3.794 \(\int \frac{\sqrt{a+i a \tan (e+f x)} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=208 \[ -\frac{2 (-4 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{105 c^3 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 (-4 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac{(-4 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{(B+i A) \sqrt{a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

[Out]

-((I*A + B)*Sqrt[a + I*a*Tan[e + f*x]])/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) - (((3*I)*A - 4*B)*Sqrt[a + I*a*Tan
[e + f*x]])/(35*c*f*(c - I*c*Tan[e + f*x])^(5/2)) - (2*((3*I)*A - 4*B)*Sqrt[a + I*a*Tan[e + f*x]])/(105*c^2*f*
(c - I*c*Tan[e + f*x])^(3/2)) - (2*((3*I)*A - 4*B)*Sqrt[a + I*a*Tan[e + f*x]])/(105*c^3*f*Sqrt[c - I*c*Tan[e +
f*x]])

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Rubi [A]  time = 0.276411, antiderivative size = 208, normalized size of antiderivative = 1., 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} \[ -\frac{2 (-4 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{105 c^3 f \sqrt{c-i c \tan (e+f x)}}-\frac{2 (-4 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac{(-4 B+3 i A) \sqrt{a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{(B+i A) \sqrt{a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

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])^(7/2),x]

[Out]

-((I*A + B)*Sqrt[a + I*a*Tan[e + f*x]])/(7*f*(c - I*c*Tan[e + f*x])^(7/2)) - (((3*I)*A - 4*B)*Sqrt[a + I*a*Tan
[e + f*x]])/(35*c*f*(c - I*c*Tan[e + f*x])^(5/2)) - (2*((3*I)*A - 4*B)*Sqrt[a + I*a*Tan[e + f*x]])/(105*c^2*f*
(c - I*c*Tan[e + f*x])^(3/2)) - (2*((3*I)*A - 4*B)*Sqrt[a + I*a*Tan[e + f*x]])/(105*c^3*f*Sqrt[c - I*c*Tan[e +
f*x]])

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]

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 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 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]

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))^{7/2}} \, dx &=\frac{(a c) \operatorname{Subst}\left (\int \frac{A+B x}{\sqrt{a+i a x} (c-i c x)^{9/2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(i A+B) \sqrt{a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}+\frac{(a (3 A+4 i B)) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+i a x} (c-i c x)^{7/2}} \, dx,x,\tan (e+f x)\right )}{7 f}\\ &=-\frac{(i A+B) \sqrt{a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac{(3 i A-4 B) \sqrt{a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}+\frac{(2 a (3 A+4 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 )}{35 c f}\\ &=-\frac{(i A+B) \sqrt{a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac{(3 i A-4 B) \sqrt{a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 (3 i A-4 B) \sqrt{a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}+\frac{(2 a (3 A+4 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 )}{105 c^2 f}\\ &=-\frac{(i A+B) \sqrt{a+i a \tan (e+f x)}}{7 f (c-i c \tan (e+f x))^{7/2}}-\frac{(3 i A-4 B) \sqrt{a+i a \tan (e+f x)}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{2 (3 i A-4 B) \sqrt{a+i a \tan (e+f x)}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac{2 (3 i A-4 B) \sqrt{a+i a \tan (e+f x)}}{105 c^3 f \sqrt{c-i c \tan (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 12.329, size = 136, normalized size = 0.65 \[ \frac{\cos (e+f x) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (\cos (4 (e+f x))+i \sin (4 (e+f x))) (-(3 A+4 i B) (7 \sin (e+f x)+15 \sin (3 (e+f x)))+7 (B-12 i A) \cos (e+f x)+15 (3 B-4 i A) \cos (3 (e+f x)))}{420 c^4 f} \]

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])^(7/2),x]

[Out]

(Cos[e + f*x]*(7*((-12*I)*A + B)*Cos[e + f*x] + 15*((-4*I)*A + 3*B)*Cos[3*(e + f*x)] - (3*A + (4*I)*B)*(7*Sin[
e + f*x] + 15*Sin[3*(e + f*x)]))*(Cos[4*(e + f*x)] + I*Sin[4*(e + f*x)])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I
*c*Tan[e + f*x]])/(420*c^4*f)

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Maple [A]  time = 0.129, size = 147, normalized size = 0.7 \begin{align*}{\frac{8\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{4}+30\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{3}+6\,A \left ( \tan \left ( fx+e \right ) \right ) ^{4}-84\,iB \left ( \tan \left ( fx+e \right ) \right ) ^{2}-40\,B \left ( \tan \left ( fx+e \right ) \right ) ^{3}-75\,iA\tan \left ( fx+e \right ) -63\,A \left ( \tan \left ( fx+e \right ) \right ) ^{2}+13\,iB+65\,B\tan \left ( fx+e \right ) +36\,A}{105\,f{c}^{4} \left ( \tan \left ( fx+e \right ) +i \right ) ^{5}}\sqrt{a \left ( 1+i\tan \left ( fx+e \right ) \right ) }\sqrt{-c \left ( -1+i\tan \left ( fx+e \right ) \right ) }} \end{align*}

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))^(7/2),x)

[Out]

1/105/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)/c^4*(8*I*B*tan(f*x+e)^4+30*I*A*tan(f*x+e)^3+6*
A*tan(f*x+e)^4-84*I*B*tan(f*x+e)^2-40*B*tan(f*x+e)^3-75*I*A*tan(f*x+e)-63*A*tan(f*x+e)^2+13*I*B+65*B*tan(f*x+e
)+36*A)/(tan(f*x+e)+I)^5

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

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))^(7/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 1.42906, size = 373, normalized size = 1.79 \begin{align*} \frac{{\left ({\left (-15 i \, A - 15 \, B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-78 i \, A - 36 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-168 i \, A + 14 \, B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-210 i \, A + 140 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 105 i \, A + 105 \, 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 )}}{840 \, c^{4} f} \end{align*}

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))^(7/2),x, algorithm="fricas")

[Out]

1/840*((-15*I*A - 15*B)*e^(8*I*f*x + 8*I*e) + (-78*I*A - 36*B)*e^(6*I*f*x + 6*I*e) + (-168*I*A + 14*B)*e^(4*I*
f*x + 4*I*e) + (-210*I*A + 140*B)*e^(2*I*f*x + 2*I*e) - 105*I*A + 105*B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqr
t(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e)/(c^4*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{7}{2}}}\,{d x} \end{align*}

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))^(7/2),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)^(7/2), x)