Optimal. Leaf size=182 \[ -\frac {i (i a-b)^{5/2} \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {i (i a+b)^{5/2} \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]
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Rubi [A]
time = 0.51, antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3646, 3730,
3697, 3696, 95, 209, 212} \begin {gather*} -\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {i (-b+i a)^{5/2} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {i (b+i a)^{5/2} \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.
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Rule 95
Rule 209
Rule 212
Rule 3646
Rule 3696
Rule 3697
Rule 3730
Rubi steps
\begin {align*} \int \frac {(a+b \tan (c+d x))^{5/2}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2}{3} \int \frac {\frac {7 a^2 b}{2}-\frac {3}{2} a \left (a^2-3 b^2\right ) \tan (c+d x)-\frac {1}{2} b \left (2 a^2-3 b^2\right ) \tan ^2(c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {4 \int \frac {\frac {3}{4} a^2 \left (a^2-3 b^2\right )+\frac {3}{4} a b \left (3 a^2-b^2\right ) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a}\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {1}{2} (a-i b)^3 \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx-\frac {1}{2} (a+i b)^3 \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx\\ &=-\frac {2 a^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {(a-i b)^3 \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 {(a+i b)^3 \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^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}-\frac {(a-i b)^3 \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 {(a+i b)^3 \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)^{5/2} \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)^{5/2} \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^2 \sqrt {a+b \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {14 a b \sqrt {a+b \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 0.91, size = 170, normalized size = 0.93 \begin {gather*} \frac {-3 (-1)^{3/4} (-a+i b)^{5/2} \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+3 (-1)^{3/4} (a+i b)^{5/2} \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 a \sqrt {a+b \tan (c+d x)} (a+7 b \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)}}{3 d} \end {gather*}
Antiderivative was successfully verified.
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 0.79, size = 16922, normalized size = 92.98
method | result | size |
default | \(\text {Expression too large to display}\) | \(16922\) |
Verification of antiderivative is not currently implemented for this CAS.
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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.
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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.
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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 {5}{2}}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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