3.779 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=291 \[ \frac{35 c^4 (-5 B+i A) \sqrt{c-i c \tan (e+f x)}}{8 a^3 f}+\frac{35 c^3 (-5 B+i A) (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac{7 c^2 (-5 B+i A) (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac{35 c^{9/2} (-5 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a^3 f}-\frac{c (-5 B+i A) (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(-B+i A) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

(-35*(I*A - 5*B)*c^(9/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(4*Sqrt[2]*a^3*f) + (35*(I*A -
5*B)*c^4*Sqrt[c - I*c*Tan[e + f*x]])/(8*a^3*f) + (35*(I*A - 5*B)*c^3*(c - I*c*Tan[e + f*x])^(3/2))/(48*a^3*f)
+ (7*(I*A - 5*B)*c^2*(c - I*c*Tan[e + f*x])^(5/2))/(16*a^3*f*(1 + I*Tan[e + f*x])) - ((I*A - 5*B)*c*(c - I*c*
Tan[e + f*x])^(7/2))/(8*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(9/2))/(6*a^3*f*(1 +
I*Tan[e + f*x])^3)

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Rubi [A]  time = 0.304555, antiderivative size = 291, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {3588, 78, 47, 50, 63, 208} \[ \frac{35 c^4 (-5 B+i A) \sqrt{c-i c \tan (e+f x)}}{8 a^3 f}+\frac{35 c^3 (-5 B+i A) (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac{7 c^2 (-5 B+i A) (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac{35 c^{9/2} (-5 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a^3 f}-\frac{c (-5 B+i A) (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(-B+i A) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(9/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(-35*(I*A - 5*B)*c^(9/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(4*Sqrt[2]*a^3*f) + (35*(I*A -
5*B)*c^4*Sqrt[c - I*c*Tan[e + f*x]])/(8*a^3*f) + (35*(I*A - 5*B)*c^3*(c - I*c*Tan[e + f*x])^(3/2))/(48*a^3*f)
+ (7*(I*A - 5*B)*c^2*(c - I*c*Tan[e + f*x])^(5/2))/(16*a^3*f*(1 + I*Tan[e + f*x])) - ((I*A - 5*B)*c*(c - I*c*
Tan[e + f*x])^(7/2))/(8*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I*A - B)*(c - I*c*Tan[e + f*x])^(9/2))/(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 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))^{9/2}}{(a+i a \tan (e+f x))^3} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{7/2}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{((A+5 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{7/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac{(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{\left (7 (A+5 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{5/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{16 a f}\\ &=\frac{7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac{(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left (35 (A+5 i B) c^3\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{32 a^2 f}\\ &=\frac{35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac{7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac{(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left (35 (A+5 i B) c^4\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}\\ &=\frac{35 (i A-5 B) c^4 \sqrt{c-i c \tan (e+f x)}}{8 a^3 f}+\frac{35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac{7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac{(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left (35 (A+5 i B) c^5\right ) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{8 a^2 f}\\ &=\frac{35 (i A-5 B) c^4 \sqrt{c-i c \tan (e+f x)}}{8 a^3 f}+\frac{35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac{7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac{(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left (35 (i A-5 B) c^4\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 )}{4 a^2 f}\\ &=-\frac{35 (i A-5 B) c^{9/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{4 \sqrt{2} a^3 f}+\frac{35 (i A-5 B) c^4 \sqrt{c-i c \tan (e+f x)}}{8 a^3 f}+\frac{35 (i A-5 B) c^3 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f}+\frac{7 (i A-5 B) c^2 (c-i c \tan (e+f x))^{5/2}}{16 a^3 f (1+i \tan (e+f x))}-\frac{(i A-5 B) c (c-i c \tan (e+f x))^{7/2}}{8 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{9/2}}{6 a^3 f (1+i \tan (e+f x))^3}\\ \end{align*}

Mathematica [F]  time = 180.006, 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])^(9/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

