3.7.21 \(\int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx\) [621]

Optimal. Leaf size=224 \[ -\frac {i (i a-b)^{3/2} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (i a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}} \]

[Out]

-I*(I*a-b)^(3/2)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+I*(I*a+b)^(3/2)*arctanh((I*a+
b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))/d+2/5*(5*a^2-b^2)*(a+b*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(1/2
)-2/5*a*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(5/2)-4/5*b*(a+b*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(3/2)

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Rubi [A]
time = 0.61, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3648, 3730, 3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {i (-b+i a)^{3/2} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}+\frac {i (b+i a)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),x]

[Out]

((-I)*(I*a - b)^(3/2)*ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d + (I*(I*a + b)^(3
/2)*ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - (2*a*Sqrt[a + b*Tan[c + d*x]])/(
5*d*Tan[c + d*x]^(5/2)) - (4*b*Sqrt[a + b*Tan[c + d*x]])/(5*d*Tan[c + d*x]^(3/2)) + (2*(5*a^2 - b^2)*Sqrt[a +
b*Tan[c + d*x]])/(5*a*d*Sqrt[Tan[c + d*x]])

Rule 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3648

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n - 1)/(f*(m + 1)*(a^2 + b^2))), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*Tan[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d
^2, 0] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*m]

Rule 3696

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^2/f, Subst[Int[(a + b*x)^m*((c + d*x)^n/(A - B*x)), x], x, Tan[e
+ f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]

Rule 3697

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 + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]

Rule 3730

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Ta
n[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2))), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] &&  !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))

Rubi steps

\begin {align*} \int \frac {(a+b \tan (c+d x))^{3/2}}{\tan ^{\frac {7}{2}}(c+d x)} \, dx &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {-3 a b+\frac {5}{2} \left (a^2-b^2\right ) \tan (c+d x)+2 a b \tan ^2(c+d x)}{\tan ^{\frac {5}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {-\frac {3}{4} a \left (5 a^2-b^2\right )-\frac {15}{2} a^2 b \tan (c+d x)-3 a b^2 \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx}{15 a}\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {8 \int \frac {\frac {15 a^3 b}{4}-\frac {15}{8} a^2 \left (a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {1}{2} \left (i (a-i b)^2\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx+\frac {1}{2} \left (i (a+i b)^2\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {\left (i (a-i b)^2\right ) \text {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac {\left (i (a+i b)^2\right ) \text {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}-\frac {\left (i (a-i b)^2\right ) \text {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {\left (i (a+i b)^2\right ) \text {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\\ &=-\frac {i (i a-b)^{3/2} \tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}+\frac {i (i a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {5}{2}}(c+d x)}-\frac {4 b \sqrt {a+b \tan (c+d x)}}{5 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (5 a^2-b^2\right ) \sqrt {a+b \tan (c+d x)}}{5 a d \sqrt {\tan (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.69, size = 197, normalized size = 0.88 \begin {gather*} \frac {-5 \sqrt [4]{-1} (-a+i b)^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+5 \sqrt [4]{-1} (a+i b)^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+\frac {2 \sqrt {a+b \tan (c+d x)} \left (-a^2-2 a b \tan (c+d x)+\left (5 a^2-b^2\right ) \tan ^2(c+d x)\right )}{a \tan ^{\frac {5}{2}}(c+d x)}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])^(3/2)/Tan[c + d*x]^(7/2),x]

[Out]

(-5*(-1)^(1/4)*(-a + I*b)^(3/2)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]
] + 5*(-1)^(1/4)*(a + I*b)^(3/2)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]
] + (2*Sqrt[a + b*Tan[c + d*x]]*(-a^2 - 2*a*b*Tan[c + d*x] + (5*a^2 - b^2)*Tan[c + d*x]^2))/(a*Tan[c + d*x]^(5
/2)))/(5*d)

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Maple [B] result has leaf size over 500,000. Avoiding possible recursion issues.
time = 1.25, size = 1346038, normalized size = 6009.10 \[\text {output too large to display}\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x)

[Out]

result too large to display

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^(3/2)/tan(d*x + c)^(7/2), x)

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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))**(3/2)/tan(d*x+c)**(7/2),x)

[Out]

Integral((a + b*tan(c + d*x))**(3/2)/tan(c + d*x)**(7/2), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))^(3/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))^(3/2)/tan(c + d*x)^(7/2),x)

[Out]

int((a + b*tan(c + d*x))^(3/2)/tan(c + d*x)^(7/2), x)

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