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

[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]))

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Rubi [A]  time = 0.239907, antiderivative size = 180, normalized size of antiderivative = 1., 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 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]

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 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 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]

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*}

Mathematica [F]  time = 180.007, size = 0, normalized size = 0. \[ \text{\$Aborted} \]

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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Maple [A]  time = 0.101, size = 150, normalized size = 0.8 \begin{align*}{\frac{2\,ic}{af} \left ({\frac{i}{3}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+3\,iBc\sqrt{c-ic\tan \left ( fx+e \right ) }+Ac\sqrt{c-ic\tan \left ( fx+e \right ) }+4\,{c}^{2} \left ({\frac{ \left ( -A/4-i/4B \right ) \sqrt{c-ic\tan \left ( fx+e \right ) }}{-c-ic\tan \left ( fx+e \right ) }}-1/8\,{\frac{ \left ( 3\,A+7\,iB \right ) \sqrt{2}}{\sqrt{c}}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{c-ic\tan \left ( fx+e \right ) }\sqrt{2}}{\sqrt{c}}} \right ) } \right ) \right ) } \end{align*}

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)+A*c*(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [B]  time = 1.1653, size = 1057, normalized size = 5.87 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

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]

Exception raised: AttributeError

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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)