3.4.35 \(\int \frac {(a+i a \tan (c+d x))^m}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [335]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 79, normalized size of antiderivative = 1.00, 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 = {3645, 129, 525, 524} \begin {gather*} -\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)}} \end {gather*}

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

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Mathematica [F]
time = 3.21, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+i a \tan (c+d x))^m}{\tan ^{\frac {3}{2}}(c+d x)} \, dx \end {gather*}

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 = 1.64, size = 0, normalized size = 0.00 \[\int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{m}}{\tan \left (d x +c \right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{m}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

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]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^m}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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