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

Optimal. Leaf size=206 \[ \frac {8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {-B-i A}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \]

[Out]

8/15*A*tan(f*x+e)/a^2/c^2/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+1/5*(-I*A-B)/f/(a+I*a*tan(f*x+e)
)^(5/2)/(c-I*c*tan(f*x+e))^(5/2)+1/5*I*A/c/f/(a+I*a*tan(f*x+e))^(5/2)/(c-I*c*tan(f*x+e))^(3/2)+4/15*A*tan(f*x+
e)/a/c/f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2)

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Rubi [A]  time = 0.28, antiderivative size = 204, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3588, 78, 45, 40, 39} \[ \frac {8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}-\frac {B+i A}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(I*A + B)/(5*f*(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(5/2)) + ((I/5)*A)/(c*f*(a + I*a*Tan[e + f
*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2)) + (4*A*Tan[e + f*x])/(15*a*c*f*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*
Tan[e + f*x])^(3/2)) + (8*A*Tan[e + f*x])/(15*a^2*c^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])

Rule 39

Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> Simp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d
*x]), x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0]

Rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(x*(a + b*x)^(m + 1)*(c + d*x)^(m +
1))/(2*a*c*(m + 1)), x] + Dist[(2*m + 3)/(2*a*c*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(m + 1), x], x] /; F
reeQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && ILtQ[m + 3/2, 0]

Rule 45

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*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

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 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]

Rubi steps

\begin {align*} \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{5/2} (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)^{7/2} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {(a A) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{7/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {(4 A) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{5/2} (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=-\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {(8 A) \operatorname {Subst}\left (\int \frac {1}{(a+i a x)^{3/2} (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{15 a c f}\\ &=-\frac {i A+B}{5 f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2}}+\frac {i A}{5 c f (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}}+\frac {4 A \tan (e+f x)}{15 a c f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {8 A \tan (e+f x)}{15 a^2 c^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 11.71, size = 151, normalized size = 0.73 \[ \frac {\sec ^2(e+f x) \sqrt {c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-150 A \sin (e+f x)-25 A \sin (3 (e+f x))-3 A \sin (5 (e+f x))+30 B \cos (e+f x)+15 B \cos (3 (e+f x))+3 B \cos (5 (e+f x)))}{240 a^2 c^3 f (\tan (e+f x)-i)^2 \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]^2*(Cos[3*(e + f*x)] + I*Sin[3*(e + f*x)])*(30*B*Cos[e + f*x] + 15*B*Cos[3*(e + f*x)] + 3*B*Cos[5
*(e + f*x)] - 150*A*Sin[e + f*x] - 25*A*Sin[3*(e + f*x)] - 3*A*Sin[5*(e + f*x)])*Sqrt[c - I*c*Tan[e + f*x]])/(
240*a^2*c^3*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

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fricas [A]  time = 0.58, size = 182, normalized size = 0.88 \[ \frac {{\left ({\left (-3 i \, A - 3 \, B\right )} e^{\left (12 i \, f x + 12 i \, e\right )} + {\left (-28 i \, A - 18 \, B\right )} e^{\left (10 i \, f x + 10 i \, e\right )} + {\left (-175 i \, A - 45 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + 96 \, B e^{\left (7 i \, f x + 7 i \, e\right )} - 60 \, B e^{\left (6 i \, f x + 6 i \, e\right )} + 96 \, B e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (175 i \, A - 45 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (28 i \, A - 18 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{480 \, a^{3} c^{3} f} \]

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

[Out]

1/480*((-3*I*A - 3*B)*e^(12*I*f*x + 12*I*e) + (-28*I*A - 18*B)*e^(10*I*f*x + 10*I*e) + (-175*I*A - 45*B)*e^(8*
I*f*x + 8*I*e) + 96*B*e^(7*I*f*x + 7*I*e) - 60*B*e^(6*I*f*x + 6*I*e) + 96*B*e^(5*I*f*x + 5*I*e) + (175*I*A - 4
5*B)*e^(4*I*f*x + 4*I*e) + (28*I*A - 18*B)*e^(2*I*f*x + 2*I*e) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)
)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-5*I*f*x - 5*I*e)/(a^3*c^3*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}\,{d x} \]

