3.832 \(\int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{\sqrt {a+i a \tan (e+f x)}} \, dx\)

Optimal. Leaf size=110 \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {2 B \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3588, 78, 63, 217, 203} \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{f \sqrt {a+i a \tan (e+f x)}}-\frac {2 B \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {a} f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/Sqrt[a + I*a*Tan[e + f*x]],x]

[Out]

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

Rule 63

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 78

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

Rule 203

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

Rule 217

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 3588

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((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)*(A + B*x), x
], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 4.49, size = 152, normalized size = 1.38 \[ \frac {c \sec (e+f x) \left (\sin \left (\frac {1}{2} (e+f x)\right )+i \cos \left (\frac {1}{2} (e+f x)\right )\right ) \left ((A+i B) \left (\cos \left (\frac {1}{2} (e+f x)\right )-i \sin \left (\frac {1}{2} (e+f x)\right )\right )+2 i B \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \tan ^{-1}(\cos (e+f x)+i \sin (e+f x))\right )}{f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/Sqrt[a + I*a*Tan[e + f*x]],x]

[Out]

(c*Sec[e + f*x]*((A + I*B)*(Cos[(e + f*x)/2] - I*Sin[(e + f*x)/2]) + (2*I)*B*ArcTan[Cos[e + f*x] + I*Sin[e + f
*x]]*(Cos[(e + f*x)/2] + I*Sin[(e + f*x)/2]))*(I*Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))/(f*Sqrt[a + I*a*Tan[e +
f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

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fricas [B]  time = 0.63, size = 351, normalized size = 3.19 \[ \frac {{\left (a f \sqrt {-\frac {B^{2} c}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt {-\frac {B^{2} c}{a f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) - a f \sqrt {-\frac {B^{2} c}{a f^{2}}} e^{\left (i \, f x + i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} - a f\right )} \sqrt {-\frac {B^{2} c}{a f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) + {\left ({\left (2 i \, A - 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{2 \, a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*tan(f*x + e) + A)*sqrt(-I*c*tan(f*x + e) + c)/sqrt(I*a*tan(f*x + e) + a), x)

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maple [B]  time = 0.47, size = 323, normalized size = 2.94 \[ -\frac {i \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (-2 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c +B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +i A \sqrt {c a}\, \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \tan \left (f x +e \right )-B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}-B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{f a \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (-\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-I*c*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(1/2),x)

[Out]

-I/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a*(-2*I*B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2
))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*tan(f*x+e)*a*c+B*ln((c*a*tan(f*x+e)+(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2
))/(c*a)^(1/2))*tan(f*x+e)^2*a*c+I*A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)-B*ln((c*a*tan(f*x+e)+
(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a)^(1/2))*a*c+I*B*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)-B*(c*a
*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2)*tan(f*x+e)+A*(c*a*(1+tan(f*x+e)^2))^(1/2)*(c*a)^(1/2))/(c*a*(1+tan(f*x+e)
^2))^(1/2)/(-tan(f*x+e)+I)^2/(c*a)^(1/2)

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maxima [A]  time = 0.83, size = 137, normalized size = 1.25 \[ -\frac {{\left (2 \, B \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 2 \, B \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + {\left (-2 i \, A + 2 \, B\right )} \cos \left (f x + e\right ) + i \, B \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - i \, B \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (A + i \, B\right )} \sin \left (f x + e\right )\right )} \sqrt {c}}{2 \, \sqrt {a} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.16, size = 250, normalized size = 2.27 \[ \frac {A\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{f\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}-\frac {4\,B\,\sqrt {c}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {a}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )}{\sqrt {a}\,f}-\frac {4\,B\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{f\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )\,\left (-\frac {a}{c}+\frac {{\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}{{\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}^2}+\frac {2\,\sqrt {a}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {c}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(f*(a + a*tan(e + f*x)*1i)^(1/2)) - (4*B*c^(1/2)*atan((c^(1/2)*((a + a*ta
n(e + f*x)*1i)^(1/2) - a^(1/2)))/(a^(1/2)*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2)))))/(a^(1/2)*f) - (4*B*((a
+ a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(f*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2))*(((a + a*tan(e + f*x)*1i)^
(1/2) - a^(1/2))^2/((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2))^2 - a/c + (2*a^(1/2)*((a + a*tan(e + f*x)*1i)^(1/
2) - a^(1/2)))/(c^(1/2)*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2)))))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(-I*c*(tan(e + f*x) + I))*(A + B*tan(e + f*x))/sqrt(I*a*(tan(e + f*x) - I)), x)

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