \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\text {Exception raised: RuntimeError} \]
Exception raised: RuntimeError >> an error occurred running a Giac command :INPUT:sage2OUTPUT:Unable to divide, perhaps due to rounding error%%%{-1,[ 0,1,0]%%%} / %%%{1,[0,0,2]%%%}+%%%{-1,[0,0,0]%%%} Error: Bad Argument Valu e
Timed out. \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int {\left (a+b\,\sqrt {c\,\mathrm {tan}\left (e+f\,x\right )}\right )}^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m \,d x \]