3.780 \(\int \frac{(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=252 \[ \frac{5 c^3 (-13 B+i A) \sqrt{c-i c \tan (e+f x)}}{16 a^3 f}+\frac{5 c^2 (-13 B+i A) (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac{5 c^{7/2} (-13 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^3 f}-\frac{c (-13 B+i A) (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

[Out]

(-5*(I*A - 13*B)*c^(7/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(8*Sqrt[2]*a^3*f) + (5*(I*A -
13*B)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(16*a^3*f) + (5*(I*A - 13*B)*c^2*(c - I*c*Tan[e + f*x])^(3/2))/(48*a^3*f
*(1 + I*Tan[e + f*x])) - ((I*A - 13*B)*c*(c - I*c*Tan[e + f*x])^(5/2))/(24*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I
*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(6*a^3*f*(1 + I*Tan[e + f*x])^3)

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Rubi [A]  time = 0.271302, antiderivative size = 252, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ \frac{5 c^3 (-13 B+i A) \sqrt{c-i c \tan (e+f x)}}{16 a^3 f}+\frac{5 c^2 (-13 B+i A) (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac{5 c^{7/2} (-13 B+i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^3 f}-\frac{c (-13 B+i A) (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(-B+i A) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(7/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

(-5*(I*A - 13*B)*c^(7/2)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(8*Sqrt[2]*a^3*f) + (5*(I*A -
13*B)*c^3*Sqrt[c - I*c*Tan[e + f*x]])/(16*a^3*f) + (5*(I*A - 13*B)*c^2*(c - I*c*Tan[e + f*x])^(3/2))/(48*a^3*f
*(1 + I*Tan[e + f*x])) - ((I*A - 13*B)*c*(c - I*c*Tan[e + f*x])^(5/2))/(24*a^3*f*(1 + I*Tan[e + f*x])^2) + ((I
*A - B)*(c - I*c*Tan[e + f*x])^(7/2))/(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))^{7/2}}{(a+i a \tan (e+f x))^3} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{(A+B x) (c-i c x)^{5/2}}{(a+i a x)^4} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{((A+13 i B) c) \operatorname{Subst}\left (\int \frac{(c-i c x)^{5/2}}{(a+i a x)^3} \, dx,x,\tan (e+f x)\right )}{12 f}\\ &=-\frac{(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}+\frac{\left (5 (A+13 i B) c^2\right ) \operatorname{Subst}\left (\int \frac{(c-i c x)^{3/2}}{(a+i a x)^2} \, dx,x,\tan (e+f x)\right )}{48 a f}\\ &=\frac{5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac{(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left (5 (A+13 i B) c^3\right ) \operatorname{Subst}\left (\int \frac{\sqrt{c-i c x}}{a+i a x} \, dx,x,\tan (e+f x)\right )}{32 a^2 f}\\ &=\frac{5 (i A-13 B) c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f}+\frac{5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac{(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left (5 (A+13 i B) c^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 a^2 f}\\ &=\frac{5 (i A-13 B) c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f}+\frac{5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac{(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}-\frac{\left (5 (i A-13 B) c^3\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 )}{8 a^2 f}\\ &=-\frac{5 (i A-13 B) c^{7/2} \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{8 \sqrt{2} a^3 f}+\frac{5 (i A-13 B) c^3 \sqrt{c-i c \tan (e+f x)}}{16 a^3 f}+\frac{5 (i A-13 B) c^2 (c-i c \tan (e+f x))^{3/2}}{48 a^3 f (1+i \tan (e+f x))}-\frac{(i A-13 B) c (c-i c \tan (e+f x))^{5/2}}{24 a^3 f (1+i \tan (e+f x))^2}+\frac{(i A-B) (c-i c \tan (e+f x))^{7/2}}{6 a^3 f (1+i \tan (e+f x))^3}\\ \end{align*}

Mathematica [F]  time = 180.003, 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])^(7/2))/(a + I*a*Tan[e + f*x])^3,x]

[Out]

$Aborted

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Maple [A]  time = 0.119, size = 167, normalized size = 0.7 \begin{align*}{\frac{2\,i{c}^{3}}{f{a}^{3}} \left ( iB\sqrt{c-ic\tan \left ( fx+e \right ) }+c \left ({\frac{1}{ \left ( -c-ic\tan \left ( fx+e \right ) \right ) ^{3}} \left ( \left ( -{\frac{47\,i}{16}}B-{\frac{11\,A}{16}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}+ \left ({\frac{29\,i}{3}}Bc+{\frac{5\,Ac}{3}} \right ) \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}+ \left ( -{\frac{33\,i}{4}}B{c}^{2}-{\frac{5\,A{c}^{2}}{4}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) } \right ) }-{\frac{ \left ( 65\,iB+5\,A \right ) \sqrt{2}}{32}{\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 ) \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))^(7/2)/(a+I*a*tan(f*x+e))^3,x)

[Out]

2*I/f/a^3*c^3*(I*B*(c-I*c*tan(f*x+e))^(1/2)+c*(((-47/16*I*B-11/16*A)*(c-I*c*tan(f*x+e))^(5/2)+(29/3*I*B*c+5/3*
A*c)*(c-I*c*tan(f*x+e))^(3/2)+(-33/4*I*B*c^2-5/4*A*c^2)*(c-I*c*tan(f*x+e))^(1/2))/(-c-I*c*tan(f*x+e))^3-5/32*(
13*I*B+A)*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))^(7/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.58213, size = 1092, normalized size = 4.33 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{{\left (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left ({\left (-5 i \, A + 65 \, B\right )} c^{4} + \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 (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{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 )}}{4 \, a^{3} f}\right ) - 3 \, \sqrt{\frac{1}{2}} a^{3} f \sqrt{-\frac{{\left (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{a^{6} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (\frac{{\left ({\left (-5 i \, A + 65 \, B\right )} c^{4} - \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 (25 \, A^{2} + 650 i \, A B - 4225 \, B^{2}\right )} c^{7}}{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 )}}{4 \, a^{3} f}\right ) + \sqrt{2}{\left ({\left (15 i \, A - 195 \, B\right )} c^{3} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (5 i \, A - 65 \, B\right )} c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-2 i \, A + 26 \, B\right )} c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} +{\left (8 i \, A - 8 \, B\right )} c^{3}\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 )}}{48 \, 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))^(7/2)/(a+I*a*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

1/48*(3*sqrt(1/2)*a^3*f*sqrt(-(25*A^2 + 650*I*A*B - 4225*B^2)*c^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4*((-5*
I*A + 65*B)*c^4 + sqrt(2)*sqrt(1/2)*(a^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-(25*A^2 + 650*I*A*B - 4225*B^2)*
c^7/(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*sqrt(-(25*A^2
+ 650*I*A*B - 4225*B^2)*c^7/(a^6*f^2))*e^(6*I*f*x + 6*I*e)*log(1/4*((-5*I*A + 65*B)*c^4 - sqrt(2)*sqrt(1/2)*(a
^3*f*e^(2*I*f*x + 2*I*e) + a^3*f)*sqrt(-(25*A^2 + 650*I*A*B - 4225*B^2)*c^7/(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)*((15*I*A - 195*B)*c^3*e^(6*I*f*x + 6*I*e) + (5*I*A - 65*B)*c^3
*e^(4*I*f*x + 4*I*e) + (-2*I*A + 26*B)*c^3*e^(2*I*f*x + 2*I*e) + (8*I*A - 8*B)*c^3)*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))**(7/2)/(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{7}{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))^(7/2)/(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)^(7/2)/(I*a*tan(f*x + e) + a)^3, x)