3.8.92 \(\int \frac {\sqrt {a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [792]

Optimal. Leaf size=102 \[ -\frac {(i A+B) \sqrt {a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}}-\frac {(i A-2 B) \sqrt {a+i a \tan (e+f x)}}{3 c f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 102, 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 = {3669, 79, 37} \begin {gather*} -\frac {(-2 B+i A) \sqrt {a+i a \tan (e+f x)}}{3 c f \sqrt {c-i c \tan (e+f x)}}-\frac {(B+i A) \sqrt {a+i a \tan (e+f x)}}{3 f (c-i c \tan (e+f x))^{3/2}} \end {gather*}

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])^(3/2),x]

[Out]

-1/3*((I*A + B)*Sqrt[a + I*a*Tan[e + f*x]])/(f*(c - I*c*Tan[e + f*x])^(3/2)) - ((I*A - 2*B)*Sqrt[a + I*a*Tan[e
+ f*x]])/(3*c*f*Sqrt[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 79

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] || L
tQ[p, n]))))

Rule 3669

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

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Mathematica [A]
time = 1.65, size = 101, normalized size = 0.99 \begin {gather*} \frac {\cos (e+f x) ((-2 i A+B) \cos (e+f x)-(A+2 i B) \sin (e+f x)) (\cos (2 (e+f x))+i \sin (2 (e+f x))) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{3 c^2 f} \end {gather*}

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])^(3/2),x]

[Out]

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

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Maple [A]
time = 0.42, size = 100, normalized size = 0.98

method	result	size
risch	\(-\frac {\sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (i A \,{\mathrm e}^{2 i \left (f x +e \right )}+B \,{\mathrm e}^{2 i \left (f x +e \right )}+3 i A -3 B \right )}{6 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(84\)
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i B \left (\tan ^{2}\left (f x +e \right )\right )+3 i A \tan \left (f x +e \right )+A \left (\tan ^{2}\left (f x +e \right )\right )-i B -3 B \tan \left (f x +e \right )-2 A \right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}\)	\(100\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 i B \left (\tan ^{2}\left (f x +e \right )\right )+3 i A \tan \left (f x +e \right )+A \left (\tan ^{2}\left (f x +e \right )\right )-i B -3 B \tan \left (f x +e \right )-2 A \right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}\)	\(100\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 7.75, size = 101, normalized size = 0.99 \begin {gather*} \frac {{\left ({\left (-i \, A - B\right )} e^{\left (5 i \, f x + 5 i \, e\right )} - 2 \, {\left (2 i \, A - B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} - 3 \, {\left (i \, A - 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}}}{6 \, c^{2} f} \end {gather*}

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))^(3/2),x, algorithm="fricas")

[Out]

1/6*((-I*A - B)*e^(5*I*f*x + 5*I*e) - 2*(2*I*A - B)*e^(3*I*f*x + 3*I*e) - 3*(I*A - 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^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {3}{2}}}\, dx \end {gather*}

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))**(3/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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))^(3/2),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)^(3/2), x)

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Mupad [B]
time = 1.24, size = 145, normalized size = 1.42 \begin {gather*} -\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\,3{}\mathrm {i}-3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+B\,\cos \left (2\,e+2\,f\,x\right )-A\,\sin \left (2\,e+2\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{6\,c\,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}}} \end {gather*}

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)^(3/2),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*3i - 3*B + A*cos(2*e + 2*
f*x)*1i + B*cos(2*e + 2*f*x) - A*sin(2*e + 2*f*x) + B*sin(2*e + 2*f*x)*1i))/(6*c*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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