3.511 \(\int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\)

Optimal. Leaf size=230 \[ -\frac {3 \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d (a-b)^{5/2}}+\frac {3 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d (a+b)^{5/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 d \left (a^2-b^2\right )^2} \]

[Out]

-3/32*(4*a^2-10*a*b+7*b^2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/(a-b)^(5/2)/d+3/32*(4*a^2+10*a*b+7*b^2)
*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))/(a+b)^(5/2)/d-1/4*sec(d*x+c)^4*(b-a*sin(d*x+c))*(a+b*sin(d*x+c))^
(1/2)/d/(a^2-b^2)-1/16*sec(d*x+c)^2*(b*(a^2-7*b^2)-6*a*(a^2-2*b^2)*sin(d*x+c))*(a+b*sin(d*x+c))^(1/2)/(a^2-b^2
)^2/d

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Rubi [A]  time = 0.39, antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2668, 741, 823, 827, 1166, 206} \[ -\frac {3 \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d (a-b)^{5/2}}+\frac {3 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 d (a+b)^{5/2}}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 d \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-3*(4*a^2 - 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]])/(32*(a - b)^(5/2)*d) + (3*(4*a^2 +
10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]])/(32*(a + b)^(5/2)*d) - (Sec[c + d*x]^4*(b - a*
Sin[c + d*x])*Sqrt[a + b*Sin[c + d*x]])/(4*(a^2 - b^2)*d) - (Sec[c + d*x]^2*Sqrt[a + b*Sin[c + d*x]]*(b*(a^2 -
7*b^2) - 6*a*(a^2 - 2*b^2)*Sin[c + d*x]))/(16*(a^2 - b^2)^2*d)

Rule 206

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

Rule 741

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(a*e + c*d*x)*(
a + c*x^2)^(p + 1))/(2*a*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^5(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx &=\frac {b^5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+x} \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}+\frac {b^3 \operatorname {Subst}\left (\int \frac {\frac {1}{2} \left (6 a^2-7 b^2\right )+\frac {5 a x}{2}}{\sqrt {a+x} \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}-\frac {b \operatorname {Subst}\left (\int \frac {-\frac {3}{4} \left (4 a^4-9 a^2 b^2+7 b^4\right )-\frac {3}{2} a \left (a^2-2 b^2\right ) x}{\sqrt {a+x} \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}-\frac {b \operatorname {Subst}\left (\int \frac {\frac {3}{2} a^2 \left (a^2-2 b^2\right )-\frac {3}{4} \left (4 a^4-9 a^2 b^2+7 b^4\right )-\frac {3}{2} a \left (a^2-2 b^2\right ) x^2}{-a^2+b^2+2 a x^2-x^4} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{4 \left (a^2-b^2\right )^2 d}\\ &=-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}-\frac {\left (3 \left (4 a^2-10 a b+7 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 (a-b)^2 d}+\frac {\left (3 \left (4 a^2+10 a b+7 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{32 (a+b)^2 d}\\ &=-\frac {3 \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 (a-b)^{5/2} d}+\frac {3 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{32 (a+b)^{5/2} d}-\frac {\sec ^4(c+d x) (b-a \sin (c+d x)) \sqrt {a+b \sin (c+d x)}}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \sqrt {a+b \sin (c+d x)} \left (b \left (a^2-7 b^2\right )-6 a \left (a^2-2 b^2\right ) \sin (c+d x)\right )}{16 \left (a^2-b^2\right )^2 d}\\ \end {align*}

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Mathematica [A]  time = 1.90, size = 244, normalized size = 1.06 \[ \frac {\sqrt {a-b} \left (3 (a-b)^2 \left (4 a^2+10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )+\sqrt {a+b} \sec ^4(c+d x) \sqrt {a+b \sin (c+d x)} \left (3 \left (a^3-2 a b^2\right ) \sin (3 (c+d x))+\left (7 b^3-a^2 b\right ) \cos (2 (c+d x))+a \left (11 a^2-14 b^2\right ) \sin (c+d x)-9 a^2 b+15 b^3\right )\right )-3 (a+b)^{5/2} \left (4 a^2-10 a b+7 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{32 d \sqrt {a-b} \sqrt {a+b} \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/Sqrt[a + b*Sin[c + d*x]],x]

[Out]

