\(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx\) [1328]

Optimal result

Integrand size = 27, antiderivative size = 247 \[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\frac {\operatorname {AppellF1}\left (\frac {5}{2},-m,1,\frac {7}{2},\frac {b (c+d \tan (e+f x))}{b c-a d},\frac {c+d \tan (e+f x)}{c-i d}\right ) (a+b \tan (e+f x))^m \left (-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )^{-m} (c+d \tan (e+f x))^{5/2}}{5 (i c+d) f}-\frac {\operatorname {AppellF1}\left (\frac {5}{2},-m,1,\frac {7}{2},\frac {b (c+d \tan (e+f x))}{b c-a d},\frac {c+d \tan (e+f x)}{c+i d}\right ) (a+b \tan (e+f x))^m \left (-\frac {d (a+b \tan (e+f x))}{b c-a d}\right )^{-m} (c+d \tan (e+f x))^{5/2}}{5 (i c-d) f} \] Output:

1/5*AppellF1(5/2,1,-m,7/2,(c+d*tan(f*x+e))/(c-I*d),b*(c+d*tan(f*x+e))/(-a* 
d+b*c))*(a+b*tan(f*x+e))^m*(c+d*tan(f*x+e))^(5/2)/(I*c+d)/f/((-d*(a+b*tan( 
f*x+e))/(-a*d+b*c))^m)-1/5*AppellF1(5/2,1,-m,7/2,(c+d*tan(f*x+e))/(c+I*d), 
b*(c+d*tan(f*x+e))/(-a*d+b*c))*(a+b*tan(f*x+e))^m*(c+d*tan(f*x+e))^(5/2)/( 
I*c-d)/f/((-d*(a+b*tan(f*x+e))/(-a*d+b*c))^m)
 

Mathematica [F]

\[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx \] Input:

Integrate[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

Integrate[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(3/2), x]
 

Rubi [A] (warning: unable to verify)

Time = 0.52 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {3042, 4058, 662, 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.

\(\displaystyle \int (c+d \tan (e+f x))^{3/2} (a+b \tan (e+f x))^m \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d \tan (e+f x))^{3/2} (a+b \tan (e+f x))^mdx\)

\(\Big \downarrow \) 4058

\(\displaystyle \frac {\int \frac {(a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2}}{\tan ^2(e+f x)+1}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 662

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {(b c-a d) \sqrt {c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {3}{2},1,m+2,-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a-i b}\right )}{2 b (m+1) (b+i a) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}-\frac {(b c-a d) \sqrt {c+d \tan (e+f x)} (a+b \tan (e+f x))^{m+1} \operatorname {AppellF1}\left (m+1,-\frac {3}{2},1,m+2,-\frac {d (a+b \tan (e+f x))}{b c-a d},\frac {a+b \tan (e+f x)}{a+i b}\right )}{2 b (m+1) (-b+i a) \sqrt {\frac {b (c+d \tan (e+f x))}{b c-a d}}}}{f}\)

Input:

Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(3/2),x]
 

Output:

(((b*c - a*d)*AppellF1[1 + m, -3/2, 1, 2 + m, -((d*(a + b*Tan[e + f*x]))/( 
b*c - a*d)), (a + b*Tan[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 + m)* 
Sqrt[c + d*Tan[e + f*x]])/(2*b*(I*a + b)*(1 + m)*Sqrt[(b*(c + d*Tan[e + f* 
x]))/(b*c - a*d)]) - ((b*c - a*d)*AppellF1[1 + m, -3/2, 1, 2 + m, -((d*(a 
+ b*Tan[e + f*x]))/(b*c - a*d)), (a + b*Tan[e + f*x])/(a + I*b)]*(a + b*Ta 
n[e + f*x])^(1 + m)*Sqrt[c + d*Tan[e + f*x]])/(2*(I*a - b)*b*(1 + m)*Sqrt[ 
(b*(c + d*Tan[e + f*x]))/(b*c - a*d)]))/f
 

Defintions of rubi rules used

rule 662

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_ 
)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^ 
2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] &&  !IntegerQ[m] &&  !Inte 
gerQ[n]
 

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 4058

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*((c + d*ff*x)^n/(1 + ff^2*x^2)), x], x, 
 Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a 
*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
 

Maple [F]

\[\int \left (a +b \tan \left (f x +e \right )\right )^{m} \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\int { {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

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

Output:

integral((d*tan(f*x + e) + c)^(3/2)*(b*tan(f*x + e) + a)^m, x)
 

Sympy [F]

\[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\int \left (a + b \tan {\left (e + f x \right )}\right )^{m} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}\, dx \] Input:

integrate((a+b*tan(f*x+e))**m*(c+d*tan(f*x+e))**(3/2),x)
 

Output:

Integral((a + b*tan(e + f*x))**m*(c + d*tan(e + f*x))**(3/2), x)
 

Maxima [F]

\[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\int { {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

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

Output:

integrate((d*tan(f*x + e) + c)^(3/2)*(b*tan(f*x + e) + a)^m, x)
 

Giac [F]

\[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\int { {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (b \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \] Input:

integrate((a+b*tan(f*x+e))^m*(c+d*tan(f*x+e))^(3/2),x, algorithm="giac")
 

Output:

integrate((d*tan(f*x + e) + c)^(3/2)*(b*tan(f*x + e) + a)^m, x)
 

Mupad [F(-1)]

Timed out. \[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\int {\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^m\,{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^{3/2} \, dx=\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \left (\tan \left (f x +e \right ) b +a \right )^{m} \tan \left (f x +e \right )d x \right ) d +\left (\int \sqrt {d \tan \left (f x +e \right )+c}\, \left (\tan \left (f x +e \right ) b +a \right )^{m}d x \right ) c \] Input:

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

Output:

int(sqrt(tan(e + f*x)*d + c)*(tan(e + f*x)*b + a)**m*tan(e + f*x),x)*d + i 
nt(sqrt(tan(e + f*x)*d + c)*(tan(e + f*x)*b + a)**m,x)*c