\(\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx\) [320]

Optimal result

Integrand size = 28, antiderivative size = 89 \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx=\frac {a \operatorname {AppellF1}\left (1+n,-\frac {1}{2},1,2+n,-i \tan (e+f x),i \tan (e+f x)\right ) (d \tan (e+f x))^{1+n} \sqrt {a+i a \tan (e+f x)}}{d f (1+n) \sqrt {1+i \tan (e+f x)}} \]

[Out]

a*AppellF1(1+n,-1/2,1,2+n,-I*tan(f*x+e),I*tan(f*x+e))*(a+I*a*tan(f*x+e))^(1/2)*(d*tan(f*x+e))^(1+n)/d/f/(1+n)/
(1+I*tan(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3645, 140, 138} \[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx=\frac {a \sqrt {a+i a \tan (e+f x)} \operatorname {AppellF1}\left (n+1,-\frac {1}{2},1,n+2,-i \tan (e+f x),i \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) \sqrt {1+i \tan (e+f x)}} \]

[In]

Int[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^(3/2),x]

[Out]

(a*AppellF1[1 + n, -1/2, 1, 2 + n, (-I)*Tan[e + f*x], I*Tan[e + f*x]]*(d*Tan[e + f*x])^(1 + n)*Sqrt[a + I*a*Ta
n[e + f*x]])/(d*f*(1 + n)*Sqrt[1 + I*Tan[e + f*x]])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (i a^2\right ) \text {Subst}\left (\int \frac {\left (-\frac {i d x}{a}\right )^n \sqrt {a+x}}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f} \\ & = \frac {\left (i a^2 \sqrt {a+i a \tan (e+f x)}\right ) \text {Subst}\left (\int \frac {\left (-\frac {i d x}{a}\right )^n \sqrt {1+\frac {x}{a}}}{-a^2+a x} \, dx,x,i a \tan (e+f x)\right )}{f \sqrt {1+i \tan (e+f x)}} \\ & = \frac {a \operatorname {AppellF1}\left (1+n,-\frac {1}{2},1,2+n,-i \tan (e+f x),i \tan (e+f x)\right ) (d \tan (e+f x))^{1+n} \sqrt {a+i a \tan (e+f x)}}{d f (1+n) \sqrt {1+i \tan (e+f x)}} \\ \end{align*}

Mathematica [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx=\int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx \]

[In]

Integrate[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^(3/2),x]

[Out]

Integrate[(d*Tan[e + f*x])^n*(a + I*a*Tan[e + f*x])^(3/2), x]

Maple [F]

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

[In]

int((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^(3/2),x)

[Out]

int((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^(3/2),x)

Fricas [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]

[In]

integrate((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(2*sqrt(2)*a*((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))^n*sqrt(a/(e^(2*I*f*x + 2*I*e
) + 1))*e^(3*I*f*x + 3*I*e)/(e^(2*I*f*x + 2*I*e) + 1), x)

Sympy [F]

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

[In]

integrate((d*tan(f*x+e))**n*(a+I*a*tan(f*x+e))**(3/2),x)

[Out]

Integral((d*tan(e + f*x))**n*(I*a*(tan(e + f*x) - I))**(3/2), x)

Maxima [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]

[In]

integrate((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((I*a*tan(f*x + e) + a)^(3/2)*(d*tan(f*x + e))^n, x)

Giac [F]

\[ \int (d \tan (e+f x))^n (a+i a \tan (e+f x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} \left (d \tan \left (f x + e\right )\right )^{n} \,d x } \]

[In]

integrate((d*tan(f*x+e))^n*(a+I*a*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(3/2)*(d*tan(f*x + e))^n, x)

Mupad [F(-1)]

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

[In]

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^(3/2),x)

[Out]

int((d*tan(e + f*x))^n*(a + a*tan(e + f*x)*1i)^(3/2), x)