3.3.0 ∫ tanm(c + dx)∘___________a + ia tan (c + dx)(A + B tan(c + dx)) dx [214]

Integrand size = 36, antiderivative size = 159 \[ \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {a (A-i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{d (1+m) \sqrt {a+i a \tan (c+d x)}}+\frac {2 B \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{d} \]

2*B*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/((-I*tan(d*x+c))^m)+a*(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*

Rubi [A] (veriﬁed)

Time = 0.44 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3682, 3645, 140, 138, 3680, 69, 67} \[ \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx=\frac {a (A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{d (m+1) \sqrt {a+i a \tan (c+d x)}}+\frac {2 B \sqrt {a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},i \tan (c+d x)+1\right )}{d} \]

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

(a*(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]]*Tan[c

+ d*x]^(1 + m))/(d*(1 + m)*Sqrt[a + I*a*Tan[c + d*x]]) + (2*B*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*((-I)*Tan[c + d*x])^m)

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

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]

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])

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]

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]

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,

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}& = -\left ((-A+i B) \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\right )+\frac {(i B) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{a} \\ & = \frac {(i a B) \text {Subst}\left (\int \frac {x^m}{\sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (a^2 (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 B (-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}+\frac {\left (a^2 (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 {a (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 B \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} \\ \end{align*}

Mathematica [F]

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

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

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

Maple [F]

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

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

Fricas [F]

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

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

integral(sqrt(2)*((A - I*B)*e^(3*I*d*x + 3*I*c) + (A + I*B)*e^(I*d*x + 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^(2*I*d*x + 2*I*c) + 1), x)

Sympy [F]

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

integrate(tan(d*x+c)**m*(a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

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

Maxima [F]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

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

Giac [F]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

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

Mupad [F(-1)]

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

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

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

\(\int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\) [214]

Optimal result