3.765 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx\)

Optimal. Leaf size=180 \[ -\frac {\sqrt {2} c^{5/2} (-7 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{a f}+\frac {c^2 (-7 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {c (-7 B+3 i A) (c-i c \tan (e+f x))^{3/2}}{6 a f}+\frac {(-B+i A) (c-i c \tan (e+f x))^{5/2}}{2 a f (1+i \tan (e+f x))} \]

[Out]

-(3*I*A-7*B)*c^(5/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a/f+(3*I*A-7*B)*c^2*(c-I*c*
tan(f*x+e))^(1/2)/a/f+1/6*(3*I*A-7*B)*c*(c-I*c*tan(f*x+e))^(3/2)/a/f+1/2*(I*A-B)*(c-I*c*tan(f*x+e))^(5/2)/a/f/
(1+I*tan(f*x+e))

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Rubi [A]  time = 0.24, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3588, 78, 50, 63, 208} \[ \frac {c^2 (-7 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{a f}-\frac {\sqrt {2} c^{5/2} (-7 B+3 i A) \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{a f}+\frac {c (-7 B+3 i A) (c-i c \tan (e+f x))^{3/2}}{6 a f}+\frac {(-B+i A) (c-i c \tan (e+f x))^{5/2}}{2 a f (1+i \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2))/(a + I*a*Tan[e + f*x]),x]

[Out]

-((Sqrt[2]*((3*I)*A - 7*B)*c^(5/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(a*f)) + (((3*I)*A -
7*B)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(a*f) + (((3*I)*A - 7*B)*c*(c - I*c*Tan[e + f*x])^(3/2))/(6*a*f) + ((I*A
- B)*(c - I*c*Tan[e + f*x])^(5/2))/(2*a*f*(1 + I*Tan[e + f*x]))

Rule 50

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 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)) (c-i c \tan (e+f x))^{5/2}}{a+i a \tan (e+f x)} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) (c-i c x)^{3/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{2 a f (1+i \tan (e+f x))}-\frac {((3 A+7 i B) c) \operatorname {Subst}\left (\int \frac {(c-i c x)^{3/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac {(3 i A-7 B) c (c-i c \tan (e+f x))^{3/2}}{6 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left ((3 A+7 i B) c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{2 f}\\ &=\frac {(3 i A-7 B) c^2 \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(3 i A-7 B) c (c-i c \tan (e+f x))^{3/2}}{6 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left ((3 A+7 i B) c^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(3 i A-7 B) c^2 \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(3 i A-7 B) c (c-i c \tan (e+f x))^{3/2}}{6 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{2 a f (1+i \tan (e+f x))}-\frac {\left (2 (3 i A-7 B) c^2\right ) \operatorname {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{f}\\ &=-\frac {\sqrt {2} (3 i A-7 B) c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{a f}+\frac {(3 i A-7 B) c^2 \sqrt {c-i c \tan (e+f x)}}{a f}+\frac {(3 i A-7 B) c (c-i c \tan (e+f x))^{3/2}}{6 a f}+\frac {(i A-B) (c-i c \tan (e+f x))^{5/2}}{2 a f (1+i \tan (e+f x))}\\ \end {align*}

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Mathematica [F]  time = 180.00, size = 0, normalized size = 0.00 \[ \text {\$Aborted} \]

Verification is Not applicable to the result.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2))/(a + I*a*Tan[e + f*x]),x]

[Out]

$Aborted

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fricas [B]  time = 1.20, size = 402, normalized size = 2.23 \[ \frac {3 \, {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {-\frac {{\left (72 \, A^{2} + 336 i \, A B - 392 \, B^{2}\right )} c^{5}}{a^{2} f^{2}}} \log \left (\frac {{\left ({\left (-12 i \, A + 28 \, B\right )} c^{3} + \sqrt {2} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {{\left (72 \, A^{2} + 336 i \, A B - 392 \, B^{2}\right )} c^{5}}{a^{2} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) - 3 \, {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {-\frac {{\left (72 \, A^{2} + 336 i \, A B - 392 \, B^{2}\right )} c^{5}}{a^{2} f^{2}}} \log \left (\frac {{\left ({\left (-12 i \, A + 28 \, B\right )} c^{3} - \sqrt {2} {\left (a f e^{\left (2 i \, f x + 2 i \, e\right )} + a f\right )} \sqrt {-\frac {{\left (72 \, A^{2} + 336 i \, A B - 392 \, B^{2}\right )} c^{5}}{a^{2} f^{2}}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-i \, f x - i \, e\right )}}{a f}\right ) + \sqrt {2} {\left ({\left (36 i \, A - 84 \, B\right )} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (48 i \, A - 112 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (12 i \, A - 12 \, B\right )} c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (a f e^{\left (4 i \, f x + 4 i \, e\right )} + a f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="fricas")

[Out]

