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

Optimal. Leaf size=223 \[ -\frac{-3 B+7 i A}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(-3 B+7 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a c^{5/2} f}-\frac{-3 B+7 i A}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{-3 B+7 i A}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{-B+i A}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

[Out]

(((7*I)*A - 3*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(16*Sqrt[2]*a*c^(5/2)*f) - ((7*I)*A -
3*B)/(20*a*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(2*a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2
)) - ((7*I)*A - 3*B)/(24*a*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((7*I)*A - 3*B)/(16*a*c^2*f*Sqrt[c - I*c*Tan[e
+ f*x]])

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Rubi [A]  time = 0.288617, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.116, Rules used = {3588, 78, 51, 63, 208} \[ -\frac{-3 B+7 i A}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(-3 B+7 i A) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a c^{5/2} f}-\frac{-3 B+7 i A}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{-3 B+7 i A}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{-B+i A}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((7*I)*A - 3*B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(16*Sqrt[2]*a*c^(5/2)*f) - ((7*I)*A -
3*B)/(20*a*f*(c - I*c*Tan[e + f*x])^(5/2)) + (I*A - B)/(2*a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2
)) - ((7*I)*A - 3*B)/(24*a*c*f*(c - I*c*Tan[e + f*x])^(3/2)) - ((7*I)*A - 3*B)/(16*a*c^2*f*Sqrt[c - I*c*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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{(a+i a x)^2 (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{((7 A+3 i B) c) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=-\frac{7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}+\frac{(7 A+3 i B) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}+\frac{(7 A+3 i B) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{16 c f}\\ &=-\frac{7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i A-3 B}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(7 A+3 i B) \operatorname{Subst}\left (\int \frac{1}{(a+i a x) \sqrt{c-i c x}} \, dx,x,\tan (e+f x)\right )}{32 c^2 f}\\ &=-\frac{7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i A-3 B}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}+\frac{(7 i A-3 B) \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 c^3 f}\\ &=\frac{(7 i A-3 B) \tanh ^{-1}\left (\frac{\sqrt{c-i c \tan (e+f x)}}{\sqrt{2} \sqrt{c}}\right )}{16 \sqrt{2} a c^{5/2} f}-\frac{7 i A-3 B}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac{i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac{7 i A-3 B}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac{7 i A-3 B}{16 a c^2 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 7.55901, size = 213, normalized size = 0.96 \[ -\frac{e^{-2 i (e+f x)} \sqrt{\frac{c}{1+e^{2 i (e+f x)}}} \left (\left (1+e^{2 i (e+f x)}\right ) \left (i A \left (116 e^{2 i (e+f x)}+32 e^{4 i (e+f x)}+6 e^{6 i (e+f x)}-15\right )+3 B \left (-8 e^{2 i (e+f x)}+4 e^{4 i (e+f x)}+2 e^{6 i (e+f x)}+5\right )\right )+15 (3 B-7 i A) e^{2 i (e+f x)} \sqrt{1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (e+f x)}}\right )\right )}{240 \sqrt{2} a c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[c/(1 + E^((2*I)*(e + f*x)))]*((1 + E^((2*I)*(e + f*x)))*(3*B*(5 - 8*E^((2*I)*(e + f*x)) + 4*E^((4*I)*(e
+ f*x)) + 2*E^((6*I)*(e + f*x))) + I*A*(-15 + 116*E^((2*I)*(e + f*x)) + 32*E^((4*I)*(e + f*x)) + 6*E^((6*I)*(
e + f*x)))) + 15*((-7*I)*A + 3*B)*E^((2*I)*(e + f*x))*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[Sqrt[1 + E^((2*I)*
(e + f*x))]]))/(240*Sqrt[2]*a*c^3*E^((2*I)*(e + f*x))*f)

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Maple [A]  time = 0.103, size = 168, normalized size = 0.8 \begin{align*}{\frac{2\,ic}{af} \left ( -{\frac{1}{16\,{c}^{3}} \left ({\frac{1}{-c-ic\tan \left ( fx+e \right ) } \left ({\frac{i}{2}}B+{\frac{A}{2}} \right ) \sqrt{c-ic\tan \left ( fx+e \right ) }}-{\frac{ \left ( 7\,A+3\,iB \right ) \sqrt{2}}{4}{\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 ) }-{\frac{A}{12\,{c}^{2}} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}-{\frac{3\,A+iB}{16\,{c}^{3}}{\frac{1}{\sqrt{c-ic\tan \left ( fx+e \right ) }}}}-{\frac{A-iB}{20\,c} \left ( c-ic\tan \left ( fx+e \right ) \right ) ^{-{\frac{5}{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*I/f/a*c*(-1/16/c^3*((1/2*I*B+1/2*A)*(c-I*c*tan(f*x+e))^(1/2)/(-c-I*c*tan(f*x+e))-1/4*(7*A+3*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)))-1/12*A/c^2/(c-I*c*tan(f*x+e))^(3/2)-1/16/c^3*(3*A+
I*B)/(c-I*c*tan(f*x+e))^(1/2)-1/20/c*(A-I*B)/(c-I*c*tan(f*x+e))^(5/2))

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.36938, size = 1112, normalized size = 4.99 \begin{align*} \frac{{\left (15 \, \sqrt{\frac{1}{2}} a c^{3} f \sqrt{-\frac{49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} + 7 i \, A - 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) - 15 \, \sqrt{\frac{1}{2}} a c^{3} f \sqrt{-\frac{49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac{{\left (\sqrt{2} \sqrt{\frac{1}{2}}{\left (a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + a c^{2} f\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt{-\frac{49 \, A^{2} + 42 i \, A B - 9 \, B^{2}}{a^{2} c^{5} f^{2}}} - 7 i \, A + 3 \, B\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) + \sqrt{2}{\left ({\left (-6 i \, A - 6 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-38 i \, A - 18 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-148 i \, A + 12 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-101 i \, A + 9 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 15 i \, A - 15 \, B\right )} \sqrt{\frac{c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{480 \, a c^{3} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*sqrt(1/2)*a*c^3*f*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/(a^2*c^5*f^2))*e^(2*I*f*x + 2*I*e)*log(1/8*(sqrt
(2)*sqrt(1/2)*(a*c^2*f*e^(2*I*f*x + 2*I*e) + a*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 + 42*I*A
*B - 9*B^2)/(a^2*c^5*f^2)) + 7*I*A - 3*B)*e^(-I*f*x - I*e)/(a*c^2*f)) - 15*sqrt(1/2)*a*c^3*f*sqrt(-(49*A^2 + 4
2*I*A*B - 9*B^2)/(a^2*c^5*f^2))*e^(2*I*f*x + 2*I*e)*log(-1/8*(sqrt(2)*sqrt(1/2)*(a*c^2*f*e^(2*I*f*x + 2*I*e) +
a*c^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(49*A^2 + 42*I*A*B - 9*B^2)/(a^2*c^5*f^2)) - 7*I*A + 3*B)*e^
(-I*f*x - I*e)/(a*c^2*f)) + sqrt(2)*((-6*I*A - 6*B)*e^(8*I*f*x + 8*I*e) + (-38*I*A - 18*B)*e^(6*I*f*x + 6*I*e)
+ (-148*I*A + 12*B)*e^(4*I*f*x + 4*I*e) + (-101*I*A + 9*B)*e^(2*I*f*x + 2*I*e) + 15*I*A - 15*B)*sqrt(c/(e^(2*
I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*c^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))/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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