\(\int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [216]

Optimal result

Integrand size = 36, antiderivative size = 285 \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {(A+i B) \tan ^{1+m}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A (5-4 m)-i (B+4 B m)) \tan ^{1+m}(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \operatorname {AppellF1}\left (1+m,\frac {1}{2},1,2+m,-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{1+m}(c+d x)}{4 a d (1+m) \sqrt {a+i a \tan (c+d x)}}+\frac {(1+2 m) (B+i A (5-4 m)+4 B m) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},1+i \tan (c+d x)\right ) (-i \tan (c+d x))^{-m} \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)}}{6 a^2 d} \]

[Out]

1/6*(1+2*m)*(B+I*A*(5-4*m)+4*B*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/a^2/d/((-I*tan(d*x+c))^m)+1/6*(A*(5-4*m)-I*(4*B*m+B))*tan(d*x+c)^(1+m)/a/d/(a+I*a*tan(d*x+c))^(1/2)+1/4*(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)/a/d/(1+m)/(a+
I*a*tan(d*x+c))^(1/2)+1/3*(A+I*B)*tan(d*x+c)^(1+m)/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3677, 3682, 3645, 140, 138, 3680, 69, 67} \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {(2 m+1) (i A (5-4 m)+4 B m+B) \sqrt {a+i a \tan (c+d x)} \tan ^m(c+d x) (-i \tan (c+d x))^{-m} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},-m,\frac {3}{2},i \tan (c+d x)+1\right )}{6 a^2 d}+\frac {(A-i B) \sqrt {1+i \tan (c+d x)} \tan ^{m+1}(c+d x) \operatorname {AppellF1}\left (m+1,\frac {1}{2},1,m+2,-i \tan (c+d x),i \tan (c+d x)\right )}{4 a d (m+1) \sqrt {a+i a \tan (c+d x)}}+\frac {(A (5-4 m)-i (4 B m+B)) \tan ^{m+1}(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

((A + I*B)*Tan[c + d*x]^(1 + m))/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((A*(5 - 4*m) - I*(B + 4*B*m))*Tan[c + d
*x]^(1 + m))/(6*a*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((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))/(4*a*d*(1 + m)*Sqrt[a + I*a*Tan[c + d*x]]) + (
(1 + 2*m)*(B + I*A*(5 - 4*m) + 4*B*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]])/(6*a^2*d*((-I)*Tan[c + d*x])^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 3677

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

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 {(A+i B) \tan ^{1+m}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\tan ^m(c+d x) \left (a (A (2-m)-i B (1+m))-\frac {1}{2} a (i A-B) (1-2 m) \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2} \\ & = \frac {(A+i B) \tan ^{1+m}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A (5-4 m)-i (B+4 B m)) \tan ^{1+m}(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{2} a^2 \left (i B \left (1-3 m-4 m^2\right )+A \left (1+3 m-4 m^2\right )\right )+\frac {1}{4} a^2 (1+2 m) (B+i A (5-4 m)+4 B m) \tan (c+d x)\right ) \, dx}{3 a^4} \\ & = \frac {(A+i B) \tan ^{1+m}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A (5-4 m)-i (B+4 B m)) \tan ^{1+m}(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(A-i B) \int \tan ^m(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}-\frac {((1+2 m) (A (5-4 m)-i (B+4 B m))) \int \tan ^m(c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{12 a^3} \\ & = \frac {(A+i B) \tan ^{1+m}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A (5-4 m)-i (B+4 B m)) \tan ^{1+m}(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(i A+B) \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 )}{4 d}-\frac {((1+2 m) (A (5-4 m)-i (B+4 B m))) \text {Subst}\left (\int \frac {x^m}{\sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{12 a d} \\ & = \frac {(A+i B) \tan ^{1+m}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A (5-4 m)-i (B+4 B m)) \tan ^{1+m}(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {\left ((1+2 m) (A (5-4 m)-i (B+4 B 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 )}{12 a d}+\frac {\left ((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 )}{4 d \sqrt {a+i a \tan (c+d x)}} \\ & = \frac {(A+i B) \tan ^{1+m}(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {(A (5-4 m)-i (B+4 B m)) \tan ^{1+m}(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {(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)}{4 a d (1+m) \sqrt {a+i a \tan (c+d x)}}+\frac {(1+2 m) (B+i A (5-4 m)+4 B 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)}}{6 a^2 d} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[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)