\(\int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [231]

Optimal result

Integrand size = 36, antiderivative size = 247 \[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}-\frac {2 (A-i B) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d}-\frac {2 (1-2 n) (3 i B-2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.83 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3679, 3682, 3645, 129, 441, 440, 3680, 68, 66} \[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 (A-i B) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right )}{d}-\frac {2 (1-2 n) (-2 A n+3 i B) \sqrt {\tan (c+d x)} (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right )}{3 d}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}-\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]

[In]

Int[((a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

(-2*A*(a + I*a*Tan[c + d*x])^n)/(3*d*Tan[c + d*x]^(3/2)) - (2*(3*B + (2*I)*A*n)*(a + I*a*Tan[c + d*x])^n)/(3*d
*Sqrt[Tan[c + d*x]]) - (2*(A - I*B)*AppellF1[1/2, 1 - n, 1, 3/2, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Sqrt[Tan[c
+ d*x]]*(a + I*a*Tan[c + d*x])^n)/(d*(1 + I*Tan[c + d*x])^n) - (2*(1 - 2*n)*((3*I)*B - 2*A*n)*Hypergeometric2
F1[1/2, 1 - n, 3/2, (-I)*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^n)/(3*d*(1 + I*Tan[c + d*x])^
n)

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-d/(b*c), 0])))

Rule 68

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), 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, x], x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Int
egerQ[n] &&  !GtQ[c, 0] &&  !GtQ[-d/(b*c), 0] && ((RationalQ[m] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0])) |
|  !RationalQ[n])

Rule 129

Int[((e_.)*(x_))^(p_)*((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> With[{k = Denominator[p]
}, Dist[k/e, Subst[Int[x^(k*(p + 1) - 1)*(a + b*(x^k/e))^m*(c + d*(x^k/e))^n, x], x, (e*x)^(1/k)], x]] /; Free
Q[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && FractionQ[p] && IntegerQ[m]

Rule 440

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 441

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^F
racPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n,
p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n, -1] &&  !(IntegerQ[p] || GtQ[a, 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 3679

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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*
(n + 1)*(c^2 + d^2))), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && 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 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \int \frac {(a+i a \tan (c+d x))^n \left (\frac {1}{2} a (3 B+2 i A n)-\frac {1}{2} a A (3-2 n) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a} \\ & = -\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}+\frac {4 \int \frac {(a+i a \tan (c+d x))^n \left (\frac {1}{4} a^2 \left (6 i B n-A \left (3-2 n+4 n^2\right )\right )-\frac {1}{4} a^2 (1-2 n) (3 B+2 i A n) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx}{3 a^2} \\ & = -\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}+(-A+i B) \int \frac {(a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx-\frac {((1-2 n) (3 i B-2 A n)) \int \frac {(a-i a \tan (c+d x)) (a+i a \tan (c+d x))^n}{\sqrt {\tan (c+d x)}} \, dx}{3 a} \\ & = -\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}-\frac {\left (a^2 (i A+B)\right ) \text {Subst}\left (\int \frac {(a+x)^{-1+n}}{\sqrt {-\frac {i x}{a}} \left (-a^2+a x\right )} \, dx,x,i a \tan (c+d x)\right )}{d}-\frac {(a (1-2 n) (3 i B-2 A n)) \text {Subst}\left (\int \frac {(a+i a x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{3 d} \\ & = -\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}+\frac {\left (2 a^3 (A-i B)\right ) \text {Subst}\left (\int \frac {\left (a+i a x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d}-\frac {\left ((1-2 n) (3 i B-2 A n) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \text {Subst}\left (\int \frac {(1+i x)^{-1+n}}{\sqrt {x}} \, dx,x,\tan (c+d x)\right )}{3 d} \\ & = -\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}-\frac {2 (1-2 n) (3 i B-2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{3 d}+\frac {\left (2 a^2 (A-i B) (1+i \tan (c+d x))^{-n} (a+i a \tan (c+d x))^n\right ) \text {Subst}\left (\int \frac {\left (1+i x^2\right )^{-1+n}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {2 A (a+i a \tan (c+d x))^n}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 (3 B+2 i A n) (a+i a \tan (c+d x))^n}{3 d \sqrt {\tan (c+d x)}}-\frac {2 (A-i B) \operatorname {AppellF1}\left (\frac {1}{2},1-n,1,\frac {3}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{d}-\frac {2 (1-2 n) (3 i B-2 A n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},1-n,\frac {3}{2},-i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-n} \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^n}{3 d} \\ \end{align*}

Mathematica [F]

\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx \]

[In]

Integrate[((a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2),x]

[Out]

Integrate[((a + I*a*Tan[c + d*x])^n*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2), x]

Maple [F]

\[\int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{n} \left (A +B \tan \left (d x +c \right )\right )}{\tan \left (d x +c \right )^{\frac {5}{2}}}d x\]

[In]

int((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)

[Out]

int((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)

Fricas [F]

\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral(((-I*A - B)*e^(6*I*d*x + 6*I*c) + (-3*I*A - B)*e^(4*I*d*x + 4*I*c) + (-3*I*A + B)*e^(2*I*d*x + 2*I*c)
- I*A + B)*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^n*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*
x + 2*I*c) + 1))/(e^(6*I*d*x + 6*I*c) - 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) - 1), x)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{n} \left (A + B \tan {\left (c + d x \right )}\right )}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((a+I*a*tan(d*x+c))**n*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)

[Out]

Integral((I*a*(tan(c + d*x) - I))**n*(A + B*tan(c + d*x))/tan(c + d*x)**(5/2), x)

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n/tan(d*x + c)^(5/2), x)

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{n}}{\tan \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^n*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^n/tan(d*x + c)^(5/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^n (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n)/tan(c + d*x)^(5/2),x)

[Out]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^n)/tan(c + d*x)^(5/2), x)