(-3*(a + b)^(5/2)*(4*a^2 - 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a - b]] + Sqrt[a - b]*(3*(a -
b)^2*(4*a^2 + 10*a*b + 7*b^2)*ArcTanh[Sqrt[a + b*Sin[c + d*x]]/Sqrt[a + b]] + Sqrt[a + b]*Sec[c + d*x]^4*Sqrt
[a + b*Sin[c + d*x]]*(-9*a^2*b + 15*b^3 + (-(a^2*b) + 7*b^3)*Cos[2*(c + d*x)] + a*(11*a^2 - 14*b^2)*Sin[c + d*
x] + 3*(a^3 - 2*a*b^2)*Sin[3*(c + d*x)])))/(32*Sqrt[a - b]*Sqrt[a + b]*(a^2 - b^2)^2*d)

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (re
sidue poly has multiple non-linear factors)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{5}}{\sqrt {b \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/sqrt(b*sin(d*x + c) + a), x)

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maple [B]  time = 1.11, size = 618, normalized size = 2.69 \[ -\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 d \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {9 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 d \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a}{16 d \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a -b \right )}-\frac {11 b^{2} \sqrt {a +b \sin \left (d x +c \right )}}{32 d \left (b \sin \left (d x +c \right )+b \right )^{2} \left (a -b \right )}+\frac {3 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2}}{8 d \left (a^{2}-2 a b +b^{2}\right ) \sqrt {-a +b}}-\frac {15 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a b}{16 d \left (a^{2}-2 a b +b^{2}\right ) \sqrt {-a +b}}+\frac {21 \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) b^{2}}{32 d \left (a^{2}-2 a b +b^{2}\right ) \sqrt {-a +b}}-\frac {3 b \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}} a}{16 d \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a^{2}+2 a b +b^{2}\right )}-\frac {9 b^{2} \left (a +b \sin \left (d x +c \right )\right )^{\frac {3}{2}}}{32 d \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a^{2}+2 a b +b^{2}\right )}+\frac {3 b \sqrt {a +b \sin \left (d x +c \right )}\, a}{16 d \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a +b \right )}+\frac {11 b^{2} \sqrt {a +b \sin \left (d x +c \right )}}{32 d \left (b \sin \left (d x +c \right )-b \right )^{2} \left (a +b \right )}+\frac {3 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2}}{8 d \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a +b}}+\frac {15 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a b}{16 d \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a +b}}+\frac {21 \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) b^{2}}{32 d \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x)

[Out]

-3/16/d/(b*sin(d*x+c)+b)^2*b/(a^2-2*a*b+b^2)*(a+b*sin(d*x+c))^(3/2)*a+9/32/d/(b*sin(d*x+c)+b)^2*b^2/(a^2-2*a*b
+b^2)*(a+b*sin(d*x+c))^(3/2)+3/16/d/(b*sin(d*x+c)+b)^2*b/(a-b)*(a+b*sin(d*x+c))^(1/2)*a-11/32/d/(b*sin(d*x+c)+
b)^2*b^2/(a-b)*(a+b*sin(d*x+c))^(1/2)+3/8/d/(a^2-2*a*b+b^2)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^
(1/2))*a^2-15/16/d/(a^2-2*a*b+b^2)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a*b+21/32/d/(a^2-2
*a*b+b^2)/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*b^2-3/16/d/(b*sin(d*x+c)-b)^2*b/(a^2+2*a*b+
b^2)*(a+b*sin(d*x+c))^(3/2)*a-9/32/d/(b*sin(d*x+c)-b)^2*b^2/(a^2+2*a*b+b^2)*(a+b*sin(d*x+c))^(3/2)+3/16/d/(b*s
in(d*x+c)-b)^2*b/(a+b)*(a+b*sin(d*x+c))^(1/2)*a+11/32/d/(b*sin(d*x+c)-b)^2*b^2/(a+b)*(a+b*sin(d*x+c))^(1/2)+3/
8/d/(a^2+2*a*b+b^2)/(a+b)^(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2+15/16/d/(a^2+2*a*b+b^2)/(a+b)^
(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a*b+21/32/d/(a^2+2*a*b+b^2)/(a+b)^(1/2)*arctanh((a+b*sin(d*x
+c))^(1/2)/(a+b)^(1/2))*b^2

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+b*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more details)Is 4*a-4*b positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^5\,\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(a + b*sin(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(a + b*sin(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+b*sin(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**5/sqrt(a + b*sin(c + d*x)), x)

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