3.11.1 \(\int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2} \, dx\) [1001]

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

[Out]

-2*I*c^(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))*a^(1/2)/f-I*c*(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 = 0.10, 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 {2 i \sqrt {a} c^{3/2} \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 c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*Sqrt[a]*c^(3/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/f -
(I*c*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 \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2} \, dx &=\frac {(a c) \text {Subst}\left (\int \frac {\sqrt {c-i c x}}{\sqrt {a+i a x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}+\frac {\left (a c^2\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 c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {\left (2 i c^2\right ) \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 c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{f}-\frac {\left (2 i c^2\right ) \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 \sqrt {a} c^{3/2} \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 c \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 = 1.51, size = 100, normalized size = 0.94 \begin {gather*} -\frac {i \sqrt {2} c e^{-i (e+f x)} \sqrt {\frac {c}{1+e^{2 i (e+f x)}}} \left (e^{i (e+f x)}+\left (1+e^{2 i (e+f x)}\right ) \text {ArcTan}\left (e^{i (e+f x)}\right )\right ) \sqrt {a+i a \tan (e+f x)}}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.34, size = 122, normalized size = 1.15

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f*(-I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)+a*c*ln((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*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*c/(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. 277 vs. \(2 (82) = 164\).
time = 0.57, size = 277, normalized size = 2.61 \begin {gather*} -\frac {{\left (2 \, {\left (c \cos \left (2 \, f x + 2 \, e\right ) + i \, c \sin \left (2 \, f x + 2 \, e\right ) + c\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (c \cos \left (2 \, f x + 2 \, e\right ) + i \, c \sin \left (2 \, f x + 2 \, e\right ) + c\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + 4 \, c \cos \left (f x + e\right ) - {\left (-i \, c \cos \left (2 \, f x + 2 \, e\right ) + c \sin \left (2 \, f x + 2 \, e\right ) - i \, c\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - {\left (i \, c \cos \left (2 \, f x + 2 \, e\right ) - c \sin \left (2 \, f x + 2 \, e\right ) + i \, c\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) + 4 i \, c \sin \left (f x + e\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((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

-(2*(c*cos(2*f*x + 2*e) + I*c*sin(2*f*x + 2*e) + c)*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 2*(c*cos(2*f*x +
2*e) + I*c*sin(2*f*x + 2*e) + c)*arctan2(cos(f*x + e), -sin(f*x + e) + 1) + 4*c*cos(f*x + e) - (-I*c*cos(2*f*
x + 2*e) + c*sin(2*f*x + 2*e) - I*c)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - (I*c*cos(2*f*
x + 2*e) - c*sin(2*f*x + 2*e) + I*c)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) + 4*I*c*sin(f*x
+ 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 = 2.06, size = 304, normalized size = 2.87 \begin {gather*} -\frac {4 i \, c \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 c^{3}}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (c e^{\left (3 i \, f x + 3 i \, e\right )} + c 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 c^{3}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{c e^{\left (2 i \, f x + 2 i \, e\right )} + c}\right ) + \sqrt {\frac {a c^{3}}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (c e^{\left (3 i \, f x + 3 i \, e\right )} + c 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 c^{3}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{c e^{\left (2 i \, f x + 2 i \, e\right )} + c}\right )}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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