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

Optimal. Leaf size=155 \[ -\frac {(-3 B+2 i A) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {(-3 B+2 i A) \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3588, 78, 45, 37} \[ -\frac {(-3 B+2 i A) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f \sqrt {c-i c \tan (e+f x)}}-\frac {(-3 B+2 i A) \sqrt {a+i a \tan (e+f x)}}{15 c f (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{5 f (c-i c \tan (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 9.84, size = 114, normalized size = 0.74 \[ \frac {\cos (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)} (\cos (3 (e+f x))+i \sin (3 (e+f x))) (-3 (2 A+3 i B) \sin (2 (e+f x))+(6 B-9 i A) \cos (2 (e+f x))-5 i A)}{30 c^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.38, size = 111, normalized size = 0.72 \[ \frac {{\left ({\left (-3 i \, A - 3 \, B\right )} e^{\left (7 i \, f x + 7 i \, e\right )} + {\left (-13 i \, A - 3 \, B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} + {\left (-25 i \, A + 15 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-15 i \, A + 15 \, B\right )} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{60 \, c^{3} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*((-3*I*A - 3*B)*e^(7*I*f*x + 7*I*e) + (-13*I*A - 3*B)*e^(5*I*f*x + 5*I*e) + (-25*I*A + 15*B)*e^(3*I*f*x +
3*I*e) + (-15*I*A + 15*B)*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)
)/(c^3*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.44, size = 125, normalized size = 0.81 \[ -\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \left (\tan ^{3}\left (f x +e \right )\right )-12 i B \left (\tan ^{2}\left (f x +e \right )\right )-3 B \left (\tan ^{3}\left (f x +e \right )\right )-13 i A \tan \left (f x +e \right )-8 A \left (\tan ^{2}\left (f x +e \right )\right )+3 i B +12 B \tan \left (f x +e \right )+7 A \right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/c^3*(2*I*A*tan(f*x+e)^3-12*I*B*tan(f*x+e)^2-
3*B*tan(f*x+e)^3-13*I*A*tan(f*x+e)-8*A*tan(f*x+e)^2+3*I*B+12*B*tan(f*x+e)+7*A)/(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+I*a*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e))/(c-I*c*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 = 9.86, size = 171, normalized size = 1.10 \[ -\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,15{}\mathrm {i}-15\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,10{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}+3\,B\,\cos \left (4\,e+4\,f\,x\right )-10\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{60\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

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)^(1/2))/(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*15i - 15*B + A*cos(2*e +
2*f*x)*10i + A*cos(4*e + 4*f*x)*3i + 3*B*cos(4*e + 4*f*x) - 10*A*sin(2*e + 2*f*x) - 3*A*sin(4*e + 4*f*x) + B*s
in(4*e + 4*f*x)*3i))/(60*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 \[ \int \frac {\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

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

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