3.1064 \(\int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=67 \[ -\frac {i (a+i a \tan (e+f x))^m \, _2F_1\left (1,m-\frac {5}{2};-\frac {3}{2};\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

-1/5*I*hypergeom([1, -5/2+m],[-3/2],1/2-1/2*I*tan(f*x+e))*(a+I*a*tan(f*x+e))^m/f/(c-I*c*tan(f*x+e))^(5/2)

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Rubi [A]  time = 0.11, antiderivative size = 88, normalized size of antiderivative = 1.31, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3523, 70, 69} \[ -\frac {i 2^m (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m \, _2F_1\left (-\frac {5}{2},1-m;-\frac {3}{2};\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^m/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((-I/5)*2^m*Hypergeometric2F1[-5/2, 1 - m, -3/2, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m)/(f*(1 + I*T
an[e + f*x])^m*(c - I*c*Tan[e + f*x])^(5/2))

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 3523

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[(a*c)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f,
m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (e+f x))^m}{(c-i c \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(a+i a x)^{-1+m}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left (2^{-1+m} c (a+i a \tan (e+f x))^m \left (\frac {a+i a \tan (e+f x)}{a}\right )^{-m}\right ) \operatorname {Subst}\left (\int \frac {\left (\frac {1}{2}+\frac {i x}{2}\right )^{-1+m}}{(c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i 2^m \, _2F_1\left (-\frac {5}{2},1-m;-\frac {3}{2};\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m}{5 f (c-i c \tan (e+f x))^{5/2}}\\ \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[(a + I*a*Tan[e + f*x])^m/(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

$Aborted

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {2} \left (\frac {2 \, a e^{\left (2 i \, f x + 2 i \, e\right )}}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{m} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )}}{8 \, c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(1/8*sqrt(2)*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(
e^(6*I*f*x + 6*I*e) + 3*e^(4*I*f*x + 4*I*e) + 3*e^(2*I*f*x + 2*I*e) + 1)/c^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^m/(-I*c*tan(f*x + e) + c)^(5/2), x)

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maple [F]  time = 3.53, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{m}}{\left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

int((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(5/2),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^m/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^m/(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^m/(c - c*tan(e + f*x)*1i)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{m}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**m/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**m/(-I*c*(tan(e + f*x) + I))**(5/2), x)

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