\(\int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^m \, dx\) [332]

Optimal result

Integrand size = 26, antiderivative size = 81 \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {2 \operatorname {AppellF1}\left (\frac {5}{2},1-m,1,\frac {7}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-m} \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^m}{5 d} \]

[Out]

2/5*AppellF1(5/2,1-m,1,7/2,-I*tan(d*x+c),I*tan(d*x+c))*tan(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^m/d/((1+I*tan(d*x+c
))^m)

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3645, 129, 525, 524} \[ \int \tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^m \, dx=\frac {2 \tan ^{\frac {5}{2}}(c+d x) (1+i \tan (c+d x))^{-m} (a+i a \tan (c+d x))^m \operatorname {AppellF1}\left (\frac {5}{2},1-m,1,\frac {7}{2},-i \tan (c+d x),i \tan (c+d x)\right )}{5 d} \]

[In]

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

[Out]

(2*AppellF1[5/2, 1 - m, 1, 7/2, (-I)*Tan[c + d*x], I*Tan[c + d*x]]*Tan[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^m
)/(5*d*(1 + I*Tan[c + d*x])^m)

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 524

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

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, 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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {\left (-\frac {i x}{a}\right )^{3/2} (a+x)^{-1+m}}{-a^2+a x} \, dx,x,i a \tan (c+d x)\right )}{d} \\ & = -\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {x^4 \left (a+i a x^2\right )^{-1+m}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {\left (2 a^2 (1+i \tan (c+d x))^{-m} (a+i a \tan (c+d x))^m\right ) \text {Subst}\left (\int \frac {x^4 \left (1+i x^2\right )^{-1+m}}{-a^2+i a^2 x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = \frac {2 \operatorname {AppellF1}\left (\frac {5}{2},1-m,1,\frac {7}{2},-i \tan (c+d x),i \tan (c+d x)\right ) (1+i \tan (c+d x))^{-m} \tan ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^m}{5 d} \\ \end{align*}

Mathematica [F]

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

[In]

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

[Out]

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

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x +
2*I*c) + 1))*(-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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