$Aborted

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Maple [A]  time = 0.116, size = 206, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{3}}{f{a}^{3}} \left ({\frac{i}{3}}B \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+7\,iBc\sqrt{c-ic\tan \left ( fx+e \right ) }+Ac\sqrt{c-ic\tan \left ( fx+e \right ) }+8\,{c}^{2} \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( \left ( -{\frac{81\,i}{64}}B-{\frac{29\,A}{64}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{5/2}+ \left ({\frac{53\,i}{12}}Bc+{\frac{17\,Ac}{12}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{3/2}+ \left ( -{\frac{63\,i}{16}}B{c}^{2}-{\frac{19\,A{c}^{2}}{16}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 35\,A+175\,iB \right ) \sqrt{2}}{128\,\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))^(9/2)/(a+I*a*tan(f*x+e))^3,x)

[Out]

2*I/f/a^3*c^3*(1/3*I*B*(c-I*c*tan(f*x+e))^(3/2)+7*I*B*c*(c-I*c*tan(f*x+e))^(1/2)+A*c*(c-I*c*tan(f*x+e))^(1/2)+
8*c^2*(((-81/64*I*B-29/64*A)*(c-I*c*tan(f*x+e))^(5/2)+(53/12*I*B*c+17/12*A*c)*(c-I*c*tan(f*x+e))^(3/2)+(-63/16
*I*B*c^2-19/16*A*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-35/128*(A+5*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))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.60265, size = 1305, normalized size = 4.48 \begin{align*} \frac{3 \, \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt{-\frac{{\left (1225 \, A^{2} + 12250 i \, A B - 30625 \, B^{2}\right )} c^{9}}{a^{6} f^{2}}} \log \left (\frac{{\left ({\left (-35 i \, A + 175 \, B\right )} c^{5} + \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{{\left (1225 \, A^{2} + 12250 i \, A B - 30625 \, B^{2}\right )} c^{9}}{a^{6} 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 )}}{2 \, a^{3} f}\right ) - 3 \, \sqrt{\frac{1}{2}}{\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 i \, e\right )}\right )} \sqrt{-\frac{{\left (1225 \, A^{2} + 12250 i \, A B - 30625 \, B^{2}\right )} c^{9}}{a^{6} f^{2}}} \log \left (\frac{{\left ({\left (-35 i \, A + 175 \, B\right )} c^{5} - \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{{\left (1225 \, A^{2} + 12250 i \, A B - 30625 \, B^{2}\right )} c^{9}}{a^{6} 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 )}}{2 \, a^{3} f}\right ) + \sqrt{2}{\left ({\left (105 i \, A - 525 \, B\right )} c^{4} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (140 i \, A - 700 \, B\right )} c^{4} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (21 i \, A - 105 \, B\right )} c^{4} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-6 i \, A + 30 \, B\right )} c^{4} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (8 i \, A - 8 \, B\right )} c^{4}\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{24 \,{\left (a^{3} f e^{\left (8 i \, f x + 8 i \, e\right )} + a^{3} f e^{\left (6 i \, f x + 6 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))^(9/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/24*(3*sqrt(1/2)*(a^3*f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*I*e))*sqrt(-(1225*A^2 + 12250*I*A*B - 3062
5*B^2)*c^9/(a^6*f^2))*log(1/2*((-35*I*A + 175*B)*c^5 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*s
qrt(-(1225*A^2 + 12250*I*A*B - 30625*B^2)*c^9/(a^6*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/(
a^3*f)) - 3*sqrt(1/2)*(a^3*f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*I*e))*sqrt(-(1225*A^2 + 12250*I*A*B -
30625*B^2)*c^9/(a^6*f^2))*log(1/2*((-35*I*A + 175*B)*c^5 - sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*
f)*sqrt(-(1225*A^2 + 12250*I*A*B - 30625*B^2)*c^9/(a^6*f^2))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*
e)/(a^3*f)) + sqrt(2)*((105*I*A - 525*B)*c^4*e^(8*I*f*x + 8*I*e) + (140*I*A - 700*B)*c^4*e^(6*I*f*x + 6*I*e) +
(21*I*A - 105*B)*c^4*e^(4*I*f*x + 4*I*e) + (-6*I*A + 30*B)*c^4*e^(2*I*f*x + 2*I*e) + (8*I*A - 8*B)*c^4)*sqrt(
c/(e^(2*I*f*x + 2*I*e) + 1)))/(a^3*f*e^(8*I*f*x + 8*I*e) + a^3*f*e^(6*I*f*x + 6*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))**(9/2)/(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{9}{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))^(9/2)/(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)^(9/2)/(I*a*tan(f*x + e) + a)^3, x)