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

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maple [A]  time = 0.34, size = 124, normalized size = 0.60 \[ \frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (8 A \left (\tan ^{7}\left (f x +e \right )\right )+28 A \left (\tan ^{5}\left (f x +e \right )\right )+35 A \left (\tan ^{3}\left (f x +e \right )\right )-3 B \left (\tan ^{2}\left (f x +e \right )\right )+15 A \tan \left (f x +e \right )-3 B \right )}{15 f \,c^{3} a^{3} \left (\tan \left (f x +e \right )+i\right )^{4} \left (-\tan \left (f x +e \right )+i\right )^{4}} \]

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

[Out]

1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/c^3/a^3*(8*A*tan(f*x+e)^7+28*A*tan(f*x+e)^5+35*
A*tan(f*x+e)^3-3*B*tan(f*x+e)^2+15*A*tan(f*x+e)-3*B)/(tan(f*x+e)+I)^4/(-tan(f*x+e)+I)^4

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maxima [B]  time = 0.79, size = 331, normalized size = 1.61 \[ -\frac {{\left ({\left (-150 i \, A + 30 \, B\right )} \cos \left (4 \, f x + 4 \, e\right ) + {\left (-25 i \, A + 15 \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) + 30 \, {\left (5 \, A + i \, B\right )} \sin \left (4 \, f x + 4 \, e\right ) + 5 \, {\left (5 \, A + 3 i \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, B\right )} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (25 i \, A + 15 \, B\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (150 i \, A + 30 \, B\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - {\left (30 \, {\left (5 \, A + i \, B\right )} \cos \left (4 \, f x + 4 \, e\right ) + 5 \, {\left (5 \, A + 3 i \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) - {\left (-150 i \, A + 30 \, B\right )} \sin \left (4 \, f x + 4 \, e\right ) - {\left (-25 i \, A + 15 \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) + 6 \, A\right )} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 5 \, {\left (5 \, A - 3 i \, B\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 30 \, {\left (5 \, A - i \, B\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )}{480 \, a^{\frac {5}{2}} c^{\frac {5}{2}} f} \]

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

[Out]

-1/480*(((-150*I*A + 30*B)*cos(4*f*x + 4*e) + (-25*I*A + 15*B)*cos(2*f*x + 2*e) + 30*(5*A + I*B)*sin(4*f*x + 4
*e) + 5*(5*A + 3*I*B)*sin(2*f*x + 2*e) + 6*B)*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (25*I*A +
15*B)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (150*I*A + 30*B)*cos(1/2*arctan2(sin(2*f*x + 2*e
), cos(2*f*x + 2*e))) - (30*(5*A + I*B)*cos(4*f*x + 4*e) + 5*(5*A + 3*I*B)*cos(2*f*x + 2*e) - (-150*I*A + 30*B
)*sin(4*f*x + 4*e) - (-25*I*A + 15*B)*sin(2*f*x + 2*e) + 6*A)*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*
e))) - 5*(5*A - 3*I*B)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 30*(5*A - I*B)*sin(1/2*arctan2(s
in(2*f*x + 2*e), cos(2*f*x + 2*e))))/(a^(5/2)*c^(5/2)*f)

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mupad [B]  time = 10.94, size = 249, normalized size = 1.21 \[ \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (175\,A\,\sin \left (2\,e+2\,f\,x\right )-30\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,125{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,22{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-45\,B\,\cos \left (2\,e+2\,f\,x\right )-18\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )-A\,150{}\mathrm {i}+28\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,12{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{480\,a^3\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(e + f*x))/((a + a*tan(e + f*x)*1i)^(5/2)*(c - c*tan(e + f*x)*1i)^(5/2)),x)

[Out]

(((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(2*e + 2*f*x)*125i - 30
*B - A*150i + A*cos(4*e + 4*f*x)*22i + A*cos(6*e + 6*f*x)*3i - 45*B*cos(2*e + 2*f*x) - 18*B*cos(4*e + 4*f*x) -
3*B*cos(6*e + 6*f*x) + 175*A*sin(2*e + 2*f*x) + 28*A*sin(4*e + 4*f*x) + 3*A*sin(6*e + 6*f*x) + B*sin(2*e + 2*
f*x)*15i + B*sin(4*e + 4*f*x)*12i + B*sin(6*e + 6*f*x)*3i))/(480*a^3*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2
*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

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))**(5/2)/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

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

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