3.3.37 \(\int \sec ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [237]

Optimal. Leaf size=45 \[ \frac {2 (d \tan (a+b x))^{5/2}}{5 b d}+\frac {2 (d \tan (a+b x))^{9/2}}{9 b d^3} \]

[Out]

2/5*(d*tan(b*x+a))^(5/2)/b/d+2/9*(d*tan(b*x+a))^(9/2)/b/d^3

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Rubi [A]
time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2687, 14} \begin {gather*} \frac {2 (d \tan (a+b x))^{9/2}}{9 b d^3}+\frac {2 (d \tan (a+b x))^{5/2}}{5 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^4*(d*Tan[a + b*x])^(3/2),x]

[Out]

(2*(d*Tan[a + b*x])^(5/2))/(5*b*d) + (2*(d*Tan[a + b*x])^(9/2))/(9*b*d^3)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2687

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rubi steps

\begin {align*} \int \sec ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {\text {Subst}\left (\int (d x)^{3/2} \left (1+x^2\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \left ((d x)^{3/2}+\frac {(d x)^{7/2}}{d^2}\right ) \, dx,x,\tan (a+b x)\right )}{b}\\ &=\frac {2 (d \tan (a+b x))^{5/2}}{5 b d}+\frac {2 (d \tan (a+b x))^{9/2}}{9 b d^3}\\ \end {align*}

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Mathematica [A]
time = 0.15, size = 42, normalized size = 0.93 \begin {gather*} \frac {2 d \left (-4-\sec ^2(a+b x)+5 \sec ^4(a+b x)\right ) \sqrt {d \tan (a+b x)}}{45 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^4*(d*Tan[a + b*x])^(3/2),x]

[Out]

(2*d*(-4 - Sec[a + b*x]^2 + 5*Sec[a + b*x]^4)*Sqrt[d*Tan[a + b*x]])/(45*b)

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Maple [A]
time = 0.41, size = 50, normalized size = 1.11

method	result	size
default	\(\frac {2 \left (4 \left (\cos ^{2}\left (b x +a \right )\right )+5\right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sin \left (b x +a \right )}{45 b \cos \left (b x +a \right )^{3}}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^4*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

2/45/b*(4*cos(b*x+a)^2+5)*(d*sin(b*x+a)/cos(b*x+a))^(3/2)*sin(b*x+a)/cos(b*x+a)^3

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Maxima [A]
time = 0.28, size = 36, normalized size = 0.80 \begin {gather*} \frac {2 \, {\left (5 \, \left (d \tan \left (b x + a\right )\right )^{\frac {9}{2}} + 9 \, \left (d \tan \left (b x + a\right )\right )^{\frac {5}{2}} d^{2}\right )}}{45 \, b d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

2/45*(5*(d*tan(b*x + a))^(9/2) + 9*(d*tan(b*x + a))^(5/2)*d^2)/(b*d^3)

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Fricas [A]
time = 0.40, size = 56, normalized size = 1.24 \begin {gather*} -\frac {2 \, {\left (4 \, d \cos \left (b x + a\right )^{4} + d \cos \left (b x + a\right )^{2} - 5 \, d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}}{45 \, b \cos \left (b x + a\right )^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*(d*tan(b*x+a))^(3/2),x, algorithm="fricas")

[Out]

-2/45*(4*d*cos(b*x + a)^4 + d*cos(b*x + a)^2 - 5*d)*sqrt(d*sin(b*x + a)/cos(b*x + a))/(b*cos(b*x + a)^4)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d \tan {\left (a + b x \right )}\right )^{\frac {3}{2}} \sec ^{4}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**4*(d*tan(b*x+a))**(3/2),x)

[Out]

Integral((d*tan(a + b*x))**(3/2)*sec(a + b*x)**4, x)

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Giac [A]
time = 0.46, size = 55, normalized size = 1.22 \begin {gather*} \frac {2 \, {\left (5 \, \sqrt {d \tan \left (b x + a\right )} d^{4} \tan \left (b x + a\right )^{4} + 9 \, \sqrt {d \tan \left (b x + a\right )} d^{4} \tan \left (b x + a\right )^{2}\right )}}{45 \, b d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^4*(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

2/45*(5*sqrt(d*tan(b*x + a))*d^4*tan(b*x + a)^4 + 9*sqrt(d*tan(b*x + a))*d^4*tan(b*x + a)^2)/(b*d^3)

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Mupad [B]
time = 6.94, size = 276, normalized size = 6.13 \begin {gather*} -\frac {8\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{45\,b}-\frac {8\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{45\,b\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}+\frac {56\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{15\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {64\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {32\,d\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1}}}{9\,b\,{\left ({\mathrm {e}}^{a\,2{}\mathrm {i}+b\,x\,2{}\mathrm {i}}+1\right )}^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(3/2)/cos(a + b*x)^4,x)

[Out]

(56*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(15*b*(exp(a*2i + b*x*2i) + 1)^2) -
(8*d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(45*b*(exp(a*2i + b*x*2i) + 1)) - (8*
d*(-(d*(exp(a*2i + b*x*2i)*1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(45*b) - (64*d*(-(d*(exp(a*2i + b*x*2i)*
1i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(9*b*(exp(a*2i + b*x*2i) + 1)^3) + (32*d*(-(d*(exp(a*2i + b*x*2i)*1
i - 1i))/(exp(a*2i + b*x*2i) + 1))^(1/2))/(9*b*(exp(a*2i + b*x*2i) + 1)^4)

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