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

Optimal. Leaf size=104 \[ \frac {(4 B+i A) (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3588, 78, 37} \[ \frac {(4 B+i A) (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 37

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] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -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)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx &=\frac {(a c) \operatorname {Subst}\left (\int \frac {(A+B x) \sqrt {c-i c x}}{(a+i a x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {((A-4 i B) c) \operatorname {Subst}\left (\int \frac {\sqrt {c-i c x}}{(a+i a x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(i A+4 B) (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 8.01, size = 92, normalized size = 0.88 \[ \frac {c (1-i \tan (e+f x)) \sqrt {c-i c \tan (e+f x)} ((A-4 i B) \tan (e+f x)-4 i A-B)}{15 a^2 f (\tan (e+f x)-i)^2 \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(1 - I*Tan[e + f*x])*((-4*I)*A - B + (A - (4*I)*B)*Tan[e + f*x])*Sqrt[c - I*c*Tan[e + f*x]])/(15*a^2*f*(-I
+ Tan[e + f*x])^2*Sqrt[a + I*a*Tan[e + f*x]])

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fricas [A]  time = 0.93, size = 97, normalized size = 0.93 \[ \frac {{\left ({\left (5 i \, A + 5 \, B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (8 i \, A + 2 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (3 i \, A - 3 \, B\right )} c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{30 \, a^{3} f} \]

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))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/30*((5*I*A + 5*B)*c*e^(4*I*f*x + 4*I*e) + (8*I*A + 2*B)*c*e^(2*I*f*x + 2*I*e) + (3*I*A - 3*B)*c)*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*f)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

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))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x + e) + a)^(5/2), x)

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maple [A]  time = 0.50, size = 92, normalized size = 0.88 \[ \frac {i \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (i A \tan \left (f x +e \right )-i B +4 B \tan \left (f x +e \right )+4 A \right )}{15 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x)

[Out]

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

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maxima [A]  time = 1.26, size = 150, normalized size = 1.44 \[ \frac {{\left (150 \, {\left (A - i \, B\right )} c \cos \left (4 \, f x + 4 \, e\right ) + 60 \, {\left (4 \, A - i \, B\right )} c \cos \left (2 \, f x + 2 \, e\right ) + {\left (150 i \, A + 150 \, B\right )} c \sin \left (4 \, f x + 4 \, e\right ) + {\left (240 i \, A + 60 \, B\right )} c \sin \left (2 \, f x + 2 \, e\right ) + 90 \, {\left (A + i \, B\right )} c\right )} \sqrt {a} \sqrt {c}}{{\left (-900 i \, a^{3} \cos \left (7 \, f x + 7 \, e\right ) - 900 i \, a^{3} \cos \left (5 \, f x + 5 \, e\right ) + 900 \, a^{3} \sin \left (7 \, f x + 7 \, e\right ) + 900 \, a^{3} \sin \left (5 \, f x + 5 \, e\right )\right )} f} \]

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))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

(150*(A - I*B)*c*cos(4*f*x + 4*e) + 60*(4*A - I*B)*c*cos(2*f*x + 2*e) + (150*I*A + 150*B)*c*sin(4*f*x + 4*e) +
(240*I*A + 60*B)*c*sin(2*f*x + 2*e) + 90*(A + I*B)*c)*sqrt(a)*sqrt(c)/((-900*I*a^3*cos(7*f*x + 7*e) - 900*I*a
^3*cos(5*f*x + 5*e) + 900*a^3*sin(7*f*x + 7*e) + 900*a^3*sin(5*f*x + 5*e))*f)

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mupad [B]  time = 11.07, size = 240, normalized size = 2.31 \[ \frac {c\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,8{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+5\,B\,\cos \left (2\,e+2\,f\,x\right )+2\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+5\,A\,\sin \left (2\,e+2\,f\,x\right )+8\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,2{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{60\,a^3\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^(3/2))/(a + a*tan(e + f*x)*1i)^(5/2),x)

[Out]

(c*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*((c*(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)*5i + A*cos(4*e + 4*f*x)*8i + A*cos(6*
e + 6*f*x)*3i + 5*B*cos(2*e + 2*f*x) + 2*B*cos(4*e + 4*f*x) - 3*B*cos(6*e + 6*f*x) + 5*A*sin(2*e + 2*f*x) + 8*
A*sin(4*e + 4*f*x) + 3*A*sin(6*e + 6*f*x) - B*sin(2*e + 2*f*x)*5i - B*sin(4*e + 4*f*x)*2i + B*sin(6*e + 6*f*x)
*3i))/(60*a^3*f)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

Integral((-I*c*(tan(e + f*x) + I))**(3/2)*(A + B*tan(e + f*x))/(I*a*(tan(e + f*x) - I))**(5/2), x)

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