3.322 \(\int \frac {(d \tan (e+f x))^n}{\sqrt {a+i a \tan (e+f x)}} \, dx\)

Optimal. Leaf size=88 \[ \frac {\sqrt {1+i \tan (e+f x)} F_1\left (n+1;\frac {3}{2},1;n+2;-i \tan (e+f x),i \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) \sqrt {a+i a \tan (e+f x)}} \]

[Out]

AppellF1(1+n,3/2,1,2+n,-I*tan(f*x+e),I*tan(f*x+e))*(1+I*tan(f*x+e))^(1/2)*(d*tan(f*x+e))^(1+n)/d/f/(1+n)/(a+I*
a*tan(f*x+e))^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 88, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {3564, 135, 133} \[ \frac {\sqrt {1+i \tan (e+f x)} F_1\left (n+1;\frac {3}{2},1;n+2;-i \tan (e+f x),i \tan (e+f x)\right ) (d \tan (e+f x))^{n+1}}{d f (n+1) \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(AppellF1[1 + n, 3/2, 1, 2 + n, (-I)*Tan[e + f*x], I*Tan[e + f*x]]*Sqrt[1 + I*Tan[e + f*x]]*(d*Tan[e + f*x])^(
1 + n))/(d*f*(1 + n)*Sqrt[a + I*a*Tan[e + f*x]])

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rule 135

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 3564

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*x)/b)^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*} \int \frac {(d \tan (e+f x))^n}{\sqrt {a+i a \tan (e+f x)}} \, dx &=\frac {\left (i a^2\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {i d x}{a}\right )^n}{(a+x)^{3/2} \left (-a^2+a x\right )} \, dx,x,i a \tan (e+f x)\right )}{f}\\ &=\frac {\left (i a \sqrt {1+i \tan (e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\left (-\frac {i d x}{a}\right )^n}{\left (1+\frac {x}{a}\right )^{3/2} \left (-a^2+a x\right )} \, dx,x,i a \tan (e+f x)\right )}{f \sqrt {a+i a \tan (e+f x)}}\\ &=\frac {F_1\left (1+n;\frac {3}{2},1;2+n;-i \tan (e+f x),i \tan (e+f x)\right ) \sqrt {1+i \tan (e+f x)} (d \tan (e+f x))^{1+n}}{d f (1+n) \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

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

[Out]

$Aborted

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fricas [F]  time = 0.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2} \left (\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{n} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x + e\right )\right )^{n}}{\sqrt {i \, a \tan \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.49, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x +e \right )\right )^{n}}{\sqrt {a +i a \tan \left (f x +e \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan \left (f x + e\right )\right )^{n}}{\sqrt {i \, a \tan \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*tan(f*x+e))^n/(a+I*a*tan(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^n}{\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d \tan {\left (e + f x \right )}\right )^{n}}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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