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

Optimal. Leaf size=79 \[ -\frac{2 (1+i \tan (c+d x))^{-m} (a+i a \tan (c+d x))^m F_1\left (-\frac{1}{2};1-m,1;\frac{1}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt{\tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.119075, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3564, 130, 511, 510} \[ -\frac{2 (1+i \tan (c+d x))^{-m} (a+i a \tan (c+d x))^m F_1\left (-\frac{1}{2};1-m,1;\frac{1}{2};-i \tan (c+d x),i \tan (c+d x)\right )}{d \sqrt{\tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 130

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 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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])

Rubi steps

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

Mathematica [F]  time = 3.15525, size = 0, normalized size = 0. \[ \int \frac{(a+i a \tan (c+d x))^m}{\tan ^{\frac{3}{2}}(c+d x)} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.212, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{m} \left ( \tan \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m}}{\tan \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\left (\frac{2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} \sqrt{\frac{-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}{e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(d*x+c))^m/tan(d*x+c)^(3/2),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))*(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2*I*c) + 1)/(e^(4*I*d*x + 4*I*c) - 2*e^(2*I*d*x + 2*I*c)
+ 1), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{m}}{\tan \left (d x + c\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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