\(\int (d \tan (e+f x))^m (a+b \sqrt {c \tan (e+f x)})^2 \, dx\) [408]

Optimal result

Integrand size = 29, antiderivative size = 212 \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\frac {\left (a^2-b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,-\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 f (1+m)}+\frac {\left (a^2+b^2 \sqrt {-c^2}\right ) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {c \tan (e+f x)}{\sqrt {-c^2}}\right ) \tan (e+f x) (d \tan (e+f x))^m}{2 f (1+m)}+\frac {4 a b \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right ) (c \tan (e+f x))^{3/2} (d \tan (e+f x))^m}{c f (3+2 m)} \] Output:

1/2*(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/f/(1+m)+1/2*(a^2+b^2*(-c^2)^(1/2))*hyper 
geom([1, 1+m],[2+m],c*tan(f*x+e)/(-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)
 

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.71 \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\frac {\tan (e+f x) (d \tan (e+f x))^m \left (\frac {a^2 \operatorname {Hypergeometric2F1}\left (1,\frac {1+m}{2},\frac {3+m}{2},-\tan ^2(e+f x)\right )}{1+m}+b \left (\frac {b c \operatorname {Hypergeometric2F1}\left (1,\frac {2+m}{2},\frac {4+m}{2},-\tan ^2(e+f x)\right ) \tan (e+f x)}{2+m}+\frac {4 a \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (3+2 m),\frac {1}{4} (7+2 m),-\tan ^2(e+f x)\right ) \sqrt {c \tan (e+f x)}}{3+2 m}\right )\right )}{f} \] Input:

Integrate[(d*Tan[e + f*x])^m*(a + b*Sqrt[c*Tan[e + f*x]])^2,x]
 

Output:

(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
 

Rubi [A] (warning: unable to verify)

Time = 0.88 (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

\(\displaystyle \frac {2 c (c \tan (e+f x))^{-m} \left (c d \tan ^2(e+f x)\right )^m \left (\frac {\left (a^2-b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,-\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1)}+\frac {\left (a^2+b^2 \sqrt {-c^2}\right ) (c \tan (e+f x))^{m+1} \operatorname {Hypergeometric2F1}\left (1,m+1,m+2,\frac {c^2 \tan ^2(e+f x)}{\sqrt {-c^2}}\right )}{4 c^2 (m+1)}+\frac {2 a b (c \tan (e+f x))^{\frac {1}{2} (2 m+3)} \operatorname {Hypergeometric2F1}\left (1,\frac {1}{4} (2 m+3),\frac {1}{4} (2 m+7),-c^2 \tan ^4(e+f x)\right )}{c^2 (2 m+3)}\right )}{f}\)

Input:

Int[(d*Tan[e + f*x])^m*(a + b*Sqrt[c*Tan[e + f*x]])^2,x]
 

Output:

(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)
 

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 2370

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
 

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]]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \] Input:

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

Output:

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)
 

Sympy [F]

\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int \left (d \tan {\left (e + f x \right )}\right )^{m} \left (a + b \sqrt {c \tan {\left (e + f x \right )}}\right )^{2}\, dx \] Input:

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

Output:

Integral((d*tan(e + f*x))**m*(a + b*sqrt(c*tan(e + f*x)))**2, x)
 

Maxima [F]

\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int { {\left (\sqrt {c \tan \left (f x + e\right )} b + a\right )}^{2} \left (d \tan \left (f x + e\right )\right )^{m} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\int {\left (a+b\,\sqrt {c\,\mathrm {tan}\left (e+f\,x\right )}\right )}^2\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^m \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (d \tan (e+f x))^m \left (a+b \sqrt {c \tan (e+f x)}\right )^2 \, dx=\frac {d^{m} \left (\tan \left (f x +e \right )^{m} b^{2} c +2 \sqrt {c}\, \left (\int \tan \left (f x +e \right )^{m +\frac {1}{2}}d x \right ) a b f m +\left (\int \tan \left (f x +e \right )^{m}d x \right ) a^{2} f m -\left (\int \frac {\tan \left (f x +e \right )^{m}}{\tan \left (f x +e \right )}d x \right ) b^{2} c f m \right )}{f m} \] Input:

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

Output:

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