3.802 \(\int \frac{(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}} \, dx\)

Optimal. Leaf size=155 \[ -\frac{(-5 B+2 i A) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac{(-5 B+2 i A) (a+i a \tan (e+f x))^{3/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.262802, antiderivative size = 155, normalized size of antiderivative = 1., 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} \[ -\frac{(-5 B+2 i A) (a+i a \tan (e+f x))^{3/2}}{105 c^2 f (c-i c \tan (e+f x))^{3/2}}-\frac{(-5 B+2 i A) (a+i a \tan (e+f x))^{3/2}}{35 c f (c-i c \tan (e+f x))^{5/2}}-\frac{(B+i A) (a+i a \tan (e+f x))^{3/2}}{7 f (c-i c \tan (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

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

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

Mathematica [A]  time = 12.8315, size = 131, normalized size = 0.85 \[ \frac{a \cos (e+f x) (\cos (f x)-i \sin (f x)) \sqrt{a+i a \tan (e+f x)} \sqrt{c-i c \tan (e+f x)} (\cos (5 e+6 f x)+i \sin (5 e+6 f x)) (-5 (2 A+5 i B) \sin (2 (e+f x))+5 (2 B-5 i A) \cos (2 (e+f x))-21 i A)}{210 c^4 f} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^(7/2),x]

[Out]

(a*Cos[e + f*x]*(Cos[f*x] - I*Sin[f*x])*((-21*I)*A + 5*((-5*I)*A + 2*B)*Cos[2*(e + f*x)] - 5*(2*A + (5*I)*B)*S
in[2*(e + f*x)])*(Cos[5*e + 6*f*x] + I*Sin[5*e + 6*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]]
)/(210*c^4*f)

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Maple [A]  time = 0.111, size = 113, normalized size = 0.7 \begin{align*}{\frac{{\frac{i}{105}}a \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) \left ( 5\,B-25\,iB\tan \left ( fx+e \right ) -5\,B \left ( \tan \left ( fx+e \right ) \right ) ^{2}-23\,iA-10\,A\tan \left ( fx+e \right ) +2\,iA \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }{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))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x)

[Out]

1/105*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(-1+I*tan(f*x+e)))^(1/2)*a/c^4*(1+tan(f*x+e)^2)*(5*B-25*I*B*tan(f*x+e
)-5*B*tan(f*x+e)^2-23*I*A-10*A*tan(f*x+e)+2*I*A*tan(f*x+e)^2)/(tan(f*x+e)+I)^5

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Maxima [A]  time = 2.51571, size = 254, normalized size = 1.64 \begin{align*} \frac{{\left (15 \,{\left (-i \, A - B\right )} a \cos \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 42 i \, A a \cos \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 35 \,{\left (-i \, A + B\right )} a \cos \left (\frac{3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (15 \, A - 15 i \, B\right )} a \sin \left (\frac{7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 42 \, A a \sin \left (\frac{5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) +{\left (35 \, A + 35 i \, B\right )} a \sin \left (\frac{3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt{a}}{420 \, c^{\frac{7}{2}} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

1/420*(15*(-I*A - B)*a*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 42*I*A*a*cos(5/2*arctan2(sin(2*f
*x + 2*e), cos(2*f*x + 2*e))) + 35*(-I*A + B)*a*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (15*A -
15*I*B)*a*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 42*A*a*sin(5/2*arctan2(sin(2*f*x + 2*e), cos
(2*f*x + 2*e))) + (35*A + 35*I*B)*a*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(7/2)*f)

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Fricas [A]  time = 1.27654, size = 355, normalized size = 2.29 \begin{align*} \frac{{\left ({\left (-15 i \, A - 15 \, B\right )} a e^{\left (8 i \, f x + 8 i \, e\right )} +{\left (-57 i \, A - 15 \, B\right )} a e^{\left (6 i \, f x + 6 i \, e\right )} +{\left (-77 i \, A + 35 \, B\right )} a e^{\left (4 i \, f x + 4 i \, e\right )} +{\left (-35 i \, A + 35 \, B\right )} a e^{\left (2 i \, f x + 2 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}} e^{\left (i \, f x + i \, e\right )}}{420 \, c^{4} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

1/420*((-15*I*A - 15*B)*a*e^(8*I*f*x + 8*I*e) + (-57*I*A - 15*B)*a*e^(6*I*f*x + 6*I*e) + (-77*I*A + 35*B)*a*e^
(4*I*f*x + 4*I*e) + (-35*I*A + 35*B)*a*e^(2*I*f*x + 2*I*e))*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)/(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))**(3/2)*(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 )}{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}{{\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))^(3/2)*(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)*(I*a*tan(f*x + e) + a)^(3/2)/(-I*c*tan(f*x + e) + c)^(7/2), x)