3.12.60 \(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx\) [1160]

Optimal. Leaf size=262 \[ -\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-13 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 c^2+50 i c d-67 d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

-1/8*I*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2
)/(c-I*d)^(1/2)+1/60*(15*c^2+50*I*c*d-67*d^2)*(c+d*tan(f*x+e))^(1/2)/a^2/(I*c-d)^3/f/(a+I*a*tan(f*x+e))^(1/2)-
1/5*(c+d*tan(f*x+e))^(1/2)/(I*c-d)/f/(a+I*a*tan(f*x+e))^(5/2)+1/30*(5*I*c-13*d)*(c+d*tan(f*x+e))^(1/2)/a/(c+I*
d)^2/f/(a+I*a*tan(f*x+e))^(3/2)

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Rubi [A]
time = 0.55, antiderivative size = 262, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3640, 3677, 12, 3625, 214} \begin {gather*} -\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f \sqrt {c-i d}}+\frac {\left (15 c^2+50 i c d-67 d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 f (-d+i c)^3 \sqrt {a+i a \tan (e+f x)}}+\frac {(-13 d+5 i c) \sqrt {c+d \tan (e+f x)}}{30 a f (c+i d)^2 (a+i a \tan (e+f x))^{3/2}}-\frac {\sqrt {c+d \tan (e+f x)}}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

((-1/4*I)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqr
t[2]*a^(5/2)*Sqrt[c - I*d]*f) - Sqrt[c + d*Tan[e + f*x]]/(5*(I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)) + (((5*I
)*c - 13*d)*Sqrt[c + d*Tan[e + f*x]])/(30*a*(c + I*d)^2*f*(a + I*a*Tan[e + f*x])^(3/2)) + ((15*c^2 + (50*I)*c*
d - 67*d^2)*Sqrt[c + d*Tan[e + f*x]])/(60*a^2*(I*c - d)^3*f*Sqrt[a + I*a*Tan[e + f*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3625

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

Rule 3640

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

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rubi steps

\begin {align*} \int \frac {1}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}} \, dx &=-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}-\frac {\int \frac {-\frac {1}{2} a (5 i c-9 d)-2 i a d \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}} \, dx}{5 a^2 (i c-d)}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-13 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}-\frac {\int \frac {-\frac {1}{4} a^2 \left (15 c^2+40 i c d-41 d^2\right )-\frac {1}{2} a^2 (5 c+13 i d) d \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \, dx}{15 a^4 (c+i d)^2}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-13 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 c^2+50 i c d-67 d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)}}-\frac {\int -\frac {15 a^3 (i c-d)^3 \sqrt {a+i a \tan (e+f x)}}{8 \sqrt {c+d \tan (e+f x)}} \, dx}{15 a^6 (i c-d)^3}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-13 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 c^2+50 i c d-67 d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)}}+\frac {\int \frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}} \, dx}{8 a^3}\\ &=-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-13 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 c^2+50 i c d-67 d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)}}-\frac {i \text {Subst}\left (\int \frac {1}{a c-i a d-2 a^2 x^2} \, dx,x,\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}}\right )}{4 a f}\\ &=-\frac {i \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} \sqrt {c-i d} f}-\frac {\sqrt {c+d \tan (e+f x)}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}}+\frac {(5 i c-13 d) \sqrt {c+d \tan (e+f x)}}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2}}+\frac {\left (15 c^2+50 i c d-67 d^2\right ) \sqrt {c+d \tan (e+f x)}}{60 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 5.60, size = 309, normalized size = 1.18 \begin {gather*} \frac {\sec ^{\frac {5}{2}}(e+f x) \left (-\frac {i \sqrt {2} e^{2 i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \log \left (2 \left (\sqrt {c-i d} e^{i (e+f x)}+\sqrt {1+e^{2 i (e+f x)}} \sqrt {c-\frac {i d \left (-1+e^{2 i (e+f x)}\right )}{1+e^{2 i (e+f x)}}}\right )\right )}{\sqrt {c-i d}}+\frac {2 i \left (11 c^2+30 i c d-19 d^2+\left (26 c^2+80 i c d-86 d^2\right ) \cos (2 (e+f x))+4 i \left (5 c^2+17 i c d-20 d^2\right ) \sin (2 (e+f x))\right ) \sqrt {c+d \tan (e+f x)}}{15 (c+i d)^3 \sqrt {\sec (e+f x)}}\right )}{8 f (a+i a \tan (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]),x]

