3.10.93 \(\int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx\) [993]

Optimal. Leaf size=106 \[ -\frac {2 i a^{3/2} \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f} \]

[Out]

-2*I*a^(3/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x+e))^(1/2))*c^(1/2)/f+I*a*(a+I*a*ta
n(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/f

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Rubi [A]
time = 0.09, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3604, 52, 65, 223, 209} \begin {gather*} \frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {2 i a^{3/2} \sqrt {c} \text {ArcTan}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

((-2*I)*a^(3/2)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f +
(I*a*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/f

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 3604

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 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {\sqrt {a+i a x}}{\sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}+\frac {\left (a^2 c\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+i a x} \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {(2 i a c) \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{f}\\ &=\frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {(2 i a c) \text {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}\right )}{f}\\ &=-\frac {2 i a^{3/2} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {i a \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}\\ \end {align*}

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Mathematica [A]
time = 2.00, size = 129, normalized size = 1.22 \begin {gather*} \frac {a c e^{-\frac {1}{2} i (4 e+f x)} \left (2 \text {ArcTan}\left (e^{i (e+f x)}\right )-\sec (e+f x)\right ) \left (-i \cos \left (\frac {3 e}{2}\right )+\sin \left (\frac {3 e}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )-i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[e + f*x]],x]

[Out]

(a*c*(2*ArcTan[E^(I*(e + f*x))] - Sec[e + f*x])*((-I)*Cos[(3*e)/2] + Sin[(3*e)/2])*(Cos[(e + f*x)/2] - I*Sin[(
e + f*x)/2])*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*E^((I/2)*(4*e + f*x))*Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*f)

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Maple [A]
time = 0.35, size = 122, normalized size = 1.15

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )\right )}{f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\)	\(122\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, a \left (i \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}+a c \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}{\sqrt {a c}}\right )\right )}{f \sqrt {a c \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {a c}}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*a*(I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+a*c*l
n((c*a*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2)))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(a*c)^(
1/2)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 481 vs. \(2 (82) = 164\).
time = 0.60, size = 481, normalized size = 4.54 \begin {gather*} -\frac {{\left (2 \, {\left (a \cos \left (2 \, f x + 2 \, e\right ) + i \, a \sin \left (2 \, f x + 2 \, e\right ) + a\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 2 \, {\left (a \cos \left (2 \, f x + 2 \, e\right ) + i \, a \sin \left (2 \, f x + 2 \, e\right ) + a\right )} \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), -\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 4 \, a \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - {\left (-i \, a \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) - i \, a\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - {\left (i \, a \cos \left (2 \, f x + 2 \, e\right ) - a \sin \left (2 \, f x + 2 \, e\right ) + i \, a\right )} \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 4 i \, a \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a} \sqrt {c}}{-2 \, f {\left (i \, \cos \left (2 \, f x + 2 \, e\right ) - \sin \left (2 \, f x + 2 \, e\right ) + i\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-(2*(a*cos(2*f*x + 2*e) + I*a*sin(2*f*x + 2*e) + a)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)
)), sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 2*(a*cos(2*f*x + 2*e) + I*a*sin(2*f*x + 2*e) +
a)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x
+ 2*e))) + 1) - 4*a*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-I*a*cos(2*f*x + 2*e) + a*sin(2*f
*x + 2*e) - I*a)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e)
, cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - (I*a*cos(2*f*x + 2*e) -
a*sin(2*f*x + 2*e) + I*a)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*
f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 4*I*a*sin(1/2
*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/(f*(-2*I*cos(2*f*x + 2*e) + 2*sin(2*f*x + 2*e)
- 2*I))

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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. 304 vs. \(2 (82) = 164\).
time = 1.52, size = 304, normalized size = 2.87 \begin {gather*} -\frac {-4 i \, a \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} - \sqrt {\frac {a^{3} c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{3} c}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) + \sqrt {\frac {a^{3} c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{3} c}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right )}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

-1/2*(-4*I*a*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - sqrt(a^3*c/
f^2)*f*log(4*(2*(a*e^(3*I*f*x + 3*I*e) + a*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f
*x + 2*I*e) + 1)) - sqrt(a^3*c/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I*f))/(a*e^(2*I*f*x + 2*I*e) + a)) + sqrt(a^3*c
/f^2)*f*log(4*(2*(a*e^(3*I*f*x + 3*I*e) + a*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*
f*x + 2*I*e) + 1)) - sqrt(a^3*c/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) + I*f))/(a*e^(2*I*f*x + 2*I*e) + a)))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-I*c*tan(f*x+e))**(1/2)*(a+I*a*tan(f*x+e))**(3/2),x)

[Out]

Integral((I*a*(tan(e + f*x) - I))**(3/2)*sqrt(-I*c*(tan(e + f*x) + I)), 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((c-I*c*tan(f*x+e))^(1/2)*(a+I*a*tan(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(f*x + e) + a)^(3/2)*sqrt(-I*c*tan(f*x + e) + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(1/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^(3/2)*(c - c*tan(e + f*x)*1i)^(1/2), x)

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