1/12*(3*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*x + 2*I*e))*sqrt(-(72*A^2 + 336*I*A*B - 392*B^2)*c^5/(a^2*f^2)
)*log(((-12*I*A + 28*B)*c^3 + sqrt(2)*(a*f*e^(2*I*f*x + 2*I*e) + a*f)*sqrt(-(72*A^2 + 336*I*A*B - 392*B^2)*c^5
/(a^2*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(a*f)) - 3*(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2
*I*f*x + 2*I*e))*sqrt(-(72*A^2 + 336*I*A*B - 392*B^2)*c^5/(a^2*f^2))*log(((-12*I*A + 28*B)*c^3 - sqrt(2)*(a*f*
e^(2*I*f*x + 2*I*e) + a*f)*sqrt(-(72*A^2 + 336*I*A*B - 392*B^2)*c^5/(a^2*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1
)))*e^(-I*f*x - I*e)/(a*f)) + sqrt(2)*((36*I*A - 84*B)*c^2*e^(4*I*f*x + 4*I*e) + (48*I*A - 112*B)*c^2*e^(2*I*f
*x + 2*I*e) + (12*I*A - 12*B)*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))/(a*f*e^(4*I*f*x + 4*I*e) + a*f*e^(2*I*f*
x + 2*I*e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{i \, a \tan \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(5/2)/(I*a*tan(f*x + e) + a), x)

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maple [A]  time = 0.63, size = 150, normalized size = 0.83 \[ \frac {2 i c \left (\frac {i B \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+3 i B c \sqrt {c -i c \tan \left (f x +e \right )}+c A \sqrt {c -i c \tan \left (f x +e \right )}+4 c^{2} \left (\frac {\left (-\frac {A}{4}-\frac {i B}{4}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{-c -i c \tan \left (f x +e \right )}-\frac {\left (7 i B +3 A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{8 \sqrt {c}}\right )\right )}{f a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x)

[Out]

2*I/f/a*c*(1/3*I*B*(c-I*c*tan(f*x+e))^(3/2)+3*I*B*c*(c-I*c*tan(f*x+e))^(1/2)+c*A*(c-I*c*tan(f*x+e))^(1/2)+4*c^
2*((-1/4*A-1/4*I*B)*(c-I*c*tan(f*x+e))^(1/2)/(-c-I*c*tan(f*x+e))-1/8*(3*A+7*I*B)*2^(1/2)/c^(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 = 0.62, size = 167, normalized size = 0.93 \[ \frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (3 \, A + 7 i \, B\right )} c^{\frac {7}{2}} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a} - \frac {12 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + i \, B\right )} c^{4}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )} a - 2 \, a c} + \frac {4 \, {\left (i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} B c^{2} + 3 \, \sqrt {-i \, c \tan \left (f x + e\right ) + c} {\left (A + 3 i \, B\right )} c^{3}\right )}}{a}\right )}}{6 \, c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e)),x, algorithm="maxima")

[Out]

1/6*I*(3*sqrt(2)*(3*A + 7*I*B)*c^(7/2)*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 - 12*sqrt(-I*c*tan(f*x + e) + c)*(A + I*B)*c^4/((-I*c*tan(f*x + e) + c)*a - 2
*a*c) + 4*(I*(-I*c*tan(f*x + e) + c)^(3/2)*B*c^2 + 3*sqrt(-I*c*tan(f*x + e) + c)*(A + 3*I*B)*c^3)/a)/(c*f)

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mupad [B]  time = 1.26, size = 245, normalized size = 1.36 \[ \frac {2\,B\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a\,f\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )-2\,a\,c\,f}+\frac {A\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a\,f}-\frac {6\,B\,c^2\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{a\,f}-\frac {2\,B\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a\,f}-\frac {\sqrt {2}\,A\,{\left (-c\right )}^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,3{}\mathrm {i}}{a\,f}-\frac {\sqrt {2}\,B\,c^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {c}}\right )\,7{}\mathrm {i}}{a\,f}+\frac {A\,c^3\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a\,f\,\left (c+c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )} \]

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)^(5/2))/(a + a*tan(e + f*x)*1i),x)

[Out]

(2*B*c^3*(c - c*tan(e + f*x)*1i)^(1/2))/(a*f*(c - c*tan(e + f*x)*1i) - 2*a*c*f) + (A*c^2*(c - c*tan(e + f*x)*1
i)^(1/2)*2i)/(a*f) - (6*B*c^2*(c - c*tan(e + f*x)*1i)^(1/2))/(a*f) - (2*B*c*(c - c*tan(e + f*x)*1i)^(3/2))/(3*
a*f) - (2^(1/2)*A*(-c)^(5/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*3i)/(a*f) - (2^(1/2)
*B*c^(5/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2)*1i)/(2*c^(1/2)))*7i)/(a*f) + (A*c^3*(c - c*tan(e + f*x)
*1i)^(1/2)*2i)/(a*f*(c + c*tan(e + f*x)*1i))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e)),x)

[Out]

-I*(Integral(A*c**2*sqrt(-I*c*tan(e + f*x) + c)/(tan(e + f*x) - I), x) + Integral(-A*c**2*sqrt(-I*c*tan(e + f*
x) + c)*tan(e + f*x)**2/(tan(e + f*x) - I), x) + Integral(B*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan
(e + f*x) - I), x) + Integral(-B*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3/(tan(e + f*x) - I), x) + Int
egral(-2*I*A*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)/(tan(e + f*x) - I), x) + Integral(-2*I*B*c**2*sqrt(
-I*c*tan(e + f*x) + c)*tan(e + f*x)**2/(tan(e + f*x) - I), x))/a

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