[Out]

(Sec[e + f*x]^(5/2)*(((-I)*Sqrt[2]*E^((2*I)*(e + f*x))*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1
+ E^((2*I)*(e + f*x))]*Log[2*(Sqrt[c - I*d]*E^(I*(e + f*x)) + Sqrt[1 + E^((2*I)*(e + f*x))]*Sqrt[c - (I*d*(-1
+ E^((2*I)*(e + f*x))))/(1 + E^((2*I)*(e + f*x)))])])/Sqrt[c - I*d] + (((2*I)/15)*(11*c^2 + (30*I)*c*d - 19*d^
2 + (26*c^2 + (80*I)*c*d - 86*d^2)*Cos[2*(e + f*x)] + (4*I)*(5*c^2 + (17*I)*c*d - 20*d^2)*Sin[2*(e + f*x)])*Sq
rt[c + d*Tan[e + f*x]])/((c + I*d)^3*Sqrt[Sec[e + f*x]])))/(8*f*(a + I*a*Tan[e + f*x])^(5/2))

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 5217 vs. \(2 (214 ) = 428\).
time = 0.82, size = 5218, normalized size = 19.92

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(5218\)
default	\(\text {Expression too large to display}\)	\(5218\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 590 vs. \(2 (210) = 420\).
time = 1.06, size = 590, normalized size = 2.25 \begin {gather*} \frac {{\left (30 \, {\left (i \, a^{3} c^{3} - 3 \, a^{3} c^{2} d - 3 i \, a^{3} c d^{2} + a^{3} d^{3}\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-4 \, {\left (i \, a^{3} c + a^{3} d\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \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 )}\right ) + 30 \, {\left (-i \, a^{3} c^{3} + 3 \, a^{3} c^{2} d + 3 i \, a^{3} c d^{2} - a^{3} d^{3}\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-4 \, {\left (-i \, a^{3} c - a^{3} d\right )} f \sqrt {\frac {i}{8 \, {\left (-i \, a^{5} c - a^{5} d\right )} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \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 )}\right ) + \sqrt {2} {\left (3 \, c^{2} + 6 i \, c d - 3 \, d^{2} + {\left (23 \, c^{2} + 74 i \, c d - 83 \, d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, {\left (17 \, c^{2} + 52 i \, c d - 51 \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (7 \, c^{2} + 18 i \, c d - 11 \, d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, {\left (-i \, a^{3} c^{3} + 3 \, a^{3} c^{2} d + 3 i \, a^{3} c d^{2} - a^{3} d^{3}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/120*(30*(I*a^3*c^3 - 3*a^3*c^2*d - 3*I*a^3*c*d^2 + a^3*d^3)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2))*e^(5*I*f*
x + 5*I*e)*log(-4*(I*a^3*c + a^3*d)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2))*e^(I*f*x + I*e) + sqrt(2)*sqrt(((c
- I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*(e^(2*I*f*x
+ 2*I*e) + 1)) + 30*(-I*a^3*c^3 + 3*a^3*c^2*d + 3*I*a^3*c*d^2 - a^3*d^3)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2
))*e^(5*I*f*x + 5*I*e)*log(-4*(-I*a^3*c - a^3*d)*f*sqrt(1/8*I/((-I*a^5*c - a^5*d)*f^2))*e^(I*f*x + I*e) + sqrt
(2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)
)*(e^(2*I*f*x + 2*I*e) + 1)) + sqrt(2)*(3*c^2 + 6*I*c*d - 3*d^2 + (23*c^2 + 74*I*c*d - 83*d^2)*e^(6*I*f*x + 6*
I*e) + 2*(17*c^2 + 52*I*c*d - 51*d^2)*e^(4*I*f*x + 4*I*e) + 2*(7*c^2 + 18*I*c*d - 11*d^2)*e^(2*I*f*x + 2*I*e))
*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*
e^(-5*I*f*x - 5*I*e)/((-I*a^3*c^3 + 3*a^3*c^2*d + 3*I*a^3*c*d^2 - a^3*d^3)*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {c + d \tan {\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))**(1/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*sqrt(c + d*tan(e + f*x))), x)

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c+d*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, need to choose a branch for the root of a polynomial with parameters. This might be wrong.Non regu
lar value [

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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