\(\int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx\) [1321]

Optimal result

Integrand size = 25, antiderivative size = 214 \[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx=\frac {d^2 (3 b c-a d) (a+b \tan (e+f x))^{1+m}}{b^2 f (1+m)}+\frac {(i c+d)^3 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a-i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a-i b) f (1+m)}-\frac {(i c-d)^3 \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a+i b}\right ) (a+b \tan (e+f x))^{1+m}}{2 (a+i b) f (1+m)}+\frac {d^3 (a+b \tan (e+f x))^{2+m}}{b^2 f (2+m)} \] Output:

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

Mathematica [A] (verified)

Time = 1.44 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.88 \[ \int (a+b \tan (e+f x))^m (c+d \tan (e+f x))^3 \, dx=\frac {(a+b \tan (e+f x))^{1+m} \left (\frac {2 d^2 (-a d+b c (5+2 m))}{b (1+m)}-\frac {i b (c-i d)^3 (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a-i b}\right )}{(a-i b) (1+m)}+\frac {i b (c+i d)^3 (2+m) \operatorname {Hypergeometric2F1}\left (1,1+m,2+m,\frac {a+b \tan (e+f x)}{a+i b}\right )}{(a+i b) (1+m)}+2 d^2 (c+d \tan (e+f x))\right )}{2 b f (2+m)} \] Input:

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

Output:

((a + b*Tan[e + f*x])^(1 + m)*((2*d^2*(-(a*d) + b*c*(5 + 2*m)))/(b*(1 + m) 
) - (I*b*(c - I*d)^3*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Tan 
[e + f*x])/(a - I*b)])/((a - I*b)*(1 + m)) + (I*b*(c + I*d)^3*(2 + m)*Hype 
rgeometric2F1[1, 1 + m, 2 + m, (a + b*Tan[e + f*x])/(a + I*b)])/((a + I*b) 
*(1 + m)) + 2*d^2*(c + d*Tan[e + f*x])))/(2*b*f*(2 + m))
 

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4049, 3042, 4113, 3042, 4022, 3042, 4020, 25, 78}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

\(\displaystyle \frac {\int (a+b \tan (e+f x))^m \left (b (m+2) c^3-d^2 (a d-b c (2 m+5)) \tan ^2(e+f x)-d^2 (a d+b c (m+1))+b d \left (3 c^2-d^2\right ) (m+2) \tan (e+f x)\right )dx}{b (m+2)}+\frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (a+b \tan (e+f x))^m \left (b (m+2) c^3-d^2 (a d-b c (2 m+5)) \tan (e+f x)^2-d^2 (a d+b c (m+1))+b d \left (3 c^2-d^2\right ) (m+2) \tan (e+f x)\right )dx}{b (m+2)}+\frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 4113

\(\displaystyle \frac {\int (a+b \tan (e+f x))^m \left (b c \left (c^2-3 d^2\right ) (m+2)+b d \left (3 c^2-d^2\right ) \tan (e+f x) (m+2)\right )dx-\frac {d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b f (m+1)}}{b (m+2)}+\frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int (a+b \tan (e+f x))^m \left (b c \left (c^2-3 d^2\right ) (m+2)+b d \left (3 c^2-d^2\right ) \tan (e+f x) (m+2)\right )dx-\frac {d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b f (m+1)}}{b (m+2)}+\frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}\)

\(\Big \downarrow \) 4022

\(\displaystyle \frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}+\frac {\frac {1}{2} b (m+2) (c+i d)^3 \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^mdx+\frac {1}{2} b (m+2) (c-i d)^3 \int (i \tan (e+f x)+1) (a+b \tan (e+f x))^mdx-\frac {d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b f (m+1)}}{b (m+2)}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}+\frac {\frac {1}{2} b (m+2) (c+i d)^3 \int (1-i \tan (e+f x)) (a+b \tan (e+f x))^mdx+\frac {1}{2} b (m+2) (c-i d)^3 \int (i \tan (e+f x)+1) (a+b \tan (e+f x))^mdx-\frac {d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b f (m+1)}}{b (m+2)}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}+\frac {\frac {i b (m+2) (c-i d)^3 \int -\frac {(a+b \tan (e+f x))^m}{1-i \tan (e+f x)}d(i \tan (e+f x))}{2 f}-\frac {i b (m+2) (c+i d)^3 \int -\frac {(a+b \tan (e+f x))^m}{i \tan (e+f x)+1}d(-i \tan (e+f x))}{2 f}-\frac {d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b f (m+1)}}{b (m+2)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {d^2 (c+d \tan (e+f x)) (a+b \tan (e+f x))^{m+1}}{b f (m+2)}+\frac {-\frac {i b (m+2) (c-i d)^3 \int \frac {(a+b \tan (e+f x))^m}{1-i \tan (e+f x)}d(i \tan (e+f x))}{2 f}+\frac {i b (m+2) (c+i d)^3 \int \frac {(a+b \tan (e+f x))^m}{i \tan (e+f x)+1}d(-i \tan (e+f x))}{2 f}-\frac {d^2 (a d-b c (2 m+5)) (a+b \tan (e+f x))^{m+1}}{b f (m+1)}}{b (m+2)}\)

\(\Big \downarrow \) 78

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

Input:

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

Output:

(d^2*(a + b*Tan[e + f*x])^(1 + m)*(c + d*Tan[e + f*x]))/(b*f*(2 + m)) + (- 
((d^2*(a*d - b*c*(5 + 2*m))*(a + b*Tan[e + f*x])^(1 + m))/(b*f*(1 + m))) - 
 ((I/2)*b*(c - I*d)^3*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b*Ta 
n[e + f*x])/(a - I*b)]*(a + b*Tan[e + f*x])^(1 + m))/((a - I*b)*f*(1 + m)) 
 + ((I/2)*b*(c + I*d)^3*(2 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (a + b* 
Tan[e + f*x])/(a + I*b)]*(a + b*Tan[e + f*x])^(1 + m))/((a + I*b)*f*(1 + m 
)))/(b*(2 + m))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 78

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b 
*c - a*d)^n*((a + b*x)^(m + 1)/(b^(n + 1)*(m + 1)))*Hypergeometric2F1[-n, m 
 + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m}, x] 
 &&  !IntegerQ[m] && IntegerQ[n]
 

rule 3042

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

rule 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4049

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c 
+ d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[1/(d*(m + n - 1)) 
 Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2, x], x], 
x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2 
, 0] && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])) 
)
 

rule 4113

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*((a + 
 b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f*x])^m*Si 
mp[A - C + B*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && 
NeQ[A*b^2 - a*b*B + a^2*C, 0] &&  !LeQ[m, -1]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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