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

Optimal. Leaf size=212 \[ \frac {2 (2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{15 a^2 c f \sqrt {a+i a \tan (e+f x)}}-\frac {B+i A}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 (2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{15 a c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{5 c f (a+i a \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 212, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ \frac {2 (2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{15 a^2 c f \sqrt {a+i a \tan (e+f x)}}-\frac {B+i A}{f (a+i a \tan (e+f x))^{5/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 (2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{15 a c f (a+i a \tan (e+f x))^{3/2}}+\frac {(2 B+3 i A) \sqrt {c-i c \tan (e+f x)}}{5 c f (a+i a \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 6.78, size = 132, normalized size = 0.62 \[ -\frac {\sec (e+f x) \sqrt {c-i c \tan (e+f x)} (-i (3 A-2 i B) (5 \sin (e+f x)-3 \sin (3 (e+f x)))+(-30 A+5 i B) \cos (e+f x)+(6 A-9 i B) \cos (3 (e+f x)))}{60 a^2 c f (\tan (e+f x)-i) \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/60*(Sec[e + f*x]*((-30*A + (5*I)*B)*Cos[e + f*x] + (6*A - (9*I)*B)*Cos[3*(e + f*x)] - I*(3*A - (2*I)*B)*(5*
Sin[e + f*x] - 3*Sin[3*(e + f*x)]))*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*c*f*(-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[
e + f*x]])

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fricas [A]  time = 1.42, size = 158, normalized size = 0.75 \[ \frac {{\left ({\left (-15 i \, A - 15 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} + {\left (-48 i \, A + 8 \, B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} + 30 i \, A e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-48 i \, A + 8 \, B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (60 i \, A + 10 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (18 i \, A - 8 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, a^{3} c f} \]

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))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

1/120*((-15*I*A - 15*B)*e^(8*I*f*x + 8*I*e) + (-48*I*A + 8*B)*e^(7*I*f*x + 7*I*e) + 30*I*A*e^(6*I*f*x + 6*I*e)
+ (-48*I*A + 8*B)*e^(5*I*f*x + 5*I*e) + (60*I*A + 10*B)*e^(4*I*f*x + 4*I*e) + (18*I*A - 8*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*f)

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

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))^(1/2)/(a+I*a*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)*sqrt(-I*c*tan(f*x + e) + c)), x)

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maple [A]  time = 0.46, size = 186, normalized size = 0.88 \[ -\frac {\sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \left (4 i B \left (\tan ^{5}\left (f x +e \right )\right )+12 i A \left (\tan ^{4}\left (f x +e \right )\right )-6 A \left (\tan ^{5}\left (f x +e \right )\right )+2 i B \left (\tan ^{3}\left (f x +e \right )\right )+8 B \left (\tan ^{4}\left (f x +e \right )\right )+18 i A \left (\tan ^{2}\left (f x +e \right )\right )-3 A \left (\tan ^{3}\left (f x +e \right )\right )-2 i B \tan \left (f x +e \right )+7 B \left (\tan ^{2}\left (f x +e \right )\right )+6 i A +3 A \tan \left (f x +e \right )-B \right )}{15 f c \,a^{3} \left (\tan \left (f x +e \right )+i\right )^{2} \left (-\tan \left (f x +e \right )+i\right )^{4}} \]

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))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x)

[Out]

-1/15/f*(-c*(-1+I*tan(f*x+e)))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/c/a^3*(4*I*B*tan(f*x+e)^5+12*I*A*tan(f*x+e)^4-
6*A*tan(f*x+e)^5+2*I*B*tan(f*x+e)^3+8*B*tan(f*x+e)^4+18*I*A*tan(f*x+e)^2-3*A*tan(f*x+e)^3-2*I*B*tan(f*x+e)+7*B
*tan(f*x+e)^2+6*I*A+3*A*tan(f*x+e)-B)/(tan(f*x+e)+I)^2/(-tan(f*x+e)+I)^4

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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))^(1/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [B]  time = 10.72, size = 246, normalized size = 1.16 \[ \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (15\,B\,\cos \left (2\,e+2\,f\,x\right )-15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,45{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,15{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}-A\,15{}\mathrm {i}-5\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+45\,A\,\sin \left (2\,e+2\,f\,x\right )+15\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,15{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,5{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{120\,a^3\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

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)^(1/2)),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)*45i - 15*
B - A*15i + A*cos(4*e + 4*f*x)*15i + A*cos(6*e + 6*f*x)*3i + 15*B*cos(2*e + 2*f*x) - 5*B*cos(4*e + 4*f*x) - 3*
B*cos(6*e + 6*f*x) + 45*A*sin(2*e + 2*f*x) + 15*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)*5i + B*sin(6*e + 6*f*x)*3i))/(120*a^3*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 \[ \int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}}\, dx \]

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))**(1/2)/(a+I*a*tan(f*x+e))**(5/2),x)

[Out]

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

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