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

Optimal. Leaf size=109 \[ \frac {2 \sqrt {a} B \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

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

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Rubi [A]  time = 0.22, antiderivative size = 109, 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} \[ \frac {2 \sqrt {a} B \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 4.45, size = 127, normalized size = 1.17 \[ \frac {\sqrt {a+i a \tan (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-i \sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )-i \cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (A+2 B \tan ^{-1}\left (e^{i (e+f x)}\right ) (\sin (e+f x)+i \cos (e+f x))-i B\right )}{f \sqrt {c-i c \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.53, size = 336, normalized size = 3.08 \[ -\frac {c f \sqrt {-\frac {B^{2} a}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt {-\frac {B^{2} a}{c f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) - c f \sqrt {-\frac {B^{2} a}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (B e^{\left (3 i \, f x + 3 i \, e\right )} + B e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} - c f\right )} \sqrt {-\frac {B^{2} a}{c f^{2}}}\right )}}{B e^{\left (2 i \, f x + 2 i \, e\right )} + B}\right ) - {\left ({\left (-2 i \, A - 2 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-2 i \, A - 2 \, B\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.61, size = 321, normalized size = 2.94 \[ -\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (-2 i B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \tan \left (f x +e \right ) a c -B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) \left (\tan ^{2}\left (f x +e \right )\right ) a c +i A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )+i B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}+B \ln \left (\frac {c a \tan \left (f x +e \right )+\sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}}{\sqrt {c a}}\right ) a c +B \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\, \tan \left (f x +e \right )-A \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \sqrt {c a}\right )}{f c \sqrt {c a \left (1+\tan ^{2}\left (f x +e \right )\right )}\, \left (\tan \left (f x +e \right )+i\right )^{2} \sqrt {c a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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mupad [B]  time = 12.41, size = 266, normalized size = 2.44 \[ \frac {4\,B\,\sqrt {a}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {a}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )}{\sqrt {c}\,f}+\frac {A\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{c\,f\,\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}-\frac {4\,B\,a\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{c\,f\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )\,\left (\frac {a}{c}-\frac {{\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}^2}{{\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}^2}+\frac {2\,\sqrt {a}\,\left (\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {a}\right )}{\sqrt {c}\,\left (\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}-\sqrt {c}\right )}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*B*a^(1/2)*atan((c^(1/2)*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(a^(1/2)*((c - c*tan(e + f*x)*1i)^(1/2)
- c^(1/2)))))/(c^(1/2)*f) + (A*(a + a*tan(e + f*x)*1i)^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(c*f*(tan(e + f*x)
+ 1i)) - (4*B*a*((a + a*tan(e + f*x)*1i)^(1/2) - a^(1/2)))/(c*f*((c - c*tan(e + f*x)*1i)^(1/2) - c^(1/2))*(a/
c - ((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 + (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 \[ \int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (A + B \tan {\left (e + f x \right )}\right )}{\sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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