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

Optimal. Leaf size=157 \[ \frac {-B+i A}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}-\frac {(-B+2 i A) \sqrt {a+i a \tan (e+f x)}}{3 a c f \sqrt {c-i c \tan (e+f x)}}-\frac {(-B+2 i A) \sqrt {a+i a \tan (e+f x)}}{3 a f (c-i c \tan (e+f x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.25, antiderivative size = 157, 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} \[ \frac {-B+i A}{f \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}-\frac {(-B+2 i A) \sqrt {a+i a \tan (e+f x)}}{3 a c f \sqrt {c-i c \tan (e+f x)}}-\frac {(-B+2 i A) \sqrt {a+i a \tan (e+f x)}}{3 a f (c-i c \tan (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 7.00, size = 103, normalized size = 0.66 \[ \frac {i \sqrt {c-i c \tan (e+f x)} (\cos (2 (e+f x))+i \sin (2 (e+f x))) ((B-2 i A) \sin (2 (e+f x))+(A+2 i B) \cos (2 (e+f x))-3 A)}{6 c^2 f \sqrt {a+i a \tan (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.97, size = 146, normalized size = 0.93 \[ \frac {{\left ({\left (-i \, A - B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + {\left (-7 i \, A - B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left (4 i \, A + 4 \, B\right )} e^{\left (3 i \, f x + 3 i \, e\right )} + {\left (-3 i \, A - 3 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (4 i \, A + 4 \, B\right )} e^{\left (i \, f x + i \, e\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-i \, f x - i \, e\right )}}{12 \, a c^{2} f} \]

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))^(1/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

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

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

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))^(1/2)/(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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maple [A]  time = 0.42, size = 151, normalized size = 0.96 \[ -\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (-1+i \tan \left (f x +e \right )\right )}\, \left (2 i A \left (\tan ^{4}\left (f x +e \right )\right )-i B \left (\tan ^{3}\left (f x +e \right )\right )-B \left (\tan ^{4}\left (f x +e \right )\right )+3 i A \left (\tan ^{2}\left (f x +e \right )\right )-2 A \left (\tan ^{3}\left (f x +e \right )\right )-i B \tan \left (f x +e \right )+i A -2 A \tan \left (f x +e \right )+B \right )}{3 f \,c^{2} a \left (\tan \left (f x +e \right )+i\right )^{3} \left (-\tan \left (f x +e \right )+i\right )^{2}} \]

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

[Out]

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

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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))/(a+I*a*tan(f*x+e))^(1/2)/(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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mupad [B]  time = 0.73, size = 146, normalized size = 0.93 \[ \frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (2\,A\,\sin \left (2\,e+2\,f\,x\right )+A\,\cos \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}-2\,B\,\cos \left (2\,e+2\,f\,x\right )-A\,3{}\mathrm {i}+B\,\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{6\,a\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]

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*cos(2*e + 2*f*x)*1i - A*3i
- 2*B*cos(2*e + 2*f*x) + 2*A*sin(2*e + 2*f*x) + B*sin(2*e + 2*f*x)*1i))/(6*a*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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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \tan {\left (e + f x \right )}}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}\, dx \]

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))**(1/2)/(c-I*c*tan(f*x+e))**(3/2),x)

[Out]

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

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