3.775 \(\int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=159 \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(5 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {\sqrt {c} (5 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f} \]

[Out]

1/32*(3*I*A+5*B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*c^(1/2)/a^2/f*2^(1/2)+1/4*(I*A-B)*(c-I*
c*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))^2+1/16*(3*I*A+5*B)*(c-I*c*tan(f*x+e))^(1/2)/a^2/f/(1+I*tan(f*x+e))

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Rubi [A]  time = 0.21, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ \frac {(-B+i A) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(5 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {\sqrt {c} (5 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)*A + 5*B)*Sqrt[c]*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(16*Sqrt[2]*a^2*f) + ((I*A - B
)*Sqrt[c - I*c*Tan[e + f*x]])/(4*a^2*f*(1 + I*Tan[e + f*x])^2) + (((3*I)*A + 5*B)*Sqrt[c - I*c*Tan[e + f*x]])/
(16*a^2*f*(1 + I*Tan[e + f*x]))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, 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 208

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

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)}}{(a+i a \tan (e+f x))^2} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {A+B x}{(a+i a x)^3 \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)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {((3 A-5 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^2 \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {((3 A-5 i B) c) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{32 a f}\\ &=\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}+\frac {(3 i A+5 B) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{16 a f}\\ &=\frac {(3 i A+5 B) \sqrt {c} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a^2 f}+\frac {(i A-B) \sqrt {c-i c \tan (e+f x)}}{4 a^2 f (1+i \tan (e+f x))^2}+\frac {(3 i A+5 B) \sqrt {c-i c \tan (e+f x)}}{16 a^2 f (1+i \tan (e+f x))}\\ \end {align*}

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Mathematica [A]  time = 3.36, size = 206, normalized size = 1.30 \[ \frac {\sec (e+f x) (\cos (f x)+i \sin (f x))^2 (A+B \tan (e+f x)) \left (2 \cos (e+f x) (\cos (2 f x)-i \sin (2 f x)) \sqrt {c-i c \tan (e+f x)} ((-3 A+5 i B) \sin (e+f x)+(B+7 i A) \cos (e+f x))+\sqrt {2} \sqrt {c} (5 B+3 i A) (\cos (2 e)+i \sin (2 e)) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )\right )}{32 f (a+i a \tan (e+f x))^2 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(Cos[f*x] + I*Sin[f*x])^2*(A + B*Tan[e + f*x])*(Sqrt[2]*((3*I)*A + 5*B)*Sqrt[c]*ArcTanh[Sqrt[c -
I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(Cos[2*e] + I*Sin[2*e]) + 2*Cos[e + f*x]*(Cos[2*f*x] - I*Sin[2*f*x])*(((
7*I)*A + B)*Cos[e + f*x] + (-3*A + (5*I)*B)*Sin[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]))/(32*f*(A*Cos[e + f*x] +
B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^2)

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fricas [B]  time = 0.84, size = 362, normalized size = 2.28 \[ \frac {{\left (\sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} + {\left (3 i \, A + 5 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) - \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (9 \, A^{2} - 30 i \, A B - 25 \, B^{2}\right )} c}{a^{4} f^{2}}} - {\left (3 i \, A + 5 \, B\right )} c\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{2} f}\right ) + \sqrt {2} {\left ({\left (5 i \, A + 3 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (7 i \, A + B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 2 i \, A - 2 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{32 \, a^{2} f} \]

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))^2,x, algorithm="fricas")

[Out]

