\(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [213]

Optimal result

Integrand size = 36, antiderivative size = 227 \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {2 a^2 (A-i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{d (1+m) \sqrt {a+i a \tan (c+d x)}}+\frac {2 a (B+(i A+B) (3+2 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)} \]

[Out]

2*a*(B+(I*A+B)*(3+2*m))*hypergeom([1/2, -m],[3/2],1+I*tan(d*x+c))*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^m/d/(3+2
*m)/((-I*tan(d*x+c))^m)+2*a^2*(A-I*B)*AppellF1(1+m,1/2,1,2+m,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1/2
)*tan(d*x+c)^(1+m)/d/(1+m)/(a+I*a*tan(d*x+c))^(1/2)+2*I*a*B*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^(1+m)/d/(3+2*m
)

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3675, 3682, 3645, 140, 138, 3680, 69, 67} \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {2 a^2 (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}+\frac {2 a (B+(2 m+3) (B+i A)) \sqrt {a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},i \tan (c+d x)+1\right )}{d (2 m+3)}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)} \tan ^{m+1}(c+d x)}{d (2 m+3)} \]

[In]

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

(2*a^2*(A - I*B)*AppellF1[1 + m, 1/2, 1, 2 + m, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[1 + I*Tan[c + d*x]]*Ta
n[c + d*x]^(1 + m))/(d*(1 + m)*Sqrt[a + I*a*Tan[c + d*x]]) + (2*a*(B + (I*A + B)*(3 + 2*m))*Hypergeometric2F1[
1/2, -m, 3/2, 1 + I*Tan[c + d*x]]*Tan[c + d*x]^m*Sqrt[a + I*a*Tan[c + d*x]])/(d*(3 + 2*m)*((-I)*Tan[c + d*x])^
m) + ((2*I)*a*B*Tan[c + d*x]^(1 + m)*Sqrt[a + I*a*Tan[c + d*x]])/(d*(3 + 2*m))

Rule 67

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)/(d*(n + 1)*(-d/(b*c))^m))
*Hypergeometric2F1[-m, n + 1, n + 2, 1 + d*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Intege
rQ[m] || GtQ[-d/(b*c), 0])

Rule 69

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[((-b)*(c/d))^IntPart[m]*((b*x)^FracPart[m]/(
(-d)*(x/c))^FracPart[m]), Int[((-d)*(x/c))^m*(c + d*x)^n, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m]
&&  !IntegerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0]

Rule 138

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[c^n*e^p*((b*x)^(m +
1)/(b*(m + 1)))*AppellF1[m + 1, -n, -p, m + 2, (-d)*(x/c), (-f)*(x/e)], x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 140

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[c^IntPart[n]*((c +
d*x)^FracPart[n]/(1 + d*(x/c))^FracPart[n]), Int[(b*x)^m*(1 + d*(x/c))^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] &&  !IntegerQ[m] &&  !IntegerQ[n] &&  !GtQ[c, 0]

Rule 3645

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

Rule 3675

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] &&  !LtQ[n, -1]

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[b*(B/f), Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]

Rule 3682

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a B \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)}+\frac {2 \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{2} a \left (2 i B (1+m)-2 A \left (\frac {3}{2}+m\right )\right )+\frac {1}{2} a (B+(i A+B) (3+2 m)) \tan (c+d x)\right ) \, dx}{3+2 m} \\ & = \frac {2 i a B \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)}+(2 a (A-i B)) \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx+\frac {(i (B+(i A+B) (3+2 m))) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{3+2 m} \\ & = \frac {2 i a B \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)}+\frac {\left (2 a^3 (i A+B)\right ) \text {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^m}{\sqrt {a+x} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}+\frac {\left (i a^2 (B+(i A+B) (3+2 m))\right ) \text {Subst}\left (\int \frac {x^m}{\sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d (3+2 m)} \\ & = \frac {2 i a B \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)}+\frac {\left (i a^2 (B+(i A+B) (3+2 m)) (-i \tan (c+d x))^{-m} \tan ^m(c+d x)\right ) \text {Subst}\left (\int \frac {(-i x)^m}{\sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d (3+2 m)}+\frac {\left (2 a^3 (i A+B) \sqrt {1+i \tan (c+d x)}\right ) \text {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^m}{\sqrt {1+\frac {x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {2 a^2 (A-i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{d (1+m) \sqrt {a+i a \tan (c+d x)}}+\frac {2 a (B+(i A+B) (3+2 m)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)}+\frac {2 i a B \tan ^{1+m}(c+d x) \sqrt {a+i a \tan (c+d x)}}{d (3+2 m)} \\ \end{align*}

Mathematica [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx \]

[In]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]

[Out]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]), x]

Maple [F]

\[\int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}} \left (A +B \tan \left (d x +c \right )\right )d x\]

[In]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x)

Fricas [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(2*sqrt(2)*((A - I*B)*a*e^(5*I*d*x + 5*I*c) + (A + I*B)*a*e^(3*I*d*x + 3*I*c))*((-I*e^(2*I*d*x + 2*I*c
) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))/(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*
I*c) + 1), x)

Sympy [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{m}{\left (c + d x \right )}\, dx \]

[In]

integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**(3/2)*(A + B*tan(c + d*x))*tan(c + d*x)**m, x)

Maxima [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^m, x)

Giac [F]

\[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{m} \,d x } \]

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^m, x)

Mupad [F(-1)]

Timed out. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

[In]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(tan(c + d*x)^m*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2), x)