3.781 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=213 \[ -\frac{c^2 (11 B+i A) \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac{c^{5/2} (11 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^3 f}+\frac{c (11 B+i A) (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

((I*A + 11*B)*c^(5/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(16*Sqrt[2]*a^3*f) - ((I*A + 11*B
)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(16*a^3*f*(1 + I*Tan[e + f*x])) + ((I*A + 11*B)*c*(c - I*c*Tan[e + f*x])^(3/
2))/(24*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(5/2))/(6*a^3*f*(1 + I*Tan[e + f*x])
^3)

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Rubi [A]  time = 0.248603, antiderivative size = 213, 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, 47, 63, 208} \[ -\frac{c^2 (11 B+i A) \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac{c^{5/2} (11 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^3 f}+\frac{c (11 B+i A) (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(-B+i A) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

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])^3,x]

[Out]

((I*A + 11*B)*c^(5/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(16*Sqrt[2]*a^3*f) - ((I*A + 11*B
)*c^2*Sqrt[c - I*c*Tan[e + f*x]])/(16*a^3*f*(1 + I*Tan[e + f*x])) + ((I*A + 11*B)*c*(c - I*c*Tan[e + f*x])^(3/
2))/(24*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(5/2))/(6*a^3*f*(1 + I*Tan[e + f*x])
^3)

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 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))^3} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{3/2}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{((A-11 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=\frac{(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left ((A-11 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=-\frac{(i A+11 B) c^2 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac{(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{\left ((A-11 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 )}{32 a^2 f}\\ &=-\frac{(i A+11 B) c^2 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac{(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{\left ((i A+11 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 )}{16 a^2 f}\\ &=\frac{(i A+11 B) c^{5/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a^3 f}-\frac{(i A+11 B) c^2 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f (1+i \tan (e+f x))}+\frac{(i A+11 B) c (c-i c \tan (e+f x))^{3/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{5/2}}{6 a^3 f (1+i \tan (e+f x))^3}\\ \end{align*}

Mathematica [A]  time = 7.38131, size = 227, normalized size = 1.07 \[ \frac{\sec ^2(e+f x) (\cos (f x)+i \sin (f x))^3 (A+B \tan (e+f x)) \left (\frac{2}{3} c^2 \cos (e+f x) (\cos (3 f x)-i \sin (3 f x)) \sqrt{c-i c \tan (e+f x)} ((11 A-25 i B) \sin (2 (e+f x))+(-41 B+5 i A) \cos (2 (e+f x))+2 i A+22 B)+\sqrt{2} c^{5/2} (11 B+i A) (\cos (3 e)+i \sin (3 e)) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )\right )}{32 f (a+i a \tan (e+f x))^3 (A \cos (e+f x)+B \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(Sec[e + f*x]^2*(Cos[f*x] + I*Sin[f*x])^3*(A + B*Tan[e + f*x])*(Sqrt[2]*(I*A + 11*B)*c^(5/2)*ArcTanh[Sqrt[c -
I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])]*(Cos[3*e] + I*Sin[3*e]) + (2*c^2*Cos[e + f*x]*(Cos[3*f*x] - I*Sin[3*f*x])
*((2*I)*A + 22*B + ((5*I)*A - 41*B)*Cos[2*(e + f*x)] + (11*A - (25*I)*B)*Sin[2*(e + f*x)])*Sqrt[c - I*c*Tan[e
+ f*x]])/3))/(32*f*(A*Cos[e + f*x] + B*Sin[e + f*x])*(a + I*a*Tan[e + f*x])^3)

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Maple [A]  time = 0.11, size = 146, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{3}}{f{a}^{3}} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( \left ( -{\frac{21\,i}{32}}B-{\frac{A}{32}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}+ \left ({\frac{11\,i}{6}}Bc-{\frac{Ac}{6}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ( -{\frac{11\,i}{8}}B{c}^{2}+{\frac{A{c}^{2}}{8}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }+{\frac{ \left ( -11\,iB+A \right ) \sqrt{2}}{64}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{c-ic\tan \left ( fx+e \right ) }{\frac{1}{\sqrt{c}}}} \right ){\frac{1}{\sqrt{c}}}} \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))^3,x)

[Out]

2*I/f/a^3*c^3*(((-21/32*I*B-1/32*A)*(c-I*c*tan(f*x+e))^(5/2)+(11/6*I*B*c-1/6*A*c)*(c-I*c*tan(f*x+e))^(3/2)+(-1
1/8*I*B*c^2+1/8*A*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3+1/64*(-11*I*B+A)*2^(1/2)/c^(1/2)*arctan
h(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))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.47521, size = 1054, normalized size = 4.95 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left ({\left (i \, A + 11 \, B\right )} c^{3} + \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left ({\left (i \, A + 11 \, B\right )} c^{3} - \sqrt{2} \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{{\left (A^{2} - 22 i \, A B - 121 \, B^{2}\right )} c^{5}}{a^{6} f^{2}}}\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a^{3} f}\right ) + \sqrt{2}{\left ({\left (-3 i \, A - 33 \, B\right )} c^{2} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-i \, A - 11 \, B\right )} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (10 i \, A + 14 \, B\right )} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (8 i \, A - 8 \, B\right )} c^{2}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{96 \, a^{3} f} \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))^3,x, algorithm="fricas")

[Out]

1/96*(3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 22*I*A*B - 121*B^2)*c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*((I*A + 11
*B)*c^3 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(A^2 -
22*I*A*B - 121*B^2)*c^5/(a^6*f^2)))*e^(-I*f*x - I*e)/(a^3*f)) - 3*sqrt(1/2)*a^3*f*sqrt(-(A^2 - 22*I*A*B - 121
*B^2)*c^5/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/8*((I*A + 11*B)*c^3 - sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I
*e) + a^3*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(A^2 - 22*I*A*B - 121*B^2)*c^5/(a^6*f^2)))*e^(-I*f*x - I*
e)/(a^3*f)) + sqrt(2)*((-3*I*A - 33*B)*c^2*e^(6*I*f*x + 6*I*e) + (-I*A - 11*B)*c^2*e^(4*I*f*x + 4*I*e) + (10*I
*A + 14*B)*c^2*e^(2*I*f*x + 2*I*e) + (8*I*A - 8*B)*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-6*I*f*x - 6*I*e
)/(a^3*f)

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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))**3,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}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3}}\,{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))^3,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)^3, x)