1/32*(sqrt(1/2)*a^2*f*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(sqrt(2)*sqrt
(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*
c/(a^4*f^2)) + (3*I*A + 5*B)*c)*e^(-I*f*x - I*e)/(a^2*f)) - sqrt(1/2)*a^2*f*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*
c/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(sqrt(2)*sqrt(1/2)*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt(c/(e^(2*
I*f*x + 2*I*e) + 1))*sqrt(-(9*A^2 - 30*I*A*B - 25*B^2)*c/(a^4*f^2)) - (3*I*A + 5*B)*c)*e^(-I*f*x - I*e)/(a^2*f
)) + sqrt(2)*((5*I*A + 3*B)*e^(4*I*f*x + 4*I*e) + (7*I*A + B)*e^(2*I*f*x + 2*I*e) + 2*I*A - 2*B)*sqrt(c/(e^(2*
I*f*x + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.60, size = 121, normalized size = 0.76 \[ -\frac {2 i c^{2} \left (\frac {\frac {\left (-5 i B +3 A \right ) \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{32 c}+\left (-\frac {5 A}{16}+\frac {3 i B}{16}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\left (-c -i c \tan \left (f x +e \right )\right )^{2}}-\frac {\left (-5 i B +3 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 c^{\frac {3}{2}}}\right )}{f \,a^{2}} \]

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))^2,x)

[Out]

-2*I/f/a^2*c^2*((1/32/c*(3*A-5*I*B)*(c-I*c*tan(f*x+e))^(3/2)+(-5/16*A+3/16*I*B)*(c-I*c*tan(f*x+e))^(1/2))/(-c-
I*c*tan(f*x+e))^2-1/64/c^(3/2)*(3*A-5*I*B)*2^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))

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maxima [A]  time = 0.66, size = 172, normalized size = 1.08 \[ -\frac {i \, {\left (\frac {\sqrt {2} {\left (3 \, A - 5 i \, B\right )} c^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{2}} + \frac {4 \, {\left ({\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (3 \, A - 5 i \, B\right )} c^{2} - 2 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (5 \, A - 3 i \, B\right )} c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} a^{2} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} a^{2} c + 4 \, a^{2} c^{2}}\right )}}{64 \, c f} \]

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))^2,x, algorithm="maxima")

[Out]

-1/64*I*(sqrt(2)*(3*A - 5*I*B)*c^(3/2)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) +
sqrt(-I*c*tan(f*x + e) + c)))/a^2 + 4*((-I*c*tan(f*x + e) + c)^(3/2)*(3*A - 5*I*B)*c^2 - 2*sqrt(-I*c*tan(f*x
+ e) + c)*(5*A - 3*I*B)*c^3)/((-I*c*tan(f*x + e) + c)^2*a^2 - 4*(-I*c*tan(f*x + e) + c)*a^2*c + 4*a^2*c^2))/(c
*f)

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mupad [B]  time = 9.18, size = 264, normalized size = 1.66 \[ -\frac {\frac {5\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{16}-\frac {3\,B\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{8}}{4\,a^2\,c^2\,f+a^2\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,a^2\,c\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}+\frac {\frac {A\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,5{}\mathrm {i}}{8\,a^2\,f}-\frac {A\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,3{}\mathrm {i}}{16\,a^2\,f}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2-4\,c\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )+4\,c^2}+\frac {\sqrt {2}\,A\,\sqrt {-c}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{32\,a^2\,f}+\frac {5\,\sqrt {2}\,B\,\sqrt {c}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{32\,a^2\,f} \]

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)^2,x)

[Out]

((A*c^2*(c - c*tan(e + f*x)*1i)^(1/2)*5i)/(8*a^2*f) - (A*c*(c - c*tan(e + f*x)*1i)^(3/2)*3i)/(16*a^2*f))/((c -
c*tan(e + f*x)*1i)^2 - 4*c*(c - c*tan(e + f*x)*1i) + 4*c^2) - ((5*B*c*(c - c*tan(e + f*x)*1i)^(3/2))/16 - (3*
B*c^2*(c - c*tan(e + f*x)*1i)^(1/2))/8)/(4*a^2*c^2*f + a^2*f*(c - c*tan(e + f*x)*1i)^2 - 4*a^2*c*f*(c - c*tan(
e + f*x)*1i)) + (2^(1/2)*A*(-c)^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*3i)/(32*a^2
*f) + (5*2^(1/2)*B*c^(1/2)*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(32*a^2*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {A \sqrt {- i c \tan {\left (e + f x \right )} + c}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx + \int \frac {B \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]

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))**2,x)

[Out]

-(Integral(A*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x) + Integral(B*sqrt(-I*c*t
an(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1), x))/a**2

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